I have one form html then I have 2 input( input text and input file)
my problem input text cannot send post to upload php.
I using jquery ajax to post data
upload.php
<?php
$nama = $_POST['name'];
$filename = $_FILES['file']['name'];
var_dump($nama);
var_dump($filename);
form
<form method='post' action='' enctype="multipart/form-data">
<input type="text" class="form-control" id="name" name="name" placeholder="Enter name" required />
Select file : <input type='file' name='file' id='file' class='form-control' ><br/>
<input type='button' class='btn btn-info' value='Upload' id='upload'>
</form>
<script type="text/javascript" >
$(function() {
$("#upload").click(function() {
var name = $('#name').val();
var fd = new FormData();
var files = $('#file')[0].files[0];
fd.append('file',files);
$.ajax({
type: "POST",
url: "upload.php",
dataType: 'text', // what to expect back from the PHP script, if anything
cache: false,
contentType: false,
processData: false,
data: fd,
success: function(){
}
});
return false;
});
});
</script>
Just remove dataType:text. Or you can use dataType:"json" as per your requirement. Well, you can read more on this dataType here also: $.ajax - dataType
And for name parameter, you are missing to append that variable in fd variable.
So, your script should be like this:
<script type="text/javascript" >
$(function() {
$("#upload").click(function() {
var name = $('#name').val();
var fd = new FormData();
var files = $('#file')[0].files[0];
fd.append('file',files);
fd.append('name',name); // you were missing this, that's why $_POST['name'] was blank
$.ajax({
type: "post",
url: "upload.php",
cache: false,
contentType: false,
processData: false,
data: fd,
success: function(){
}
});
return false;
});
});
</script>
Now try by printing below code into your upload.php file:
<?php
echo $nama = $_POST['name'];
echo $filename = $_FILES['file']['name'];
?>
Related
This is my HTML which I'm generating dynamically using drag and drop functionality.
<form method="POST" id="contact" name="13" class="form-horizontal wpc_contact" novalidate="novalidate" enctype="multipart/form-data">
<fieldset>
<div id="legend" class="">
<legend class="">file demoe 1</legend>
<div id="alert-message" class="alert hidden"></div>
</div>
<div class="control-group">
<!-- Text input-->
<label class="control-label" for="input01">Text input</label>
<div class="controls">
<input type="text" placeholder="placeholder" class="input-xlarge" name="name">
<p class="help-block" style="display:none;">text_input</p>
</div>
<div class="control-group"> </div>
<label class="control-label">File Button</label>
<!-- File Upload -->
<div class="controls">
<input class="input-file" id="fileInput" type="file" name="file">
</div>
</div>
<div class="control-group">
<!-- Button -->
<div class="controls">
<button class="btn btn-success">Button</button>
</div>
</div>
</fieldset>
</form>
This is my JavaScript code:
<script>
$('.wpc_contact').submit(function(event){
var formname = $('.wpc_contact').attr('name');
var form = $('.wpc_contact').serialize();
var FormData = new FormData($(form)[1]);
$.ajax({
url : '<?php echo plugins_url(); ?>'+'/wpc-contact-form/resources/js/tinymce.php',
data : {form:form,formname:formname,ipadd:ipadd,FormData:FormData},
type : 'POST',
processData: false,
contentType: false,
success : function(data){
alert(data);
}
});
}
For correct form data usage you need to do 2 steps.
Preparations
You can give your whole form to FormData() for processing
var form = $('form')[0]; // You need to use standard javascript object here
var formData = new FormData(form);
or specify exact data for FormData()
var formData = new FormData();
formData.append('section', 'general');
formData.append('action', 'previewImg');
// Attach file
formData.append('image', $('input[type=file]')[0].files[0]);
Sending form
Ajax request with jquery will looks like this:
$.ajax({
url: 'Your url here',
data: formData,
type: 'POST',
contentType: false, // NEEDED, DON'T OMIT THIS (requires jQuery 1.6+)
processData: false, // NEEDED, DON'T OMIT THIS
// ... Other options like success and etc
});
After this it will send ajax request like you submit regular form with enctype="multipart/form-data"
Update: This request cannot work without type:"POST" in options since all files must be sent via POST request.
