How do I send images in an ajax request? [duplicate] - javascript

I want to implement a simple file upload in my intranet-page, with the smallest setup possible.
This is my HTML part:
<input id="sortpicture" type="file" name="sortpic" />
<button id="upload">Upload</button>
and this is my JS jquery script:
$("#upload").on("click", function() {
var file_data = $("#sortpicture").prop("files")[0];
var form_data = new FormData();
form_data.append("file", file_data);
alert(form_data);
$.ajax({
url: "/uploads",
dataType: 'script',
cache: false,
contentType: false,
processData: false,
data: form_data,
type: 'post',
success: function(){
alert("works");
}
});
});
There is a folder named "uploads" in the root directory of the website, with change permissions for "users" and "IIS_users".
When I select a file with the file-form and press the upload button, the first alert returns "[object FormData]". the second alert doesn't get called and the"uploads" folder is empty too!?
Can someone help my finding out whats wrong?
Also the next step should be, to rename the file with a server side generated name. Maybe someone can give me a solution for this, too.

You need a script that runs on the server to move the file to the uploads directory. The jQuery ajax method (running on the client in the browser) sends the form data to the server, then a script running on the server handles the upload.
Your HTML is fine, but update your JS jQuery script to look like this:
(Look for comments after // <-- )
$('#upload').on('click', function() {
var file_data = $('#sortpicture').prop('files')[0];
var form_data = new FormData();
form_data.append('file', file_data);
alert(form_data);
$.ajax({
url: 'upload.php', // <-- point to server-side PHP script
dataType: 'text', // <-- what to expect back from the PHP script, if anything
cache: false,
contentType: false,
processData: false,
data: form_data,
type: 'post',
success: function(php_script_response){
alert(php_script_response); // <-- display response from the PHP script, if any
}
});
});
And now for the server-side script, using PHP in this case.
upload.php: a PHP script that is located and runs on the server, and directs the file to the uploads directory:
<?php
if ( 0 < $_FILES['file']['error'] ) {
echo 'Error: ' . $_FILES['file']['error'] . '<br>';
}
else {
move_uploaded_file($_FILES['file']['tmp_name'], 'uploads/' . $_FILES['file']['name']);
}
?>
Also, a couple things about the destination directory:
Make sure you have the correct server path, i.e., starting at the PHP script location what is the path to the uploads directory, and
Make sure it's writeable.
And a little bit about the PHP function move_uploaded_file, used in the upload.php script:
move_uploaded_file(
// this is where the file is temporarily stored on the server when uploaded
// do not change this
$_FILES['file']['tmp_name'],
// this is where you want to put the file and what you want to name it
// in this case we are putting in a directory called "uploads"
// and giving it the original filename
'uploads/' . $_FILES['file']['name']
);
$_FILES['file']['name'] is the name of the file as it is uploaded. You don't have to use that. You can give the file any name (server filesystem compatible) you want:
move_uploaded_file(
$_FILES['file']['tmp_name'],
'uploads/my_new_filename.whatever'
);
And finally, be aware of your PHP upload_max_filesize AND post_max_size configuration values, and be sure your test files do not exceed either. Here's some help how you check PHP configuration and how you set max filesize and post settings.

**1. index.php**
<body>
<span id="msg" style="color:red"></span><br/>
<input type="file" id="photo"><br/>
<script type="text/javascript" src="jquery-3.2.1.min.js"></script>
<script type="text/javascript">
$(document).ready(function(){
$(document).on('change','#photo',function(){
var property = document.getElementById('photo').files[0];
var image_name = property.name;
var image_extension = image_name.split('.').pop().toLowerCase();
if(jQuery.inArray(image_extension,['gif','jpg','jpeg','']) == -1){
alert("Invalid image file");
}
var form_data = new FormData();
form_data.append("file",property);
$.ajax({
url:'upload.php',
method:'POST',
data:form_data,
contentType:false,
cache:false,
processData:false,
beforeSend:function(){
$('#msg').html('Loading......');
},
success:function(data){
console.log(data);
$('#msg').html(data);
}
});
});
});
</script>
</body>
**2.upload.php**
<?php
if($_FILES['file']['name'] != ''){
$test = explode('.', $_FILES['file']['name']);
$extension = end($test);
$name = rand(100,999).'.'.$extension;
$location = 'uploads/'.$name;
move_uploaded_file($_FILES['file']['tmp_name'], $location);
echo '<img src="'.$location.'" height="100" width="100" />';
}

Use pure js
async function saveFile()
{
let formData = new FormData();
formData.append("file", sortpicture.files[0]);
await fetch('/uploads', {method: "POST", body: formData});
alert('works');
}
<input id="sortpicture" type="file" name="sortpic" />
<button id="upload" onclick="saveFile()">Upload</button>
<br>Before click upload look on chrome>console>network (in this snipped we will see 404)
The filename is automatically included to request and server can read it, the 'content-type' is automatically set to 'multipart/form-data'. Here is more developed example with error handling and additional json sending
async function saveFile(inp)
{
let user = { name:'john', age:34 };
let formData = new FormData();
let photo = inp.files[0];
formData.append("photo", photo);
formData.append("user", JSON.stringify(user));
try {
let r = await fetch('/upload/image', {method: "POST", body: formData});
console.log('HTTP response code:',r.status);
alert('success');
} catch(e) {
console.log('Huston we have problem...:', e);
}
}
<input type="file" onchange="saveFile(this)" >
<br><br>
Before selecting the file Open chrome console > network tab to see the request details.
<br><br>
<small>Because in this example we send request to https://stacksnippets.net/upload/image the response code will be 404 ofcourse...</small>

