In my page I had only one html file input :
input type="file" name="file" files-model="service.file" id="filo" accept=".zip" />
And I used to get my file to be uploaded in a multi-part file request like this :
var fd = new FormData();
fd.append('file', filo.files[0]);//filo is the id of the input
//and send my http request having fd as a parameter
The problem is that now I have auto generated forms in my page with angular ng-repeat so I can no more get my file using the input ID, how can I do this now ?
Cheers,
It will be a bit tricky with javascript. this library worked for me. see example below.
<a class="btn btn-sm btn-primary"
type="file"
ngf-select="receiveSlectedFiles($file)">
Select Logo
</a>
whenever you select file, you will get it in receiveSlectedFiles($file) function. when user select valid file it will be in $file if not the $file will be null.
example in controller:
var arrayOfFiles = [];
$scope.receiveSlectedFiles = function (file) {
if (file) {
arrayOfFiles.push(file);
}
};
In service, inject Upload and then use below code to upload image:
this.uploadMultiple = function (files) {
var uri = yourApiUrl;
return Upload.upload({
url: uri,
arrayKey: '',
data: {
files: files
}
})
};
in controller, you just need to pass array of files to this service.
Related
I have the following.
<form method="post" action="/send" enctype="multipart/form-data">
<input type="file" name="filename" id="AttachFile">
</form>
I want to change the name of the file the user uploads.
If the user selects "Document.docx" I want to change it to "Bank - Document.docx".
I still want to read the file the user selected, not some other file, just use a different name for it when sending to the server.
I'm working within bounds of an application which doesn't allow control of the server side, so ideally I need to do this on the client. Furthermore I need this to work within the confines of a form.
I have tried variations of the following without success:
document.getElementById("AttachFile").name = "test.txt"
document.getElementById("AttachFile").files = "test.txt"
document.getElementById("AttachFile").value ="test.txt"
You can do it through the File API. We can also use the Blob API to be compatible with Microsoft edge.
var file = document.getElementById("AttachFile").files[0];
var newFile = new File([file], "Bank - Document.docx", {
type: file.type,
});
Here's a complete example — see comments:
HTML:
<input type="file" id="AttachFile">
<input type="button" id="BtnSend" value="Send">
JavaScript:
document.getElementById("BtnSend").addEventListener("click", function() {
// Get the file the user picked
var files = document.getElementById("AttachFile").files;
if (!files.length) {
return;
}
var file = files[0];
// Create a new one with the data but a new name
var newFile = new File([file], "Bank - Document.docx", {
type: file.type,
});
// Build the FormData to send
var data = new FormData();
data.set("AttachFile", newFile);
// Send it
fetch("/some/url", {
method: "POST",
body: data
})
.then(response => {
if (!response.ok) {
throw new Error("HTTP error " + response.status);
}
return response.text(); // or response.json() or whatever
})
.then(response => {
// Do something with the response
})
.catch(error => {
// Do something with the error
});
});
You can't rename the file using a standard form submission. The name of the file being uploaded is read-only. To do this, you'd have to do it server-side. (The designers of file uploads seem to have either not considered this rename-on-upload use case or not felt it needed to be addressed by the API.)
However, you can prevent the default form submission and instead submit it programmatically via ajax, which does allow you to rename the file; see man tou's answer.
If you cannot work on the server side then you have to either rename the file BEFORE upload or AFTER download. How you present the name for the user is you to decide.
We're trying to upload a song (.mp3) file from a JSP frontend written in HTML / Javascript. We need to upload to our Java backend using websockets. Does anyone have any suggestions on how we would could go about doing this?
Currently we are doing something like this on our JSP file:
<h1>Please Choose a Song file</h1>
<form name = "JSONUploadForm">
<input type = "file" name="file" accept = ".mp3"/> <br/>
<input type = "button" value = "Click to upload!" name = "button" onClick = "submitSong();"/>
</form>
Then we have our javascript function submitSong()
function submitSong(){
var songStuffs = document.getElementById("file");
console.log(songStuffs); --> we get "null" here
sendMessage(songStuffs);
alert("song sent");
}
function sendMessage(val, string) {
socket.send(string);
return false;
}
Also, here is our connect to server function. However, this functions correctly.
function connectToServer() {
socket = new
WebSocket("ws://localhost:8080/Project/socket");
socket.onopen = function(event) {
console.log("connected!");
}
You can also see our server side (.java file):
#OnMessage
public void onMessage(String message, Session session) throws IOException, EncodeException {
FileWriter fw = new FileWriter(new File(songName + ".mp3"));
fw.write(song);
BufferedReader fr = new BufferedReader(new FileReader(songName + ".mp3"));
String data = fr.readLine();
System.out.println("Song: " + data); --> Here we get "song: null"
}
Any suggestions would be greatly appreciated!
In your code you have an error
"var songStuffs = document.getElementById("file");"
Your file input without id.
this will work "var songStuffs = document.querySelector("[name=file]");"
I prefer using querySelector, because it mo flexeble and works exactly like jquery query selectors)))
You do not need any form, for upload files.