Note: contentType: false only available from jQuery 1.6 onwards
I can't add a comment above as I do not have enough reputation, but the above answer was nearly perfect for me, except I had to add
type: "POST"
to the .ajax call. I was scratching my head for a few minutes trying to figure out what I had done wrong, that's all it needed and works a treat. So this is the whole snippet:
Full credit to the answer above me, this is just a small tweak to that. This is just in case anyone else gets stuck and can't see the obvious.
$.ajax({
url: 'Your url here',
data: formData,
type: "POST", //ADDED THIS LINE
// THIS MUST BE DONE FOR FILE UPLOADING
contentType: false,
processData: false,
// ... Other options like success and etc
})
<form id="upload_form" enctype="multipart/form-data">
jQuery with CodeIgniter file upload:
var formData = new FormData($('#upload_form')[0]);
formData.append('tax_file', $('input[type=file]')[0].files[0]);
$.ajax({
type: "POST",
url: base_url + "member/upload/",
data: formData,
//use contentType, processData for sure.
contentType: false,
processData: false,
beforeSend: function() {
$('.modal .ajax_data').prepend('<img src="' +
base_url +
'"asset/images/ajax-loader.gif" />');
//$(".modal .ajax_data").html("<pre>Hold on...</pre>");
$(".modal").modal("show");
},
success: function(msg) {
$(".modal .ajax_data").html("<pre>" + msg +
"</pre>");
$('#close').hide();
},
error: function() {
$(".modal .ajax_data").html(
"<pre>Sorry! Couldn't process your request.</pre>"
); //
$('#done').hide();
}
});
you can use.
var form = $('form')[0];
var formData = new FormData(form);
formData.append('tax_file', $('input[type=file]')[0].files[0]);
or
var formData = new FormData($('#upload_form')[0]);
formData.append('tax_file', $('input[type=file]')[0].files[0]);
Both will work.
$(document).ready(function () {
$(".submit_btn").click(function (event) {
event.preventDefault();
var form = $('#fileUploadForm')[0];
var data = new FormData(form);
data.append("CustomField", "This is some extra data, testing");
$("#btnSubmit").prop("disabled", true);
$.ajax({
type: "POST",
enctype: 'multipart/form-data',
url: "upload.php",
data: data,
processData: false,
contentType: false,
cache: false,
timeout: 600000,
success: function (data) {
console.log();
},
});
});
});
Better to use the native javascript to find the element by id like: document.getElementById("yourFormElementID").
$.ajax( {
url: "http://yourlocationtopost/",
type: 'POST',
data: new FormData(document.getElementById("yourFormElementID")),
processData: false,
contentType: false
} ).done(function(d) {
console.log('done');
});
$('#form-withdraw').submit(function(event) {
//prevent the form from submitting by default
event.preventDefault();
var formData = new FormData($(this)[0]);
$.ajax({
url: 'function/ajax/topup.php',
type: 'POST',
data: formData,
async: false,
cache: false,
contentType: false,
processData: false,
success: function (returndata) {
if(returndata == 'success')
{
swal({
title: "Great",
text: "Your Form has Been Transfer, We will comfirm the amount you reload in 3 hours",
type: "success",
showCancelButton: false,
confirmButtonColor: "#DD6B55",
confirmButtonText: "OK",
closeOnConfirm: false
},
function(){
window.location.href = '/transaction.php';
});
}
else if(returndata == 'Offline')
{
sweetAlert("Offline", "Please use other payment method", "error");
}
}
});
});
Actually The documentation shows that you can use XMLHttpRequest().send()
to simply send multiform data
in case jquery sucks
View:
<label class="btn btn-info btn-file">
Import <input type="file" style="display: none;">
</label>
<Script>
$(document).ready(function () {
$(document).on('change', ':file', function () {
var fileUpload = $(this).get(0);
var files = fileUpload.files;
var bid = 0;
if (files.length != 0) {
var data = new FormData();
for (var i = 0; i < files.length ; i++) {
data.append(files[i].name, files[i]);
}
$.ajax({
xhr: function () {
var xhr = $.ajaxSettings.xhr();
xhr.upload.onprogress = function (e) {
console.log(Math.floor(e.loaded / e.total * 100) + '%');
};
return xhr;
},
contentType: false,
processData: false,
type: 'POST',
data: data,
url: '/ControllerX/' + bid,
success: function (response) {
location.href = 'xxx/Index/';
}
});
}
});
});
</Script>
Controller:
[HttpPost]
public ActionResult ControllerX(string id)
{
var files = Request.Form.Files;
...