var formData = new FormData($("#YOUR_FORM_ID")[0]);
$.ajax({
url: "upload.php",
type: "POST",
data : formData,
processData: false,
contentType: false,
beforeSend: function() {
},
success: function(data){
},
error: function(xhr, ajaxOptions, thrownError) {
console.log(thrownError + "\r\n" + xhr.statusText + "\r\n" + xhr.responseText);
}
});

and this is the php file to receive the uplaoded files
<?
$data = array();
//check with your logic
if (isset($_FILES)) {
$error = false;
$files = array();
$uploaddir = $target_dir;
foreach ($_FILES as $file) {
if (move_uploaded_file($file['tmp_name'], $uploaddir . basename( $file['name']))) {
$files[] = $uploaddir . $file['name'];
} else {
$error = true;
}
}
$data = ($error) ? array('error' => 'There was an error uploading your files') : array('files' => $files);
} else {
$data = array('success' => 'NO FILES ARE SENT','formData' => $_REQUEST);
}
echo json_encode($data);
?>

Related

upload image from multiple form with ajax and php

in PHP page with multiple form tag to register user information.
using ajax to collect data and post to register PHP page now i want to upload image to server folder but i failed.
my html code:
<label for="upimage" class="btn btn-sm btn-primary mb-75 mr-75">Upload Image</label>
<input type="file" id="upimage" hidden accept="image/*" name="image"/>
Javascript Code:
let data1 = document.getElementById('data1').value,
data2 = document.getElementById('data1').value,
data3 = document.getElementById('data1').value,
upimage = document.getElementById('upimage').value;
$.ajax({
url:"././newregister.php",
method:"POST",
data:{action:'newregister', data1:data1, data2:data2,
data3:data3, upimage:upimage},
success:function(data){
alert(data);
}
});
newregister php code:
$uploads_dir = './uploads';
if ($_FILES["file"]["error"] == UPLOAD_ERR_OK) {
$tmp_name = $_FILES["file"]["tmp_name"];
$name = $_FILES["file"]["name"];
move_uploaded_file($tmp_name, "$uploads_dir/$name");
echo "Sucsess";
}
else
{
echo "Error" . $_FILES["file"]["error"] ;
}
ERR: Undefined index: file in .... on line ....
Given by POST method uploads
Be sure your file upload form has attribute enctype="multipart/form-data" otherwise the file upload will not work.
Your current solution lacks enctype, that's why your file is not getting uploaded to the server and therefore isn't in the superglobal variable $_FILES.
As ferikeem already said. Wrap your data in a FormData Object and send it that way.
See: https://stackoverflow.com/a/5976031/10887013
JavaScript
let fd = new FormData();
fd.append("you_file_key_here", $("#upimage")[0].files[0]);
fd.append("data1", $("#data1")[0].value);
fd.append("data2", $("#data2")[0].value);
fd.append("data3", $("#data3")[0].value);
$.ajax({
url: "././newregister.php",
method: "POST",
data: fd,
processData: false,
contentType: false,
success: function(data){
alert(data);
}
});
$_FILES["file"], here is your problem.
In $_FILES array key "file" doesn't exists. As I see you need to change $_FILES["file"] with $_FILES["upimage"].

jQuery AJAX - send file and variable to PHP in single request

I have the following HTML form for uploading a file (picture):
<input id="sortpicture" type="file" name="sortpic" />
<button id="upload">Upload</button>
I use the following JavaScript/jQuery to send form data and a variable value to PHP script:
var variable1= 1;
$('#upload').on('click', function() {
var file_data = $('#sortpicture').prop('files')[0];
var form_data = new FormData();
form_data.append('file', file_data);
$.ajax({
url: 'upload.php', // point to server-side PHP script
dataType: 'text', // what to expect back from the PHP script, if anything
cache: false,
contentType: false,
processData: false,
data: {form_data, number: variable1},
type: 'post',
success: function(php_script_response){
alert(php_script_response); // display response from the PHP script, if any
}
});
});
The PHP script (upload.php) looks like this:
$filename;
if ( 0 < $_FILES['file']['error'] ) {
echo 'Error: ' . $_FILES['file']['error'] . '<br>';
}
else {
move_uploaded_file($_FILES['file']['tmp_name'], 'uploads/' . $_FILES['file']['name']);
$filename = $_FILES['file']['name'];
}
echo $filename;
When I run this code, PHP throws some undefined index errors. If I use only "data: form_data" in ajax request (without variable1), then the file uploads successfully, but I need to send the variable too. How can I do this?
You can also append the number key => value to to the form data as
form_data.append('number', variable1);
Then in Ajax call
data: form_data,
why don't you append this value like this
form_data.append('number', variable1);
Append the variable to form_data
var form_data = new FormData();
form_data.append('file', file_data);
form_data.append('number', variable1); // append variable
In ajax
processData: false,
data: form_data, // data
type: 'post',
I am following your code because I need FILES image data and multiple variables in PHP script for insertion operation in the database. But I am unable to access any single variable in the php script. I did below to access your number variable. Please correct me.
//upload.php
if(isset($_GET['number'])){
$num = $_GET['number'];
echo $num;
}else{
echo 'not set';
}

File uploade with additional Data (Javascript, PHP)