Please read this article https://www.html5rocks.com/en/tutorials/websockets/basics/,
it will be useful for you (search words "blob" at the page)
Html
<input id="file" type = "file" name="file" accept = ".mp3"/>
Code
var fileInput = document.querySelector("#file");
fileInput.addEventListener("change",function(){
connection.send(fileInput.files[0]);
});
If you need to send file and fields, you have 3 variants
create JSON {"field1":"value1","field2":"value1",
"file":"base64"}
manualy create formdata and parse form data at the
server with some webform stuff (example
https://stackoverflow.com/a/47279216/5138198)
Firstly send JSON
data, secondly send a file
Try with this, If you have to upload file you should add enctype in form.
<form enctype="multipart/form-data">
<input type = "file" name="file" id="song" accept = ".mp3"/>
</form>
Update:
You can use simply WebSocketFileTransfer plugin to send your file. This plugin helps in with many features like Auth, progress status, file/blob compatibility.
var songStuffs = document.getElementById("song").files[0];
var transfer = new WebSocketFileTransfer({
url: 'ws://ip:port/path/to/upload_web_socket_server',
file: songStuffs,
progress: function(event) {
// Update the progress bar
},
success: function(event) {
// Do something
}
});
transfer.start();
I have a php file which takes a zip file and unpacks it then places it at the desired path on my server.
It works great with a typical form that calls on the php file in the action. I am trying to make this work with AJAX but I have tried every piece of code I can find without any luck.
Is there something here I am missing? Surely this can be done?
Form for uploading the zip file,
<div id="response"></div>
<form enctype="multipart/form-data" method="post" action="">
<label>Choose a zip file to upload: <input type="file" name="zip_file" id="zip_file" /></label>
<br />
<input type="submit" name="submit" value="Upload" onClick="uploadZip()" />
</form>
Current JS - I get no errors, the page actually reloads with my current script..
<script>
function uploadZip() {
formdata = new FormData();
if (formdata) {
$('.main-content').html('<img src="LoaderIcon.gif" />');
$.ajax({
url: "assets/upload-plugin.php",
type: "POST",
data: formdata,
processData: false,
contentType: false,
success: function (res){
document.getElementById("response").innerHTML = res;
}
});
}
}
</script>
php script which handles uploading the zip and unzipping it before placing it on the server.
function rmdir_recursive($dir) {
foreach(scandir($dir) as $file) {
if ('.' === $file || '..' === $file) continue;
if (is_dir("$dir/$file")) rmdir_recursive("$dir/$file");
else unlink("$dir/$file");
}
rmdir($dir);
}
if($_FILES["zip_file"]["name"]) {
$filename = $_FILES["zip_file"]["name"];
$source = $_FILES["zip_file"]["tmp_name"];
$type = $_FILES["zip_file"]["type"];
$name = explode(".", $filename);
$accepted_types = array('application/zip', 'application/x-zip-compressed', 'multipart/x-zip', 'application/x-compressed');
foreach($accepted_types as $mime_type) {
if($mime_type == $type) {
$okay = true;
break;
}
}
$continue = strtolower($name[1]) == 'zip' ? true : false;
if(!$continue) {
$message = "The file you are trying to upload is not a .zip file. Please try again.";
}
/* PHP current path */
$path = '../plugins/'; // absolute path to the directory where zipper.php is in
$filenoext = basename ($filename, '.zip'); // absolute path to the directory where zipper.php is in (lowercase)
$filenoext = basename ($filenoext, '.ZIP'); // absolute path to the directory where zipper.php is in (when uppercase)
$targetdir = $path . $filenoext; // target directory
$targetzip = $path . $filename; // target zip file
/* create directory if not exists', otherwise overwrite */
/* target directory is same as filename without extension */
if (is_dir($targetdir)) rmdir_recursive ( $targetdir);
mkdir($targetdir, 0777);
/* here it is really happening */
if(move_uploaded_file($source, $targetzip)) {
$zip = new ZipArchive();
$x = $zip->open($targetzip); // open the zip file to extract
if ($x === true) {
$zip->extractTo($targetdir); // place in the directory with same name
$zip->close();
unlink($targetzip);
}
$message = "Your .zip file was uploaded and unpacked.";
} else {
$message = "There was a problem with the upload. Please try again.";
}
}
This php function works great as long as I do this with the form action. So I am sure my problem exist in the AJAX function.
Thanks for any help you can provide.
formdata = new FormData();
You've created a FormData object but you never put any data into it.
The easiest approach is to specify the form:
formdata = new FormData(document.forms[0]);
You also need to stop the submit button from actually submitting the form so that the JS can do something.
