问题
How can I post file and input string data with FormData()? For instance, I have many other hidden input data that I need them to be sent to the server,
html,
<form action="image.php" method="post" enctype="multipart/form-data">
<input type="file" name="file[]" multiple="" />
<input type="hidden" name="page_id" value="<?php echo $page_id;?>"/>
<input type="hidden" name="category_id" value="<?php echo $item_category->category_id;?>"/>
<input type="hidden" name="method" value="upload"/>
<input type="hidden" name="required[category_id]" value="Category ID"/>
</form>
With this code below I only manage to send the file data but not the hidden input data.
jquery,
// HTML5 form data object.
var fd = new FormData();
var file_data = object.get(0).files[i];
var other_data = $('form').serialize(); // page_id=&category_id=15&method=upload&required%5Bcategory_id%5D=Category+ID
fd.append("file", file_data);
$.ajax({
url: 'add.php',
data: fd,
contentType: false,
processData: false,
type: 'POST',
success: function(data){
alert(data);
}
});
server.php
print_r($_FILES);
print_r($_POST);
result,
Array
(
[file] => Array
(
[name] => xxx.doc
[type] => application/msword
[tmp_name] => C:\wamp\tmp\php7C24.tmp
[error] => 0
[size] => 11776
)
)
I would like to get this as my result though,
Array
(
[file] => Array
(
[name] => xxx.doc
[type] => application/msword
[tmp_name] => C:\wamp\tmp\php7C24.tmp
[error] => 0
[size] => 11776
)
)
Array
(
[page_id] => 1000
[category_id] => 12
[method] => upload
...
)
Is it possible?
回答1:
var fd = new FormData();
var file_data = $('input[type="file"]')[0].files; // for multiple files
for(var i = 0;i<file_data.length;i++){
fd.append("file_"+i, file_data[i]);
}
var other_data = $('form').serializeArray();
$.each(other_data,function(key,input){
fd.append(input.name,input.value);
});
$.ajax({
url: 'test.php',
data: fd,
contentType: false,
processData: false,
type: 'POST',
success: function(data){
console.log(data);
}
});
Added a for loop and changed .serialize() to .serializeArray() for object reference in a .each() to append to the FormData.
回答2:
well, as an easier alternative and shorter, you could do this too!!
var fd = new FormData();
var file_data = object.get(0).files[i];
var other_data = $('form').serialize(); //page_id=&category_id=15&method=upload&required%5Bcategory_id%5D=Category+ID
fd.append("file", file_data);
$.ajax({
url: 'add.php?'+ other_data, //<== just add it to the end of url ***
data: fd,
contentType: false,
processData: false,
type: 'POST',
success: function(data){
alert(data);
}
});
回答3:
I always use this.It send form data using ajax
$(document).on("submit", "form", function(event)
{
event.preventDefault();
var url=$(this).attr("action");
$.ajax({
url: url,
type: 'POST',
data: new FormData(this),
processData: false,
contentType: false,
success: function (data, status)
{
}
});
});
回答4:
I try to contribute my code collaboration with my friend . modification from this forum.
$('#upload').on('click', function() {
var fd = new FormData();
var c=0;
var file_data,arr;
$('input[type="file"]').each(function(){
file_data = $('input[type="file"]')[c].files; // get multiple files from input file
console.log(file_data);
for(var i = 0;i<file_data.length;i++){
fd.append('arr[]', file_data[i]); // we can put more than 1 image file
}
c++;
});
$.ajax({
url: 'test.php',
data: fd,
contentType: false,
processData: false,
type: 'POST',
success: function(data){
console.log(data);
}
});
});
this my html file
<form name="form" id="form" method="post" enctype="multipart/form-data">
<input type="file" name="file[]"multiple>
<input type="button" name="submit" value="upload" id="upload">
this php code file
<?php
$count = count($_FILES['arr']['name']); // arr from fd.append('arr[]')
var_dump($count);
echo $count;
var_dump($_FILES['arr']);
if ( $count == 0 ) {
echo 'Error: ' . $_FILES['arr']['error'][0] . '<br>';
}
else {
$i = 0;
for ($i = 0; $i < $count; $i++) {
move_uploaded_file($_FILES['arr']['tmp_name'][$i], 'uploads/' . $_FILES['arr']['name'][$i]);
}
}
?>
I hope people with same problem , can fast solve this problem. i got headache because multiple upload image.
