I've been trying to figure out how to upload a file through ajax for the past several hours and nothing.
Here's the code:
HTML:
<form action="" method="post" id="uploadForm" enctype="multipart/form-data">
<input type="file" name="image" id="image">
<input type="submit">
</form>
JS:
<script>
jQuery(document).ready(function(){
jQuery('form#uploadForm').on('submit', function(e){
e.preventDefault();
var file = jQuery('#image')[0].files[0];
var form_data = new FormData( jQuery("form#uploadForm")[0] );
form_data.append( 'image', file );
jQuery.ajax({
url: 'index.php?a=do',
type: 'POST',
processData: false,
contentType: false,
cache: false,
data: form_data,
success: function(response) {
console.log(response);
},
});
return false;
});
});
</script>
PHP:
<?php
$a = isset($_GET['a']) ? $_GET['a'] : '';
if($a <> '') {
echo "result - ";
var_dump($_POST);
die();
}
?>
As a result I get an empty array, however if I leave the file field empty, then I get:
result - array(1) {
["image"]=>
string(9) "undefined"
}
I've tried serialize(), serializeObject(), serializeArray(), $.param and every damn time I get "undefined function" error in console.
I went through dozens of similar questions on stackoverflow and nothing helped. I tried doing $.post instead of $.ajax and the "data" field which contains form_data is empty.
I need this for a Wordpress plugin and I'm trying to avoid using 3rd party JS plugins for the upload part.
$_FILES is where you check for uploaded files not $_POST.
Also in your code you actually upload the file twice as it is in the form you instantiated the form data object with then you add it again with append.
Do either
var form_data = new FormData( jQuery("form#uploadForm")[0] );
or
var form_data = new FormData();
form_data.append( 'image', file );
<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: "ajax_php_file.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>
Related
For past few days i have been struggling to submit a form with jQuery and AJAX. The problem i'm facing is to upload the image in the form field.
My form is something like this:
<form action="#" method="GET" role="form" enctype="multipart/form-data">
<input type="text" placeholder="Name" name="name">
<input type="file" name="img" multiple>
<button type="submit">Submit </button>
</form>
and my jQuery script for getting the form value is something like this:
$("form").submit(function (event) {
$.dataArray = $(this).serializeArray(); // array of form data
console.log($.dataArray);
event.preventDefault();
});
But this returns all the field values except image one, in case of image is return null.
How do I store in the dataarray?
I want to store so I can send the value to the server through the AJAX.
For uploading single image its like this
<html>
<head>
<meta charset="UTF-8">
<title>AJAX image upload with, jQuery</title>
<script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.12.4/jquery.min.js"></script>
<script type="text/javascript">
$(document).ready(function (e) {
$('#upload').on('click', function () {
var file_data = $('#file').prop('files')[0];
var form_data = new FormData();
form_data.append('file', file_data);
$.ajax({
url: 'http://localhost/ci/index.php/welcome/upload', // point to server-side controller method
dataType: 'text', // what to expect back from the server
cache: false,
contentType: false,
processData: false,
data: form_data,
type: 'post',
success: function (response) {
$('#msg').html(response); // display success response from the server
},
error: function (response) {
$('#msg').html(response); // display error response from the server
}
});
});
});
</script>
</head>
<body>
<p id="msg"></p>
<input type="file" id="file" name="file" multiple />
<button id="upload">Upload</button>
</body>
</html>
For multiple images u will have to loop its kinda different
I have found a similar question, hope it will help you.
Upload image using jquery
Another option to consider is to use some sort of jQuery plugin to upload images like Cloudinary and include it in your HTML pages :
<script src='jquery.min.js' type='text/javascript'></script>
<script src='jquery.cloudinary.js' type='text/javascript'></script>
and then include all required jQuery files:
<script src='jquery.min.js' type='text/javascript'></script>
<script src='jquery.ui.widget.js' type='text/javascript'></script>
<script src='jquery.iframe-transport.js' type='text/javascript'></script>
<script src='jquery.fileupload.js' type='text/javascript'></script>
<script src='jquery.cloudinary.js' type='text/javascript'></script>
try this code, it's work for me.
$("form").submit(function (event) {
var form_data = new FormData($(this));
$.ajax({
url : url,
type : 'POST',
data : form_data,
processData: false, // tell jQuery not to process the data
contentType: false,
success : function(resp){
}
});
});
Try this code. using formData()
$("form").submit(function (event) {
var formData = new FormData($(this));
$.ajax({
url: url,
type: 'POST',
data: formData,
async: false,
success: function (data) {
//success callback
},
cache: false,
contentType: false,
processData: false
});
});
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<form action="#" method="GET" role="form" enctype="multipart/form-data">
<input type="text" placeholder="Name" name="name">
<input type="file" name="img" multiple>
<button type="submit">Submit </button>
</form>
serialize() method not able to post file data.