Good morning.
I was have the same problem with upload of multiple images. Solution was more simple than I had imagined: include [] in the name field.
<input type="file" name="files[]" multiple>
I did not make any modification on FormData.
This is my HTML which I'm generating dynamically using drag and drop functionality.
<form method="POST" id="contact" name="13" class="form-horizontal wpc_contact" novalidate="novalidate" enctype="multipart/form-data">
<fieldset>
<div id="legend" class="">
<legend class="">file demoe 1</legend>
<div id="alert-message" class="alert hidden"></div>
</div>
<div class="control-group">
<!-- Text input-->
<label class="control-label" for="input01">Text input</label>
<div class="controls">
<input type="text" placeholder="placeholder" class="input-xlarge" name="name">
<p class="help-block" style="display:none;">text_input</p>
</div>
<div class="control-group"> </div>
<label class="control-label">File Button</label>
<!-- File Upload -->
<div class="controls">
<input class="input-file" id="fileInput" type="file" name="file">
</div>
</div>
<div class="control-group">
<!-- Button -->
<div class="controls">
<button class="btn btn-success">Button</button>
</div>
</div>
</fieldset>
</form>
This is my JavaScript code:
<script>
$('.wpc_contact').submit(function(event){
var formname = $('.wpc_contact').attr('name');
var form = $('.wpc_contact').serialize();
var FormData = new FormData($(form)[1]);
$.ajax({
url : '<?php echo plugins_url(); ?>'+'/wpc-contact-form/resources/js/tinymce.php',
data : {form:form,formname:formname,ipadd:ipadd,FormData:FormData},
type : 'POST',
processData: false,
contentType: false,
success : function(data){
alert(data);
}
});
}
For correct form data usage you need to do 2 steps.
Preparations
You can give your whole form to FormData() for processing
var form = $('form')[0]; // You need to use standard javascript object here
var formData = new FormData(form);
or specify exact data for FormData()
var formData = new FormData();
formData.append('section', 'general');
formData.append('action', 'previewImg');
// Attach file
formData.append('image', $('input[type=file]')[0].files[0]);
Sending form
Ajax request with jquery will looks like this:
$.ajax({
url: 'Your url here',
data: formData,
type: 'POST',
contentType: false, // NEEDED, DON'T OMIT THIS (requires jQuery 1.6+)
processData: false, // NEEDED, DON'T OMIT THIS
// ... Other options like success and etc
});
After this it will send ajax request like you submit regular form with enctype="multipart/form-data"
Update: This request cannot work without type:"POST" in options since all files must be sent via POST request.
Note: contentType: false only available from jQuery 1.6 onwards
I can't add a comment above as I do not have enough reputation, but the above answer was nearly perfect for me, except I had to add
type: "POST"
to the .ajax call. I was scratching my head for a few minutes trying to figure out what I had done wrong, that's all it needed and works a treat. So this is the whole snippet:
Full credit to the answer above me, this is just a small tweak to that. This is just in case anyone else gets stuck and can't see the obvious.
$.ajax({
url: 'Your url here',
data: formData,
type: "POST", //ADDED THIS LINE
// THIS MUST BE DONE FOR FILE UPLOADING
contentType: false,
processData: false,
// ... Other options like success and etc
})
<form id="upload_form" enctype="multipart/form-data">
jQuery with CodeIgniter file upload:
var formData = new FormData($('#upload_form')[0]);
formData.append('tax_file', $('input[type=file]')[0].files[0]);
$.ajax({
type: "POST",
url: base_url + "member/upload/",
data: formData,
//use contentType, processData for sure.
contentType: false,
processData: false,
beforeSend: function() {
$('.modal .ajax_data').prepend('<img src="' +
base_url +
'"asset/images/ajax-loader.gif" />');
//$(".modal .ajax_data").html("<pre>Hold on...</pre>");
$(".modal").modal("show");
},
success: function(msg) {
$(".modal .ajax_data").html("<pre>" + msg +
"</pre>");
$('#close').hide();
},
error: function() {
$(".modal .ajax_data").html(
"<pre>Sorry! Couldn't process your request.</pre>"
); //
$('#done').hide();
}
});
you can use.
var form = $('form')[0];
var formData = new FormData(form);
formData.append('tax_file', $('input[type=file]')[0].files[0]);
or
var formData = new FormData($('#upload_form')[0]);
formData.append('tax_file', $('input[type=file]')[0].files[0]);
Both will work.