回答5:
From what I understand you would like to send the images and the values of the inputs together. This code works well for me, I hope it helps someone in the future.
<form id="my-form" method="post" enctype="multipart/form-data">
<input type="file" name="file[]" multiple="" />
<input type="hidden" name="page_id" value="<?php echo $page_id;?>"/>
<input type="hidden" name="category_id" value="<?php echo $item_category->category_id;?>"/>
<input type="hidden" name="method" value="upload"/>
<input type="hidden" name="required[category_id]" value="Category ID"/>
</form>
-
jQuery.ajax({
url: 'post.php',
data: new FormData($('#my-form')[0]),
cache: false,
contentType: false,
processData: false,
type: 'POST',
success: function(data){
console.log(data);
}});
Take a look at my short code for ajax multiple upload with preview.
https://santanamic.github.io/ajax-multiple-upload/
回答6:
You can try this:
var fd = new FormData();
var data = []; //<---------------declare array here
var file_data = object.get(0).files[i];
var other_data = $('form').serialize();
data.push(file_data); //<----------------push the data here
data.push(other_data); //<----------------and this data too
fd.append("file", data); //<---------then append this data
回答7:
var fd = new FormData();
//Get Form Values
var other_data = $('#form1').serializeArray();
$.each(other_data, function (key, input) {
fd.append(input.name, input.value);
});
//Get File Value
var $file = jq("#photoUpload").get(0);
if ($file.files.length > 0) {
for (var i = 0; i < $file.files.length; i++) {
fd.append('Photograph' + i, $file.files[i]);
}
}
$.ajax({
url: 'test.php',
data: fd,
contentType: false,
processData: false,
type: 'POST',
success: function(data){
console.log(data);
}
});
回答8:
For Multiple file input : Try this code :
<form name="form" id="form" method="post" enctype="multipart/form-data">
<input type="file" name="file[]">
<input type="file" name="file[]" >
<input type="text" name="name" id="name">
<input type="text" name="name1" id="name1">
<input type="button" name="submit" value="upload" id="upload">
</form>
$('#upload').on('click', function() {
var fd = new FormData();
var c=0;
var file_data;
$('input[type="file"]').each(function(){
file_data = $('input[type="file"]')[c].files; // for multiple files
for(var i = 0;i<file_data.length;i++){
fd.append("file_"+c, file_data[i]);
}
c++;
});
var other_data = $('form').serializeArray();
$.each(other_data,function(key,input){
fd.append(input.name,input.value);
});
$.ajax({
url: 'work.php',
data: fd,
contentType: false,
processData: false,
type: 'POST',
success: function(data){
console.log(data);
}
});
});
回答9:
I found that, if somehow(like your ModelState is false on server.) and page post again to server then it was taking old value to the server. So i found that solution for that.
var data = new FormData();
$.each($form.serializeArray(), function (key, input) {
if (data.has(input.name)) {
data.set(input.name, input.value);
} else {
data.append(input.name, input.value);
}
});
回答10:
For multiple files in ajax try this
var url = "your_url";
var data = $('#form').serialize();
var form_data = new FormData();
//get the length of file inputs
var length = $('input[type="file"]').length;
for(var i = 0;i<length;i++){
file_data = $('input[type="file"]')[i].files;
form_data.append("file_"+i, file_data[0]);
}
// for other data
form_data.append("data",data);
$.ajax({
url: url,
type: "POST",
data: form_data,
cache: false,
contentType: false, //important
processData: false, //important
success: function (data) {
//do something
}
})
In php
parse_str($_POST['data'], $_POST);
for($i=0;$i<count($_FILES);$i++){
if(isset($_FILES['file_'.$i])){
$file = $_FILES['file_'.$i];
$file_name = $file['name'];
$file_type = $file ['type'];
$file_size = $file ['size'];
$file_path = $file ['tmp_name'];
}
}
来源：https://stackoverflow.com/questions/21060247/send-formdata-and-string-data-together-through-jquery-ajax