For sending file using ajax use FormData instead of serializing
HTML5 introduces FormData to allow developers to build forms objects dynamically, and then to send this form object via AJAX.
your Html
<form action="upload_image.php" id="form_img" method="GET" role="form" enctype="multipart/form-data">
<input type="text" placeholder="Name" name="name">
<input type="file" name="img" multiple>
<button type="submit">Submit </button>
</form>
AJAX call
<script type="text/javascript" src="https://code.jquery.com/jquery-3.2.1.min.js"></script>
<script type="text/javascript">
$(document).ready(function() {
$("#form_img").submit(function(e){
e.preventDefault();
var formData = new FormData($("#form_img")[0]);
$.ajax({
url : $("#form_img").attr('action'),
type : 'POST',
data : formData,
contentType : false,
processData : false,
success: function(resp) {
console.log(resp);
}
});
});
});
</script>
upload_image.php
print_r($_FILES) //check you get file data or not
Try this way.Hope it will help you
Please check the follow the code, which i am using to upload image.
$.ajax({
url: UPLOADURL, // Url to which the request is send
type: "POST", // Type of request to be send, called as method
data: new FormData(this),// Data sent to server, a set of key/value pairs representing form fields and values
contentType: false,// The content type used when sending data to the server. Default is: "application/x-www-form-urlencoded"
cache: false,// To unable request pages to be cached
processData:false,// To send DOMDocument or non processed data file it is set to false (i.e. data should not be in the form of string)
success: function(data)// A function to be called if request succeeds
{
data = JSON.parse(data);
console.log(data);
if(data.status == "Success"){
attachmentListing();
//$("#mailerMessage").html(data.data.mailStatus);
//$("#mailerMessage").fadeIn();
setTimeout(function () {
$("#mailerMessage").fadeOut();
},5000);
}else{
toastr.warning(data.status);
}
$("#ajaxloader").addClass("hideajaxLoader");
},
error: function (jqXHR, errdata, errorThrown) {
log("error");
$("#ajaxloader").addClass("hideajaxLoader");
}
});
I have three type of inputs those are file input, text input and variable. I want to upload those inputs by sending data to PHP file by using Ajax JSON. Also I want to know how to capture these data in PHP file.
I am using HTML code without form syntax. variable data name as a val1 in JQuery code.
HTML Code:
<div class="container" id="post">
<textarea id="posttext" autocomplete="off"></textarea>
<input type="file" name="file" id="file" multiple accept=".mp4, .mov, .m4v, .MPEG-4, .gif, .jpg, .png"/>
<button type="button" id="submitpost">Submit</button>
</div>
JQuery Code:
$(document).on("click", "#submitpost", function(){
var val1 = "Some Datas";
$.ajax({
url: post.php,
type: 'POST',
data: VALUES,
async: false,
success: function (data) {
alert(data)
},
cache: false,
contentType: false,
processData: false
});
});
PHP Code:
<?php
if (!isset($_POST['VALUES']) && !empty($_POST['VALUES'])) {
$params = $_POST['VALUES'];
}
?>
How to get each values in PHP to Upload files and insert text and variable data to database.
Use this code:
Html:
<div class="container" id="post">
<form enctype="multipart/form-data" method="POST" id="myform">
<textarea id="posttext" name="posttext" autocomplete="off"></textarea>
<input type="file" name="file" id="file" multiple accept=".mp4, .mov, .m4v, .MPEG-4, .gif, .jpg, .png"/>
<button type="submit" name="submitpost" id="submitpost">Submit</button>
</form>
</div>
Jquery:
$(document).on("click", "#submitpost", function(e){
$("form#myform").submit();
});
$("form#myform").on('submit', function(e){
e.preventDefault();
var val1 = "Some Data";
var file = this.files[0];
var form = new FormData();
form.append('file', file);
form.append('val1', val1);
form.append('posttext', $('#posttext').val());
$.ajax({
url : "post.php",
type: "POST",
cache: false,
async: false,
contentType: false,
processData: false,
data : form,
success: function(response){
alert(response);
}
});
});
PHP Code:
<?php
if (isset($_POST) && !empty($_POST)) {
print_r($_POST);
print_r($_FILES);
}
?>
As the title says, i have try many times to get it working, but without success... the alert window show always the entire source of html part.
Where am i wrong? Please, help me.
Thanks.