I want to uploade a file and send additional data (an array) to the php script.
Here is the HTML and the Javascript Code:
<input type="file" name="fileinput" id="fileinput">
<button class="btn btn-primary" id="btnupload">upload</button>
<script>
$('#btnupload').click(function(){
var formData = new FormData();
formData.append("MacsToChangeImage", selectedMACs.toString()); //selectedMACs is the array which i want to send
formData.append("userfile", $('#fileinput').files[0]);
$.ajax({
url: "ChangeImagesFunction.php",
type:'POST',
data: formData,
processData: false,
cententType: false,
success: function(data)
{
alert('success' + data);
},
error:function(response){
alert(JSON.stringify(response));
}
});
});
</script>
And here ist the PHP Code:
if(!empty($_FILES)) {
$target_dir = "img/epd_uploads/";
$temporaryFile = $_FILES['userfile']['tmp_name'];
$targetFile = $target_dir . $_FILES['userfile']['name'];
if(!move_uploaded_file($temporaryFile,$targetFile)) {
echo "Error occurred while uploading the file to server!";
}else{
if (isset($_POST['MacsToChangeImage'])){
//Name of the Uploaded File
$ChangeImageToPHP = $_FILES['userfile']['name'];
//Get the Array
$MacsToChangeImageStr = $_POST['MacsToChangeImage'];
$MacsToChangeImagePHP = explode(",",$MacsToChangeImageStr);
}
}
}
I think the HTML and Javascript part is fine so far. But i am uncertain how to handle the PHP part. What am i doing worng/missing?
Are there other ways to achieve what im plannig to do?
Thank you in advance

Sending file with an -F curl request via Javascript/Ajax [duplicate]