$(document).ready(function () {
$(".submit_btn").click(function (event) {
event.preventDefault();
var form = $('#fileUploadForm')[0];
var data = new FormData(form);
data.append("CustomField", "This is some extra data, testing");
$("#btnSubmit").prop("disabled", true);
$.ajax({
type: "POST",
enctype: 'multipart/form-data',
url: "upload.php",
data: data,
processData: false,
contentType: false,
cache: false,
timeout: 600000,
success: function (data) {
console.log();
},
});
});
});
Better to use the native javascript to find the element by id like: document.getElementById("yourFormElementID").
$.ajax( {
url: "http://yourlocationtopost/",
type: 'POST',
data: new FormData(document.getElementById("yourFormElementID")),
processData: false,
contentType: false
} ).done(function(d) {
console.log('done');
});
$('#form-withdraw').submit(function(event) {
//prevent the form from submitting by default
event.preventDefault();
var formData = new FormData($(this)[0]);
$.ajax({
url: 'function/ajax/topup.php',
type: 'POST',
data: formData,
async: false,
cache: false,
contentType: false,
processData: false,
success: function (returndata) {
if(returndata == 'success')
{
swal({
title: "Great",
text: "Your Form has Been Transfer, We will comfirm the amount you reload in 3 hours",
type: "success",
showCancelButton: false,
confirmButtonColor: "#DD6B55",
confirmButtonText: "OK",
closeOnConfirm: false
},
function(){
window.location.href = '/transaction.php';
});
}
else if(returndata == 'Offline')
{
sweetAlert("Offline", "Please use other payment method", "error");
}
}
});
});
Actually The documentation shows that you can use XMLHttpRequest().send()
to simply send multiform data
in case jquery sucks
View:
<label class="btn btn-info btn-file">
Import <input type="file" style="display: none;">
</label>
<Script>
$(document).ready(function () {
$(document).on('change', ':file', function () {
var fileUpload = $(this).get(0);
var files = fileUpload.files;
var bid = 0;
if (files.length != 0) {
var data = new FormData();
for (var i = 0; i < files.length ; i++) {
data.append(files[i].name, files[i]);
}
$.ajax({
xhr: function () {
var xhr = $.ajaxSettings.xhr();
xhr.upload.onprogress = function (e) {
console.log(Math.floor(e.loaded / e.total * 100) + '%');
};
return xhr;
},
contentType: false,
processData: false,
type: 'POST',
data: data,
url: '/ControllerX/' + bid,
success: function (response) {
location.href = 'xxx/Index/';
}
});
}
});
});
</Script>
Controller:
[HttpPost]
public ActionResult ControllerX(string id)
{
var files = Request.Form.Files;
...
Good morning.
I was have the same problem with upload of multiple images. Solution was more simple than I had imagined: include [] in the name field.
<input type="file" name="files[]" multiple>
I did not make any modification on FormData.
I have html form:
<form enctype="multipart/form-data" id="formUpload">
<input type="file" name="file" />
<button type="submit" id="sub">Click</button>
</form>
My script:
$(function() {
$("#sub").on("click",function () {
var form = $("#formUpload")[0];
var formData = new FormData(form);
$.ajax({
url: 'upload.php',
data: formData,
type: 'POST',
contentType: false,
processData: false,
dataType: 'html',
success: function(data) {
$("#res").html(data);
}
});
});
});
I try to upload file in upload.php, but all is wrong.
How can I get $_FILES variable?
Every time you click on submit button, your page will get refreshed and you won't see any output from this statement $("#res").html(data);. Use event.preventDefault() method to prevent your form from being submitted in the first place. So your Javascript code should be like this:
$(function() {
$("#sub").on("click",function (event){
event.preventDefault();
var form = $("#formUpload")[0];
var formData = new FormData(form);
$.ajax({
url: 'upload.php',
data: formData,
type: 'POST',
contentType: false,
processData: false,
dataType: 'html',
success: function(data) {
$("#res").html(data);
}
});
});
});
And in the backend upload.php page, you can access the uploaded file data using $_FILES, like this:
<?php
if(is_uploaded_file($_FILES['file']['tmp_name'])){
$name = $_FILES['file']['name'];
$type = $_FILES['file']['type'];
$size = $_FILES['file']['size'];
// ...