PHP:
<?php
if (isset($_POST['send'])) {
$file = $_POST['fblink'];
$contents = file_get_contents($file);
echo $_POST['fblink'];
exit;
}
?>
HTML:
<html>
<head>
<script type="text/javascript" src="https://code.jquery.com/jquery-1.8.3.min.js"></script>
<script>
$(document).ready(function() {
$("input#invia").click(function(e) {
if( !confirm('Are you sure?')) {
return false;
}
var fbvideo = $("#videolink").val();
$.ajax({
type: 'POST',
data: fbvideo ,
cache: false,
//dataType: "html",
success: function(test){
alert(test);
}
});
e.preventDefault();
});
});
</script>
</head>
<div style="position:relative; margin-top:2000px;">
<form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post">
<input id="videolink" type="text" name="fblink" style="width:500px;">
<br>
<input id="invia" type="submit" name="send" value="Get Link!">
</form>
</div>
</html>
Your Problem is that you think, that your form fields are automatic send with ajax. But you must define each one into it.
Try this code:
<script>
$(document).ready(function() {
$("input#invia").click(function(e) {
if( !confirm('Are you sure?')) {
return false;
}
var fbvideo = $("#videolink").val();
$.ajax({
type: 'POST',
data: {
send: 1,
fblink: fbvideo
},
cache: false,
//dataType: "html",
success: function(test){
alert(test);
}
});
e.preventDefault();
});
});
</script>
Instead of define each input for itself, jQuery has the method .serialize(), with this method you can easily read all input of your form.
Look at the docs.
And maybe You use .submit() instead of click the submit button. Because the user have multiple ways the submit the form.
$("input#invia").closest('form').submit(function(e) {
You must specify the url to where you're going to send the data.
It can be manual or you can get the action attribute of your form tag.
If you need some additional as the send value, that's not as input in the form you can add it to the serialized form values with formSerializedValues += "&item" + value;' where formSerializedValues is already defined previously as formSerializedValues = <form>.serialize() (<form> is your current form).
<html>
<head>
<script type="text/javascript" src="https://code.jquery.com/jquery-1.8.3.min.js"></script>
<script>
$(document).ready(function() {
$("#invia").click(function(e) {
e.preventDefault();
if (!confirm('Are you sure?')) {
return false;
}
// Now you're getting the data in the form to send as object
let fbvideo = $("#videolink").parent().serialize();
// Better if you give it an id or a class to identify it
let formAction = $("#videolink").parent().attr('action');
// If you need any additional value that's not as input in the form
// fbvideo += '&item' + value;
$.ajax({
type: 'POST',
data: fbvideo ,
cache: false,
// dataType: "html",
// url optional in this case
// url: formAction,
success: function(test){
alert(test);
}
});
});
});
</script>
</head>
<body>
<div style="position:relative; margin-top:2000px;">
<form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post">
<input id="videolink" type="text" name="fblink" style="width:500px;">
<br>
<input id="invia" type="submit" name="send" value="Get Link!">
</form>
</div>
</body>
I have a form which is for sending mails.It contain fields such as to_name,subject,message and attachment button.I will create a file input field on clicking the button with class .file_add_btn.
//click event for add files
$(".file_add_btn").click(function(){
if($("#file_div").html() == '')
{
$("#file_div").append('<div class="file_btn_div" id="file_btn_div_first"><input type="file" class="btn_browse" name="file_uploads[]">'+
'<input type="button" class="del_file" value="X"></div>');
}
else
{
if($(document).find('.btn_browse:last').get(0).files.length !==0)
{
$("#file_div").append('<div class="file_btn_div"><input type="file" class="btn_browse" name="file_uploads[]">'+
'<input type="button" class="del_file" value="X"></div>');
}
}
});
I write the following function to include file inputs into formData.
$.fn.serializefiles = function() {
var obj = $(this);
var form_data = new FormData(this[0]);
$.each($(obj).find('.btn_browse'), function(i, tag) {
$.each(tag.files, function(i, file) {
console.log(tag.name+' '+file.name)//this is printing in console
form_data.append(tag.name, file);
});
});
var params = $(obj).serializeArray();
$.each(params, function (i, val) {
console.log(val.name+'<br/>');
console.log(val.value+'<br/>');
**//here file names are not coming.All other elements are coming.They are not adding to form_data object**
form_data.append(val.name, val.value);
});
return form_data;
};
My ajax call is like the following:
$.ajax({
type: "POST",
url: 'process.php',
data: $("#compose_message").serializefiles() ,//formID=#compose_message
asyn: true,
cache: false,
processData: false,
contentType: false,
success:function()
....
I am not able to append the inputs into the form_data object.In console,I see [object FormData] inside the POST on button click.
Edit: My initial comment above was correct. You don't need to do anything fancy except pass the form object into FormData constructor as shown in my example below.
console.log(JSON.stringify(formData)); will never show the values of the FormData. However, if you look in your browser's network tab, the request that gets sent will show the values being passed.