Can I use the following jQuery code to perform file upload using POST method of an ajax request ?
$.ajax({
type: "POST",
timeout: 50000,
url: url,
data: dataString,
success: function (data) {
alert('success');
return false;
}
});
If it is possible, do I need to fill data part? Is it the correct way? I only POST the file to the server side.
I have been googling around, but what I found was a plugin while in my plan I do not want to use it. At least for the moment.
File upload is not possible through AJAX.
You can upload file, without refreshing page by using IFrame.
You can check further details here.
UPDATE
With XHR2, File upload through AJAX is supported. E.g. through FormData object, but unfortunately it is not supported by all/old browsers.
FormData support starts from following desktop browsers versions.
IE 10+
Firefox 4.0+
Chrome 7+
Safari 5+
Opera 12+
For more detail, see MDN link.
Iframes is no longer needed for uploading files through ajax. I've recently done it by myself. Check out these pages:
Using HTML5 file uploads with AJAX and jQuery
http://dev.w3.org/2006/webapi/FileAPI/#FileReader-interface
Updated the answer and cleaned it up. Use the getSize function to check size or use getType function to check types.
Added progressbar html and css code.
var Upload = function (file) {
this.file = file;
};
Upload.prototype.getType = function() {
return this.file.type;
};
Upload.prototype.getSize = function() {
return this.file.size;
};
Upload.prototype.getName = function() {
return this.file.name;
};
Upload.prototype.doUpload = function () {
var that = this;
var formData = new FormData();
// add assoc key values, this will be posts values
formData.append("file", this.file, this.getName());
formData.append("upload_file", true);
$.ajax({
type: "POST",
url: "script",
xhr: function () {
var myXhr = $.ajaxSettings.xhr();
if (myXhr.upload) {
myXhr.upload.addEventListener('progress', that.progressHandling, false);
}
return myXhr;
},
success: function (data) {
// your callback here
},
error: function (error) {
// handle error
},
async: true,
data: formData,
cache: false,
contentType: false,
processData: false,
timeout: 60000
});
};
Upload.prototype.progressHandling = function (event) {
var percent = 0;
var position = event.loaded || event.position;
var total = event.total;
var progress_bar_id = "#progress-wrp";
if (event.lengthComputable) {
percent = Math.ceil(position / total * 100);
}
// update progressbars classes so it fits your code
$(progress_bar_id + " .progress-bar").css("width", +percent + "%");
$(progress_bar_id + " .status").text(percent + "%");
};
How to use the Upload class
//Change id to your id
$("#ingredient_file").on("change", function (e) {
var file = $(this)[0].files[0];
var upload = new Upload(file);
// maby check size or type here with upload.getSize() and upload.getType()
// execute upload
upload.doUpload();
});
Progressbar html code
<div id="progress-wrp">
<div class="progress-bar"></div>
<div class="status">0%</div>
</div>
Progressbar css code
#progress-wrp {
border: 1px solid #0099CC;
padding: 1px;
position: relative;
height: 30px;
border-radius: 3px;
margin: 10px;
text-align: left;
background: #fff;
box-shadow: inset 1px 3px 6px rgba(0, 0, 0, 0.12);
}
#progress-wrp .progress-bar {
height: 100%;
border-radius: 3px;
background-color: #f39ac7;
width: 0;
box-shadow: inset 1px 1px 10px rgba(0, 0, 0, 0.11);
}
#progress-wrp .status {
top: 3px;
left: 50%;
position: absolute;
display: inline-block;
color: #000000;
}
Ajax post and upload file is possible. I'm using jQuery $.ajax function to load my files. I tried to use the XHR object but could not get results on the server side with PHP.
var formData = new FormData();
formData.append('file', $('#file')[0].files[0]);
$.ajax({
url : 'upload.php',
type : 'POST',
data : formData,
processData: false, // tell jQuery not to process the data
contentType: false, // tell jQuery not to set contentType
success : function(data) {
console.log(data);
alert(data);
}
});
As you can see, you must create a FormData object, empty or from (serialized? - $('#yourForm').serialize()) existing form, and then attach the input file.
Here is more information:
- How to upload a file using jQuery.ajax and FormData
- Uploading files via jQuery, object FormData is provided and no file name, GET request
For the PHP process you can use something like this:
//print_r($_FILES);
$fileName = $_FILES['file']['name'];
$fileType = $_FILES['file']['type'];
$fileError = $_FILES['file']['error'];
$fileContent = file_get_contents($_FILES['file']['tmp_name']);
if($fileError == UPLOAD_ERR_OK){
//Processes your file here
}else{
switch($fileError){
case UPLOAD_ERR_INI_SIZE:
$message = 'Error al intentar subir un archivo que excede el tamaño permitido.';
break;
case UPLOAD_ERR_FORM_SIZE:
$message = 'Error al intentar subir un archivo que excede el tamaño permitido.';
break;
case UPLOAD_ERR_PARTIAL:
$message = 'Error: no terminó la acción de subir el archivo.';
break;
case UPLOAD_ERR_NO_FILE:
$message = 'Error: ningún archivo fue subido.';
break;
case UPLOAD_ERR_NO_TMP_DIR:
$message = 'Error: servidor no configurado para carga de archivos.';
break;
case UPLOAD_ERR_CANT_WRITE:
$message= 'Error: posible falla al grabar el archivo.';
break;
case UPLOAD_ERR_EXTENSION:
$message = 'Error: carga de archivo no completada.';
break;
default: $message = 'Error: carga de archivo no completada.';
break;
}
echo json_encode(array(
'error' => true,
'message' => $message
));
}
Simple Upload Form
<script>
//form Submit
$("form").submit(function(evt){
evt.preventDefault();
var formData = new FormData($(this)[0]);
$.ajax({
url: 'fileUpload',
type: 'POST',
data: formData,
async: false,
cache: false,
contentType: false,
enctype: 'multipart/form-data',
processData: false,
success: function (response) {
alert(response);
}
});
return false;
});
</script>
<!--Upload Form-->
<form>
<table>
<tr>
<td colspan="2">File Upload</td>
</tr>
<tr>
<th>Select File </th>
<td><input id="csv" name="csv" type="file" /></td>
</tr>
<tr>
<td colspan="2">
<input type="submit" value="submit"/>
</td>
</tr>
</table>
</form>
I'm pretty late for this but I was looking for an ajax based image uploading solution and the answer I was looking for was kinda scattered throughout this post. The solution I settled on involved the FormData object. I assembled a basic form of the code I put together. You can see it demonstrates how to add a custom field to the form with fd.append() as well as how to handle response data when the ajax request is done.
Upload html:
<!DOCTYPE html>
<html>
<head>
<title>Image Upload Form</title>
<script src="//code.jquery.com/jquery-1.9.1.js"></script>
<script type="text/javascript">
function submitForm() {
console.log("submit event");
var fd = new FormData(document.getElementById("fileinfo"));
fd.append("label", "WEBUPLOAD");
$.ajax({
url: "upload.php",
type: "POST",
data: fd,
processData: false, // tell jQuery not to process the data
contentType: false // tell jQuery not to set contentType
}).done(function( data ) {
console.log("PHP Output:");
console.log( data );
});
return false;
}
</script>
</head>
<body>
<form method="post" id="fileinfo" name="fileinfo" onsubmit="return submitForm();">
<label>Select a file:</label><br>
<input type="file" name="file" required />