A cleaner approach would be to:
Stop using intrinsic event attributes
Use the submit handler for the form
Get the form from the event
<input type="submit" name="submit" value="Upload" onClick="uploadZip()" />
Becomes:
<input type="submit" name="submit" value="Upload">
function uploadZip() {
formdata = new FormData();
becomes:
function uploadZip(event) {
var formdata = new FormData(this);
// Rest of function
event.preventDefault();
}
and you add:
jQuery("form").on("submit", uploadZip);
We are taking over a site that is using Angular js for a form for data and multi-file upload (up to 4 files).
We are posting to a .php file. I have been able to separate and capture the field data - but can only seem to 'see' one file. If I upload multiple files I 'see' only the last one added to the form.
Can someone please help me to get the files using PHP?
Form HTML (for files):
<ul class="attachments">
<li ng-repeat="attachment in attachments" class="attachment">
{{attachment.name}}
<i class="btn-submit-att-x-hover-off"></i>
</li>
</ul>
<button type="button" class="attachments-button btn btn-default" ngf-select ngf-change="onFileSelect($files)" ngf-multiple="true" ng-if="attachments.length < 4">
<i class="btn-submit-attch"></i>Upload Attachments
</button>
Angular JS Code - for the files as I see:
$scope.attachments = [];
//onFileSelect adds file attachments to the $scope up to the
//configured maximum value.
$scope.onFileSelect = function ($files) {
var MAX_ATTACHMENTS = 4;
for(var i=0; i < $files.length; i++){
if($scope.attachments.length >= MAX_ATTACHMENTS){
break;
}
$scope.attachments.push($files[i]);
}
};
$scope.removeAttachment = function (attachment) {
var index = $scope.attachments.indexOf(attachment);
if (index > -1) {
$scope.attachments.splice(index, 1);
}
};
//POST form data
$scope.upload = Upload.upload({
url: 'forms/pressrelease.html',
method: 'POST',
data: $scope.model,
file: $scope.attachments,
fileFormDataName: "files"
})
.success(function (data, status, headers, config) {
alertSuccess("Your press release has been submitted successfully!");
$scope.showSuccess = true;
}
The only PHP code we've used that has given us anything back - but again - only 1, or the last file name is:
if(isset($_FILES['files'])){
$files = $_FILES['files'];
$sentfilename = $_FILES['files']['name'];
$email_body .= "I see files: $sentfilename<br />";
$tmpsentfile = $_FILES['files']['tmp_name'];
$i=0;
$getfile = $files[0];
$sentfilename = $getfile;
$sentfilesize = $_FILES['files']['size'];
$errormessage = $_FILES['files']['error'];
$email_body .= "I see files: $sentfilename<br />";
}
If anyone one can show us the PHP code we need to use to capture these uploaded files.
Thanks
Create a names array and post the dynamic file names:
var names = []
foreach file:
names.push(files[i].name)
in upload:
method: 'POST',
data: $scope.model,
file: $scope.attachments,
fileFormDataName: names
I have some trouble sending the following html input type to my php script through ajax. I'm guessing that I have to define the file in tje js code hoverver how to do so when I have more variables that are taking information from the same file (se php code)?
<input id="imagefile" class="file" type="file" name="image" />
through this code
$("#addmedia").click(function(ev) {
ev.preventDefault();
var p = $("#p").val();
var mediatype = $("#mediatype option:selected").val();
var addmediatype = $("#mediatype option:selected").val();
var title = $("#title").val();
var video = $("#medialink").val();
var imagefile = $("#imagefile").val();
$.post("lib/action.php", {
mediatype: mediatype,
addmediatype: addmediatype,
title: title,
video: video,
addmedia: true
}, function(data) {
$("#notify").hide().html("<h1>!</h1><h2>" + data + "</h2>").slideDown(500);
setTimeout(function() { $("#notify").slideUp(500) }, 2500);
});
});
so that it works with my php upload script.
In my php code i use following variables to get infro from the file
if( $_POST['p'] == 1 ) {
$name = $_FILES['image']['name'];
$temp = $_FILES['image']['tmp_name'];
$type = $_FILES['image']['type'];
$size = $_FILES['image']['size'];
(...)
}
When you use $().val for a file field, you're only getting the filename because of security restrictions.
One solution (for IE 10+, Chrome, FF) is to read the file contents using https://developer.mozilla.org/en-US/docs/Web/API/FileReader, base64 encode it and upload it. See Reading client side text file using Javascript
document.getElementById('file').addEventListener('change', readFile, false);
function readFile (evt) {
var files = evt.target.files;
var file = files[0];
var reader = new FileReader();
reader.onload = function() {
console.log(this.result);
}
reader.readAsText(file)
}
Note that there are too many gotchas when uploading files through AJAX and can't possibly address with a concise answer as StackOverflow answers should be.
The easiest workaround is to not send it using AJAX, use a regular form upload, but target a hidden iframe so your page doesn't reload.
See:
Sending binary data in javascript over HTTP
Multiupload with PHP/JavaScript
http://www.html5rocks.com/en/tutorials/file/xhr2/
http://blueimp.github.io/jQuery-File-Upload/