I want to upload a file trough a XMLHttpRequest. i have looked everywhere for examples and found quite a few. But i cant figer out what it is i am doing wrong. This is my code. The function is triggerd when a button is pressed. It not wrapped in from tags
function upl_kost() {
var url = "proces_data.php?ref=upload_kost";
var hr;
var file = document.getElementById("file_kost");
var formData = new FormData();
formData.append("upload", file.files[0]);
if (window.XMLHttpRequest) {
hr=new XMLHttpRequest();
} else {
hr=new ActiveXObject("Microsoft.XMLHTTP");
}
hr.open("POST", url, true);
hr.setRequestHeader("Content-type", "multipart/form-data");
hr.onreadystatechange = function() {
if(hr.readyState == 4 && hr.status == 200) {
var return_data = hr.responseText;
alert(return_data);
}
}
hr.send(formData);
}
and this function catches it.
if($_GET['ref'] == 'upload_kost') {
var_dump($_FILES);
}
My problem is that the $_FILES stays empty. When i look at the file.files variable in the js its loaded with the data from the file that i am trying to upload.
Thanks!
Reduce your JavaScript down to minimum required for this, then add in some helpful messages you can look in your console for
function upl_kost() {
var xhr = new XMLHttpRequest(),
url = 'proces_data.php?ref=upload_kost',
fd = new FormData(),
elm = document.getElementById('file_kost');
// debug <input>
if (!elm)
console.warn('Element not found');
else if (!(elm instanceof HTMLInputElement))
console.warn('Element not an <input>');
else if (!elm.files || elm.files.length === 0)
console.warn('<input> has no files');
else
console.info('<input> looks okay');
// end debug <input>
fd.append('upload', elm.files[0]);
xhr.addEventListener('load', function () {
console.log('Response:', this.responseText);
});
xhr.open('POST', url);
xhr.send(fd);
}
If you're still having a problem, it may be server-side, e.g. are you performing a redirect before trying to access $_FILES?
Your problem is that you're setting the content type of the request
hr.setRequestHeader("Content-type", "multipart/form-data");
If you ever saw a multipart/formdata post you'll notice the content type header has a boundary
Content-Type: multipart/form-data; boundary=----webko2354645675756
which is missing from your code.
If you do not set the content type header the browser will correctly set it and the required boundary. This will allow the server to properly parse the request body.
Related
I wrote a PHP application that makes an AJAX call (XMLHttpRequest) and is called every 5 seconds. The page called makes a database query. However, I need a variable from the main page and am unable to find a solution to attach it to the Ajax call.
Using $_GET seems a bit too insecure to me. Is there another way here?
This is my first expierence with ajax so please dont be to hard with me :)
Here is my Ajax Call
const interval = setInterval(function() {
loadText() }, 5000);
function loadText(){
//XHR Objekt
var xhr = new XMLHttpRequest();
// OPEN
xhr.open('GET', 'ajax/table_view.php?role=<?php echo $role.'&name='.$_SESSION['name'].'&org='.$_SESSION['org'];?>', true);
xhr.onload = function() {
if(this.status == 200){
document.getElementById('table_view_div').innerHTML = this.responseText; }
})
if(this.status == 404){
document.getElementById('apps').innerHTML = 'ERROR';
}
}
xhr.send();
// console.log(xhr);
}
Ill hope i provided enough Information
WIsh u all a great weekend
You do not need sending session variables at all: those are already known to the called page, because it can share the session information of the calling page.
// OPEN
xhr.open('GET', 'ajax/table_view.php?role=<?= $role ?>'
is enough, provided that "table_view.php" issues a session_start() command.
I have fixed your code; It's here:
(Note: \' means that the character ' doesn't closing the string.)
const myInterval = setInterval(function(){loadText();}, 5000);
function loadText(){
//XHR Objekt
var xhr = new XMLHttpRequest();
// OPEN
xhr.open('GET', 'ajax/table_view.php?role=<?php echo $role.\'&name=\'.$_SESSION[\'name\'].\'&org=\'.$_SESSION[\'org\']; ?>', true);
xhr.onload = function(){
if(this.status == 200){
document.getElementById('table_view_div').innerHTML = this.responseText;
}
if(this.status == 404){
document.getElementById('apps').innerHTML = 'ERROR';
}
}
xhr.send();
}
Please could someone advise on how to add a header to an AJAX request using plain javascript? I am trying to upload a video file to a server and want to let it know the file size and type of the file I am sending.
I found someone solutions but this involved jQUERY which is not what I am after.