<input type="submit" value="Upload" />
</form>
<div id="output"></div>
</body>
</html>
In case you are working with php here's a way to handle the upload that includes making use of both of the custom fields demonstrated in the above html.
Upload.php
<?php
if ($_POST["label"]) {
$label = $_POST["label"];
}
$allowedExts = array("gif", "jpeg", "jpg", "png");
$temp = explode(".", $_FILES["file"]["name"]);
$extension = end($temp);
if ((($_FILES["file"]["type"] == "image/gif")
|| ($_FILES["file"]["type"] == "image/jpeg")
|| ($_FILES["file"]["type"] == "image/jpg")
|| ($_FILES["file"]["type"] == "image/pjpeg")
|| ($_FILES["file"]["type"] == "image/x-png")
|| ($_FILES["file"]["type"] == "image/png"))
&& ($_FILES["file"]["size"] < 200000)
&& in_array($extension, $allowedExts)) {
if ($_FILES["file"]["error"] > 0) {
echo "Return Code: " . $_FILES["file"]["error"] . "<br>";
} else {
$filename = $label.$_FILES["file"]["name"];
echo "Upload: " . $_FILES["file"]["name"] . "<br>";
echo "Type: " . $_FILES["file"]["type"] . "<br>";
echo "Size: " . ($_FILES["file"]["size"] / 1024) . " kB<br>";
echo "Temp file: " . $_FILES["file"]["tmp_name"] . "<br>";
if (file_exists("uploads/" . $filename)) {
echo $filename . " already exists. ";
} else {
move_uploaded_file($_FILES["file"]["tmp_name"],
"uploads/" . $filename);
echo "Stored in: " . "uploads/" . $filename;
}
}
} else {
echo "Invalid file";
}
?>
An AJAX upload is indeed possible with XMLHttpRequest(). No iframes necessary. Upload progress can be shown.
For details see: Answer https://stackoverflow.com/a/4943774/873282 to question jQuery Upload Progress and AJAX file upload.
Here's how I got this working:
HTML
<input type="file" id="file">
<button id='process-file-button'>Process</button>
JS
$('#process-file-button').on('click', function (e) {
let files = new FormData(), // you can consider this as 'data bag'
url = 'yourUrl';
files.append('fileName', $('#file')[0].files[0]); // append selected file to the bag named 'file'
$.ajax({
type: 'post',
url: url,
processData: false,
contentType: false,
data: files,
success: function (response) {
console.log(response);
},
error: function (err) {
console.log(err);
}
});
});
PHP
if (isset($_FILES) && !empty($_FILES)) {
$file = $_FILES['fileName'];
$name = $file['name'];
$path = $file['tmp_name'];
// process your file
}
Using pure js it is easier
async function saveFile(inp)
{
let formData = new FormData();
formData.append("file", inp.files[0]);
await fetch('/upload/somedata', {method: "POST", body: formData});
alert('success');
}
<input type="file" onchange="saveFile(this)" >
In server side you can read original file name (and other info) which is automatically included to request.
You do NOT need to set header "Content-Type" to "multipart/form-data" browser will set it automatically
This solutions should work on all major browsers.
Here is more developed snippet with error handling, timeout and additional json sending
async function saveFile(inp)
{
let user = { name:'john', age:34 };
let formData = new FormData();
let photo = inp.files[0];
formData.append("photo", photo);
formData.append("user", JSON.stringify(user));
const ctrl = new AbortController() // timeout
setTimeout(() => ctrl.abort(), 50000);
try {
let r = await fetch('/upload/image',
{method: "POST", body: formData, signal: ctrl.signal});
console.log('HTTP response code:',r.status);
alert('success');
} catch(e) {
console.log('Huston we have problem...:', e);
}
}
<input type="file" onchange="saveFile(this)" >
<br><br>
Before selecting the file Open chrome console > network tab to see the request details.
<br><br>
<small>Because in this example we send request to https://stacksnippets.net/upload/image the response code will be 404 ofcourse...</small>
In case you want to do it like that:
$.upload( form.action, new FormData( myForm))
.progress( function( progressEvent, upload) {
if( progressEvent.lengthComputable) {
var percent = Math.round( progressEvent.loaded * 100 / progressEvent.total) + '%';
if( upload) {
console.log( percent + ' uploaded');
} else {
console.log( percent + ' downloaded');
}
}
})
.done( function() {
console.log( 'Finished upload');
});
than
https://github.com/lgersman/jquery.orangevolt-ampere/blob/master/src/jquery.upload.js
might be your solution.
Use FormData. It works really well :-) ...
var jform = new FormData();
jform.append('user',$('#user').val());
jform.append('image',$('#image').get(0).files[0]); // Here's the important bit
$.ajax({
url: '/your-form-processing-page-url-here',
type: 'POST',
data: jform,
dataType: 'json',
mimeType: 'multipart/form-data', // this too
contentType: false,
cache: false,
processData: false,
success: function(data, status, jqXHR){
alert('Hooray! All is well.');
console.log(data);
console.log(status);
console.log(jqXHR);
},
error: function(jqXHR,status,error){
// Hopefully we should never reach here
console.log(jqXHR);
console.log(status);
console.log(error);
}
});
$("#submit_car").click(function() {
var formData = new FormData($('#car_cost_form')[0]);
$.ajax({
url: 'car_costs.php',
data: formData,
contentType: false,
processData: false,
cache: false,
type: 'POST',
success: function(data) {
// ...
},
});
});
edit: Note contentype and process data
You can simply use this to upload files via Ajax...... submit input cannot be outside form element :)
2019 update:
html
<form class="fr" method='POST' enctype="multipart/form-data"> {% csrf_token %}
<textarea name='text'>
<input name='example_image'>
<button type="submit">
</form>
js
$(document).on('submit', '.fr', function(){
$.ajax({
type: 'post',
url: url, <--- you insert proper URL path to call your views.py function here.
enctype: 'multipart/form-data',
processData: false,
contentType: false,
data: new FormData(this) ,
success: function(data) {
console.log(data);
}
});
return false;
});
views.py
form = ThisForm(request.POST, request.FILES)
if form.is_valid():
text = form.cleaned_data.get("text")
example_image = request.FILES['example_image']
Use a hidden iframe and set your form's target to that iframe's name. This way, when the form is submitted, only the iframe will be refreshed.
Have an event handler registered for the iframe's load event to parse the response.
I have handled these in a simple code. You can download a working demo from here
For your case, these very possible. I will take you step by step how you can upload a file to the server using AJAX jquery.
First let's us create an HTML file to add the following form file element as shown below.
<form action="" id="formContent" method="post" enctype="multipart/form-data" >
<input type="file" name="file" required id="upload">
<button class="submitI" >Upload Image</button>
</form>
Secondly create a jquery.js file and add the following code to handle our file submission to the server
$("#formContent").submit(function(e){
e.preventDefault();
var formdata = new FormData(this);
$.ajax({
url: "ajax_upload_image.php",
type: "POST",
data: formdata,
mimeTypes:"multipart/form-data",
contentType: false,
cache: false,
processData: false,
success: function(){
alert("file successfully submitted");
},error: function(){
alert("okey");
}
});
});
});
There you are done . View more