If you want to check the data before it is being passed, you could use this answer.
<!DOCTYPE html>
<html>
<head>
<script src="https://code.jquery.com/jquery-2.1.4.js"></script>
<meta charset="utf-8">
<meta name="viewport" content="width=device-width">
<title>JS Bin</title>
<style>
input {
float: left;
clear: left;
}
</style>
</head>
<body>
<form method="post" enctype="multipart/form-data">
<input type="text" name="text" value="text">
<input type="hidden" name="hidden" value="hidden">
<input type="file" name="file_uploads[]" value="">
<input type="file" name="file_uploads[]" value="">
<input type="file" name="file_uploads[]" value="">
<input type="button" value="Add">
<input type="submit" value="Submit">
</form>
<script>
$(function () {
$('form').on('submit', function (e) {
e.preventDefault();
var formData = new FormData(this);
console.log(JSON.stringify(formData)); // will always be {}
$.ajax({ url: '404', type: 'post', data: formData, processData: false, contentType: false });
});
$('[type=button]').on('click', function () {
$(this).before('<input type="file" name="file_uploads[]" value="">');
});
});
</script>
</body>
</html>
specify exact data for formdata
var formData = new FormData();
formData.append('section', 'general');
formData.append('action', 'previewImg');
// Main magic with files here
formData.append('image', $('input[type=file]')[0].files[0]);
Ajax request with jquery will looks like this:
$.ajax({
url: 'Your url here',
data: formData,
// THIS MUST BE DONE FOR FILE UPLOADING
contentType: false,
processData: false,
// ... Other options like success and etc
})
try this
var form = $("form")[0]; // You need to use standard javascript object here
var formData = new FormData(form);
formData.append('section', 'general');
formData.append('action', 'previewImg');
// Main magic with files here
formData.append('image', $('input[type=file]')[0].files[0]);
I have a file upload form:
<form class="uploadFile" name="uploadFile" method="post" action="process/uploadBuildImages.php" enctype="multipart/form-data">
<div class="uploadImageButton">Upload Image</div>
<input type="file" id="fileUpload" class="fileUpload" name="picture">
</form>
I am posting this form via ajax and formData like so:
var form = $('.uploadFile')[0];
var formData = new FormData(form);
The issue is that I have this same form output up to 4 times on the same page. I am making a CMS with blog posts and the upload image form is in each of the blog posts.
How can I target and post only from the currently posted form? I know its along the lines of using 'this.form etc but always struggle with .next / closest etc. Will look more into it soon.
If I only have one instance of this form on the page then it works fine, but otherwise I get a no file chosen error.
Thanks!
For reference, full JS:
$(document).ready(function(){
$('.uploadFile').submit(function(){
var form = $('.uploadFile')[0];
var formData = new FormData(form);
$.ajax({
type: "POST",
url: "process/uploadBuildImages.php",
data: formData,
dataType: "json",
contentType: false,
processData: false,
success: function(response){
// actions
}
}
});
}):
And there is at least 4 of the same div structures on the page which follow the same format as:
<div class="blogtest" id="'.$postID.'">
<div class="text">
<div class="postoverlay"></div>
<div class="buildtext" id="'.$postTextID.'">'.$convertedtext.'</div>
<form class="uploadFile" name="uploadFile" style="display:$showUploadForm;" method="post" action="process/uploadBuildImages.php" enctype="multipart/form-data">
<input type="hidden" name="MAX_FILE_SIZE" value="30000000" />
<div class="uploadImageButton">Upload Image</div>
<input type="file" id="fileUpload" class="fileUpload" name="picture">
</form>
<form class="updatepost" id="'.$postContentID.'" method="post">
<div class="editor">
<textarea name="muffin" id="'.$ckEID.'" class="ckeditor">'.$textFiltered.'</textarea>
</div>
</form>
</div>
</div>
Echoing the same diagnosis I have made in my comments, you lack context in your function call — your code basically submits the form data in all forms with the class .uploadFile regardless of which one is submitted:
var form = $('.uploadFile')[0]; // Problematic line
You should use $(this) to give it context upon the firing of the submit context, so that you only submit the form data from the form where the event originated from, and not all forms with the class instead. Therefore, you substitute:
var form = $('uploadFile')[0];
with:
// Alternative 1
var form = $(this)[0];
// Alternative 2
var form = this;
In other words:
$(document).ready(function(){
$('.uploadFile').submit(function(){
var form = this; // Fixed line
var formData = new FormData(form);
$.ajax({
type: "POST",
url: "process/uploadBuildImages.php",
data: formData,
dataType: "json",
contentType: false,
processData: false,
success: function(response){
// actions
}
}
});
});
You can learn more what $(this) does in jQuery here and here.