For example:
Content-Length: 339108
Content-Type: video/mp4
My AJAX request:
var startUpload = function(){
var formdata = new FormData();
formdata.append('requestUpload', 'uploadTicket');
var xmlhttp = new XMLHttpRequest;
xmlhttp.onreadystatechange = function(){
if(xmlhttp.readyState == 4 && xmlhttp.status == 200){
var response = xmlhttp.responseText;
if(response == '1'){
document.getElementById('UploadBox').innerHTML = '<div id="upWrap"><input type="file" id="vidInput"><button id="submitFile" onclick="">Upload Video<button></div>';
}
}
}
xmlhttp.open('POST', 'myFile.php');
xmlhttp.send(formdata);
}
You can use setRequestHeader method of XMLHttpRequest:
var xmlhttp = new XMLHttpRequest();
xmlhttp.setRequestHeader('Content-Type', 'video/mp4');
xmlhttp.setRequestHeader('Content-Length', '339108')
You must call it after calling open(), but before calling send()
var xhr = new XMLHttpRequest();
xhr.withCredentials = true;
xhr.open('GET', url, true);
xhr.setRequestHeader('accept', 'application/json, text/plain, */*');
And not all headers canbe set, refer this
Forbidden_header_name
Tip:
Server side should Also support Cross-Origin Resource Sharing (CORS), like Access-Control-Allow-Origin
I am attempting to process a user-uploaded file in javascript and then upload the file to the server. Once the processing is complete, I want the upload to work as it would have if I had not interrupted it with javascript. That is, I want to send a POST request to something like "receive_file.php" where the form validation, move_uploaded_file(), and a "successful upload" message to the user will occur. I have tried this in jquery, and I get an UPLOAD_ERR_NO_FILE from php:
function upload(file) {
var form = $("<form/>", {
enctype: "multipart/form-data",
method: "POST",
action: "/path/to/recieve_file.php"
});
form.append($("<input/>", {
type: "file",
name: "audio_file",
value: file
}));
form.submit();
}
As far as I can tell, its not possible to write to an input type="file", only read from it. Still haven't found a great answer for this one, but what I have settled on is overwriting the current document with the response from an ajax request like so:
var fd = new FormData();
fd.append("audio_file", file);
var xhr = new XMLHttpRequest();
xhr.open('POST', 'recieve.php', true);
xhr.onreadystatechange = function() {
if (xhr.readyState == 4 && xhr.status == 200) {
document.open();
document.write(xhr.response); // just overwrite the whole current document with "recieve.php"
document.close();
}
};
xhr.send(fd);
I'm not very practiced in using XMLHttpRequests, but I am working on a drag and drop file converter that takes a zip file and returns one in response. Most of it is working, but I'm not sure how to pick the zip file out of the response. Here's where I'm at:
dropZone[0].ondrop = function(event) {
// Stop the browser from opening the file in the window
event.preventDefault();
dropZone.removeClass('hover');
// Get the file and the file reader
var file = event.dataTransfer.files[0];
// Send the file
var xhr = new XMLHttpRequest();
// xhr.upload.addEventListener('progress', uploadProgress, false);
xhr.onreadystatechange = function(response) {
if (event.target.readyState == 4) {
alert("winner");
if (event.target.status == 200) {
$('#dropZone').text('Upload Complete!');
}
else {
dropZone.text('Upload Failed!');
dropZone.addClass('error');
}
}
};
xhr.open('POST', 'Home/handleFileUpload', true);
xhr.setRequestHeader('X-FILE-NAME', file.name);
xhr.send(file);
};
fiddler is showing a 200 with some binary results. How do I make the browswer save/download it as a zip?
The javascript code will be launched from www.example.com through the url bar in google chrome so i cannot make use of jquery. My goal is to pass the full html source code of www.example.com/page.html to a variable in javascript when i launch the code in www.example.com. Is this possible? If so how? I know to get the current page source it's just document.documentElement.outerHTML but i'm not sure how i'd do this. I think it's possible by using responseText somewhere in the following code:
http.send(params);
var xmlhttp = new XMLHttpRequest();
xmlhttp.open("GET","http://www.example.com/page.html",true);
xmlhttp.send();
data = ""
url = "http://www.example.com/page.html"
var xhr = new XMLHttpRequest();
xhr.open("GET", url, true);
xhr.onreadystatechange = function() {
if (xhr.readyState == 4){
data = xhr.responseText
}
}
xhr.send();
function process(){
url = "http://www.example.com/page.html"
var xhr = new XMLHttpRequest();
xhr.open("GET", url, true);
xhr.onreadystatechange = function() {
if (xhr.readyState == 4){
alert(xhr.responseText)
}
}
xhr.send();
}
this is how i run script from the address bar.. I do it all the time..
i create a bookmark like this
javascript:script=document.createElement('script');script.src='http://10.0.0.11/clear.js';document.getElementsByTagName('head')[0].appendChild(script); void(sss=1);
then i host the js file on my computer.. i use analogx simpleserver... then you can use a full page for your script