Using FormData is the way to go as indicated by many answers. here is a bit of code that works great for this purpose. I also agree with the comment of nesting ajax blocks to complete complex circumstances. By including e.PreventDefault(); in my experience makes the code more cross browser compatible.
$('#UploadB1').click(function(e){
e.preventDefault();
if (!fileupload.valid()) {
return false;
}
var myformData = new FormData();
myformData.append('file', $('#uploadFile')[0].files[0]);
$("#UpdateMessage5").html("Uploading file ....");
$("#UpdateMessage5").css("background","url(../include/images/loaderIcon.gif) no-repeat right");
myformData.append('mode', 'fileUpload');
myformData.append('myid', $('#myid').val());
myformData.append('type', $('#fileType').val());
//formData.append('myfile', file, file.name);
$.ajax({
url: 'include/fetch.php',
method: 'post',
processData: false,
contentType: false,
cache: false,
data: myformData,
enctype: 'multipart/form-data',
success: function(response){
$("#UpdateMessage5").html(response); //.delay(2000).hide(1);
$("#UpdateMessage5").css("background","");
console.log("file successfully submitted");
},error: function(){
console.log("not okay");
}
});
});
I have implemented a multiple file select with instant preview and upload after removing unwanted files from preview via ajax.
Detailed documentation can be found here: http://anasthecoder.blogspot.ae/2014/12/multi-file-select-preview-without.html
Demo: http://jsfiddle.net/anas/6v8Kz/7/embedded/result/
jsFiddle: http://jsfiddle.net/anas/6v8Kz/7/
Javascript:
$(document).ready(function(){
$('form').submit(function(ev){
$('.overlay').show();
$(window).scrollTop(0);
return upload_images_selected(ev, ev.target);
})
})
function add_new_file_uploader(addBtn) {
var currentRow = $(addBtn).parent().parent();
var newRow = $(currentRow).clone();
$(newRow).find('.previewImage, .imagePreviewTable').hide();
$(newRow).find('.removeButton').show();
$(newRow).find('table.imagePreviewTable').find('tr').remove();
$(newRow).find('input.multipleImageFileInput').val('');
$(addBtn).parent().parent().parent().append(newRow);
}
function remove_file_uploader(removeBtn) {
$(removeBtn).parent().parent().remove();
}
function show_image_preview(file_selector) {
//files selected using current file selector
var files = file_selector.files;
//Container of image previews
var imageContainer = $(file_selector).next('table.imagePreviewTable');
//Number of images selected
var number_of_images = files.length;
//Build image preview row
var imagePreviewRow = $('<tr class="imagePreviewRow_0"><td valign=top style="width: 510px;"></td>' +
'<td valign=top><input type="button" value="X" title="Remove Image" class="removeImageButton" imageIndex="0" onclick="remove_selected_image(this)" /></td>' +
'</tr> ');
//Add image preview row
$(imageContainer).html(imagePreviewRow);
if (number_of_images > 1) {
for (var i =1; i<number_of_images; i++) {
/**
*Generate class name of the respective image container appending index of selected images,
*sothat we can match images selected and the one which is previewed
*/
var newImagePreviewRow = $(imagePreviewRow).clone().removeClass('imagePreviewRow_0').addClass('imagePreviewRow_'+i);
$(newImagePreviewRow).find('input[type="button"]').attr('imageIndex', i);
$(imageContainer).append(newImagePreviewRow);
}
}
for (var i = 0; i < files.length; i++) {
var file = files[i];
/**
* Allow only images
*/
var imageType = /image.*/;
if (!file.type.match(imageType)) {
continue;
}
/**
* Create an image dom object dynamically
*/
var img = document.createElement("img");
/**
* Get preview area of the image
*/
var preview = $(imageContainer).find('tr.imagePreviewRow_'+i).find('td:first');
/**
* Append preview of selected image to the corresponding container
*/
preview.append(img);
/**
* Set style of appended preview(Can be done via css also)
*/
preview.find('img').addClass('previewImage').css({'max-width': '500px', 'max-height': '500px'});
/**
* Initialize file reader
*/
var reader = new FileReader();
/**
* Onload event of file reader assign target image to the preview
*/
reader.onload = (function(aImg) { return function(e) { aImg.src = e.target.result; }; })(img);
/**
* Initiate read
*/
reader.readAsDataURL(file);
}
/**
* Show preview
*/
$(imageContainer).show();
}
function remove_selected_image(close_button)
{
/**
* Remove this image from preview
*/
var imageIndex = $(close_button).attr('imageindex');
$(close_button).parents('.imagePreviewRow_' + imageIndex).remove();
}
function upload_images_selected(event, formObj)
{
event.preventDefault();
//Get number of images
var imageCount = $('.previewImage').length;
//Get all multi select inputs
var fileInputs = document.querySelectorAll('.multipleImageFileInput');
//Url where the image is to be uploaded
var url= "/upload-directory/";
//Get number of inputs
var number_of_inputs = $(fileInputs).length;
var inputCount = 0;
//Iterate through each file selector input
$(fileInputs).each(function(index, input){
fileList = input.files;
// Create a new FormData object.
var formData = new FormData();
//Extra parameters can be added to the form data object
formData.append('bulk_upload', '1');
formData.append('username', $('input[name="username"]').val());
//Iterate throug each images selected by each file selector and find if the image is present in the preview
for (var i = 0; i < fileList.length; i++) {
if ($(input).next('.imagePreviewTable').find('.imagePreviewRow_'+i).length != 0) {
var file = fileList[i];
// Check the file type.
if (!file.type.match('image.*')) {
continue;
}
// Add the file to the request.
formData.append('image_uploader_multiple[' +(inputCount++)+ ']', file, file.name);
}
}
// Set up the request.
var xhr = new XMLHttpRequest();
xhr.open('POST', url, true);
xhr.onload = function () {
if (xhr.status === 200) {
var jsonResponse = JSON.parse(xhr.responseText);
if (jsonResponse.status == 1) {
$(jsonResponse.file_info).each(function(){
//Iterate through response and find data corresponding to each file uploaded
var uploaded_file_name = this.original;
var saved_file_name = this.target;
var file_name_input = '<input type="hidden" class="image_name" name="image_names[]" value="' +saved_file_name+ '" />';
file_info_container.append(file_name_input);
imageCount--;
})
//Decrement count of inputs to find all images selected by all multi select are uploaded
number_of_inputs--;
if(number_of_inputs == 0) {
//All images selected by each file selector is uploaded
//Do necessary acteion post upload
$('.overlay').hide();
}
} else {
if (typeof jsonResponse.error_field_name != 'undefined') {
//Do appropriate error action
} else {
alert(jsonResponse.message);
}
$('.overlay').hide();
event.preventDefault();
return false;
}
} else {
/*alert('Something went wrong!');*/
$('.overlay').hide();
event.preventDefault();
}
};
xhr.send(formData);
})
return false;
}
Yes you can, just use javascript to get the file, making sure you read the file as a data URL. Parse out the stuff before base64 to actually get the base 64 encoded data and then if you are using php or really any back end language you can decode the base 64 data and save into a file like shown below
Javascript:
var reader = new FileReader();
reader.onloadend = function ()
{
dataToBeSent = reader.result.split("base64,")[1];
$.post(url, {data:dataToBeSent});
}
reader.readAsDataURL(this.files[0]);
PHP:
file_put_contents('my.pdf', base64_decode($_POST["data"]));