}
?>
I wrote script that alows users to submit comment text with image, and I did it with ajax.
yes - upload image with ajax. it's not drag & drop...
I would like to expand this scrpit and alow users to upload more than one file/image. my code don't support that (server side gets only 1 file). can you help me to fix it so I could upload array of files?
HTML:
<form class="form-horizontal" action='#' method="post" enctype="multipart/form-data">
<input type='hidden' name='comment[pageName]' value='yard' class="pagename-field" />
<input type='hidden' name='comment[refID]' value='0' class="refid-field" />
<input type='hidden' name='allowReply' value='1' class="allowReply-field" />
<textarea class="form-control comment-field" name="comment[text]" rows="1" placeholder="כתוב/י תגובה"></textarea>
<input type='file' name='file[]' class='form-control file-field hideBox' maxlength='1' accept="image/jpg,image/png,image/jpeg,image/gif" id="uploadFile0"/>
<button class="btn btn-primary submit" >SUBMIT</button>
<img src="images/loading.gif" align="absmiddle" class="loader" id="loader">
</form>
JS:
$(function() {
$(document).on('submit','form',function(e) {
e.preventDefault();
var $form = $(this);
var act = 'add';
var file_data = $form.find('.file-field').prop('files')[0];
var form_data = new FormData();
form_data.append('act', act);
form_data.append('comment[text]', $form.find('.comment-field').val());
form_data.append('comment[pageName]', $form.find('.pagename-field').val());
form_data.append('comment[refID]', $form.find('.refid-field').val());
form_data.append('allowReply', $form.find('.allowReply-field').val());
form_data.append('feel', $form.find('.feel-field').val());
form_data.append('file[]', file_data);
$("#loader").show();
// alert(form_data);
$.ajax({
type: "POST",
url: "ajax/addComment.php",
dataType: 'text',
cache: false,
contentType: false,
processData: false,
async: false,
data: form_data,
success: function(data)
{
$("#loader").hide();
$('#commentsBox'+$form.find('.refid-field').val()).prepend(data);
$("form").reset();
}
});
return false; // avoid to execute the actual submit of the form.
});
});
You can do it by FormData() object of jQuery. Refer the code below.
HTML:
<input type="file" name="multi_upload[]" multiple />
JQuery:
$(function() {
$(document).on('submit','form',function(e) {
var files = e.target.files;
var data = new FormData();
$.each(files, function (key, value)
{
data.append(key, value);
});
$.ajax({
url: "your-url",
type: 'POST',
data: data,
cache: false,
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)
{
alert("success");
}
});
});
});
PHP:
$tempfiles = $_FILES;
foreach ($tempfiles as $files) {
$_FILES['multi_upload'] = $files;
move_uploaded_file($_FILES['multi_upload']['tmp_name'],$_FILES['multi_upload']['name']);
}
i'm a amateur programer. I'm coding upload and insert to database function with php and jquery ajax but it not work
my form
<form>
<input type="file" id='iputfile1' />
</form>
my jquery script
iputfile1 = $("#iputfile1").val();
jQuery.ajax({
type:"POST",
url:"ex.php", //goi toi file ajax.php
data:"filename"=filename+"&+"&iputfile1="
+iputfile1,
success:function(html){
jQuery("#responseDiv").show();
jQuery("#responseDiv").html(html);
}
});
my ex.php file
$iputfile1 = $_REQUEST['iputfile1'];
print_r($iputfile1)
after select file and submit my ex.php file not recivice $_file['tmp']
<input type="file" class="file">
$(".file").on("change",function(){
var file = new FormData();
file.append('file',$('.file')[0].files[0]);
$.ajax({
url: "upload.php",
type: "POST",
data: file,
processData: false,
contentType: false,
beforeSend:function(){
$(".result").text("Loading ...");
},
success:function(data){
$(".result").html(data);
}
});
<div class="result"></div>
in upload.php
<?php
include("database.php");
$name = $_FILES["file"]["name"];
if(move_uploaded_file($_FILES["file"]["tmp_name"], "DESTINATION/".$name)){
// insert to data base
echo '<img src="DESTINATION/'.$name.'">';
}
?>