Of course you will probably want to do some validation like checking which file type you are dealing with and stuff like that but this is the idea.
To get all your form inputs, including the type="file" you need to use FormData object.
you will be able to see the formData content in the debugger -> network ->Headers after you will submit the form.
var url = "YOUR_URL";
var form = $('#YOUR_FORM_ID')[0];
var formData = new FormData(form);
$.ajax(url, {
method: 'post',
processData: false,
contentType: false,
data: formData
}).done(function(data){
if (data.success){
alert("Files uploaded");
} else {
alert("Error while uploading the files");
}
}).fail(function(data){
console.log(data);
alert("Error while uploading the files");
});
<html>
<head>
<title>Ajax file upload</title>
<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js"></script>
<script>
$(document).ready(function (e) {
$("#uploadimage").on('submit', (function(e) {
e.preventDefault();
$.ajax({
url: "upload.php", // Url to which the request is send
type: "POST", // Type of request to be send, called as method
data: new FormData(this), // Data sent to server, a set of key/value pairs (i.e. form fields and values)
contentType: false, // The content type used when sending data to the server.
cache: false, // To unable request pages to be cached
processData:false, // To send DOMDocument or non processed data file it is set to false
success: function(data) // A function to be called if request succeeds
{
alert(data);
}
});
}));
</script>
</head>
<body>
<div class="main">
<h1>Ajax Image Upload</h1><br/>
<hr>
<form id="uploadimage" action="" method="post" enctype="multipart/form-data">
<div id="image_preview"><img id="previewing" src="noimage.png" /></div>
<hr id="line">
<div id="selectImage">
<label>Select Your Image</label><br/>
<input type="file" name="file" id="file" required />
<input type="submit" value="Upload" class="submit" />
</div>
</form>
</div>
</body>
</html>
to upload a file which is submitted by user as a part of form using jquery please follow the below code :
var formData = new FormData();
formData.append("userfile", fileInputElement.files[0]);
Then send the form data object to server.
We can also append a File or Blob directly to the FormData object.
data.append("myfile", myBlob, "filename.txt");
You can use method ajaxSubmit as follow :)
when you select a file that need upload to server, form be submit to server :)
$(document).ready(function () {
var options = {
target: '#output', // target element(s) to be updated with server response
timeout: 30000,
error: function (jqXHR, textStatus) {
$('#output').html('have any error');
return false;
}
},
success: afterSuccess, // post-submit callback
resetForm: true
// reset the form after successful submit
};
$('#idOfInputFile').on('change', function () {
$('#idOfForm').ajaxSubmit(options);
// always return false to prevent standard browser submit and page navigation
return false;
});
});
If you want to upload file using AJAX here is code which you can use for file uploading.
$(document).ready(function() {
var options = {
beforeSubmit: showRequest,
success: showResponse,
dataType: 'json'
};
$('body').delegate('#image','change', function(){
$('#upload').ajaxForm(options).submit();
});
});
function showRequest(formData, jqForm, options) {
$("#validation-errors").hide().empty();
$("#output").css('display','none');
return true;
}
function showResponse(response, statusText, xhr, $form) {
if(response.success == false)
{
var arr = response.errors;
$.each(arr, function(index, value)
{
if (value.length != 0)
{
$("#validation-errors").append('<div class="alert alert-error"><strong>'+ value +'</strong><div>');
}
});
$("#validation-errors").show();
} else {
$("#output").html("<img src='"+response.file+"' />");
$("#output").css('display','block');
}
}
Here is the HTML for Upload the file
<form class="form-horizontal" id="upload" enctype="multipart/form-data" method="post" action="upload/image'" autocomplete="off">
<input type="file" name="image" id="image" />
</form>
var dataform = new FormData($("#myform")[0]);
//console.log(dataform);
$.ajax({
url: 'url',
type: 'POST',
data: dataform,
async: false,
success: function(res) {
response data;
},
cache: false,
contentType: false,
processData: false
});
<input class="form-control cu-b-border" type="file" id="formFile">
<img id="myImg" src="#">
In js
<script>
var formData = new FormData();
formData.append('file', $('#formFile')[0].files[0]);
$.ajax({
type: "POST",
url: '/GetData/UploadImage',
data: formData,
processData: false, // tell jQuery not to process the data
contentType: false, // tell jQuery not to set contentType
success: function (data) {
console.log(data);
$('#myImg').attr('src', data);
},
error: function (xhr, ajaxOptions, thrownError) {
}
})
</script>
In controller
public ActionResult UploadImage(HttpPostedFileBase file)
{
string filePath = "";
if (file != null)
{
string path = "/uploads/Temp/";
if (!Directory.Exists(Server.MapPath("~" + path)))
{
Directory.CreateDirectory(Server.MapPath("~" + path));
}
filePath = FileUpload.SaveUploadedFile(file, path);
}
return Json(filePath, JsonRequestBehavior.AllowGet);
}
Here was an idea I was thinking of:
Have an iframe on page and have a referencer.
Have a form in which you move the input type file element to.
Form: A processing page AND a target of the FRAME.
The result will post to the iframe, and then you can just send the fetched data up a level to the image tag you want with something like:
data:image/png;base64,asdfasdfasdfasdfa
and the page loads.
I believe it works for me, and depending you might be able to do something like:
.aftersubmit(function(){
stopPropagation(); // or some other code which would prevent a refresh.
});
This is my code that it works
var formData = new FormData();
var files = $('input[type=file]');
for (var i = 0; i < files.length; i++) {
if (files[i].value == "" || files[i].value == null) {
return false;
}
else {
formData.append(files[i].name, files[i].files[0]);
}
}
var formSerializeArray = $("#Form").serializeArray();
for (var i = 0; i < formSerializeArray.length; i++) {
formData.append(formSerializeArray[i].name, formSerializeArray[i].value)
}
$.ajax({
type: 'POST',
data: formData,
contentType: false,
processData: false,
cache: false,
url: '/Controller/Action',
success: function (response) {
if (response.Success == true) {
return true;
}
else {
return false;
}
},
error: function () {
return false;
},
failure: function () {
return false;
}
});
$("#form-id").submit(function (e) {
e.preventDefault();
});
$("#form-id").submit(function (e) {
var formObj = $(this);
var formURL = formObj.attr("action");
var formData = new FormData(this);
$.ajax({
url: formURL,
type: 'POST',
data: formData,
processData: false,
contentType: false,
async: true,
cache: false,
enctype: "multipart/form-data",
dataType: "json",
success: function (data) {
if (data.success) {
alert(data.success)
}
if (data.error) {
alert(data.error)
}
}
});
});
<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/1.7.1/jquery.min.js"></script>
<form class="form-horizontal" id="form-id" action="masterFileController" enctype="multipart/form-data">
<button class="btn-success btn" type="submit" id="btn-save" >Submit</button>
</form>
servlet responce as "out.print("your responce");"

Upload Avatar To Server Using AJAX

I have looked into the best way to do this and keep getting conflicting information and advice on the various demonstrations.
My code is as follows...
html
<img src="http://www.gravatar.com/avatar/205e460b479e2e5b48aec07710c08d50?s=265&d=identicon&r=PG" style="border: thin solid #999999;"/>
<p>Change<span class="pull-right">Powered by Gravatar</span></p>
<input type="file" name="avatar-uploader" id="avatar-uploader" style="display: none;" />
javascript
$('input[type=file]').on('change', function(){
$.ajax({
url: "/ajax/upload-new-avatar.ajax.php", // Url to which the request is send
type: "POST", // Type of request to be send, called as method
data: new FormData(this), // Data sent to server, a set of key/value pairs (i.e. form fields and values)
contentType: false, // The content type used when sending data to the server.
cache: false, // To unable request pages to be cached
processData:false, // To send DOMDocument or non processed data file it is set to false
success: function(data) // A function to be called if request succeeds
{
alert("Success");
}
});
});
PHP: /ajax/upload-new-avatar.ajax.php
error_reporting(E_ALL);
ini_set('display_errors', 1);
session_start();
$sourcePath = $_FILES['avatar-uploader']['tmp_name']; // Storing source path of the file in a variable
$targetPath = "".$_FILES['avatar-uploader']['name']; // Target path where file is to be stored
move_uploaded_file($sourcePath,$targetPath) ; // Moving Uploaded file
I'm sure there is something simple that I am missing here and i'm going to feel pretty stupid afterwards but could someone explain to me why the image isn't being uploaded to the server and saved in the AJAX directory for further processing.
What I need it to do is when the user clicks on the "change" hyperlink below the image it opens a file upload dialog (working), once an image has been selected it automatically uploads to the server over an AJAX connection (possibly working, logging shows the PHP file is being triggered), and then the image file needs to be saved in the AJAX directory to be further processed later in the code for it to be uploaded to the avatar service.
Thanks in advance.
Have managed to get it working...
Here is my amended code...
Javascript
$('input[type=file]').on('change', function(event){
files = event.target.files;
event.stopPropagation(); // Stop stuff happening
event.preventDefault(); // Totally stop stuff happening
$("#avatar-status").text("Loading new avatar...");
$("#avatar").css("opacity", "0.4");
$("#avatar").css("filter", "alpha(opacity=40);");
//Create a formdata object and add the files
var data = new FormData();
$.each(files, function(key, value) {
data.append(key, value);
});
$.ajax({
url: '/ajax/upload-new-avatar.ajax.php?files',
type: 'POST',
data: data,
cache: false,
dataType: 'json',
processData: false, // Don't process the files
contentType: false, // Set content type to false as jQuery will tell the server its a query string request
success: function(data, textStatus, jqXHR) {
if(typeof data.error === 'undefined') {
//Success so call function to process the form
//submitForm(event, data);
$("#avatar-status").text("Powered by Gravatar");
$("#avatar").css("opacity", "");
$("#avatar").css("filter", "");
} else {
//Handle errors here
alert('ERRORS: ' + textStatus);
}
},
error: function(jqXHR, textStatus, errorThrown) {
//Handle errors here
alert('ERRORS: ' + textStatus);
}
});
});
PHP
session_start();
require_once("../libraries/logging.lib.php");
new LogEntry("AJAX Upload Started - UploadNewAvatar", Log::DEBUG, "AvatarUpload");
sleep(3);
$data = array();
if(isset($_GET['files'])) {
$error = false;
$files = array();
$uploaddir = '../tmp/';
foreach($_FILES as $file) {
if(move_uploaded_file($file['tmp_name'], $uploaddir .basename($file['name']))) {
$files[] = $uploaddir .$file['name'];
new LogEntry("UploadNewAvatar - Upload Successful", Log::DEBUG, "AvatarUpload");
} else {
$error = true;
new LogEntry("UploadNewAvatar - Errors Occured", Log::ERROR, "AvatarUpload");
}
}
$data = ($error) ? array('error' => 'There was an error uploading your files') : array('files' => $files);
} else {
$data = array('success' => 'Form was submitted', 'formData' => $_POST);
new LogEntry("UploadNewAvatar - Form was submitted successfully", Log::DEBUG, "AvatarUpload");
}
echo json_encode($data);
HTML
<img id="avatar" src="http://www.gravatar.com/avatar/205e460b479e2e5b48aec07710c08d50?s=265&d=identicon&r=PG" style="border: thin solid #999999;"/>
<p>Change<span id="avatar-status" class="pull-right">Powered by Gravatar</span></p>
<input type="file" name="upfile" id="upfile" style="display: none;" />

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