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问题:
I have designed a simple form which allows the user to upload files to the server. Initially the form contains one 'browse' button. If the user wants to upload multiple files, he needs to click on the "Add More Files" button which adds another 'browse' button in the form. When the form is submitted, the file upload process is handled in 'upload.php' file. It works perfectly fine for uploading multiple files. Now I need to submit the form by using jQuery's '.submit()' and send a ajax ['.ajax()'] request to the 'upload.php' file to handle the file upload.
Here is my HTML form :
<form enctype="multipart/form-data" action="upload.php" method="post">
<input name="file[]" type="file" />
<button class="add_more">Add More Files</button>
<input type="button" id="upload" value="Upload File" />
</form>
Here is the JavaScript :
$(document).ready(function(){
$('.add_more').click(function(e){
e.preventDefault();
$(this).before("<input name='file[]' type='file' />");
});
});
Here is the code for processing file upload :
for($i=0; $i<count($_FILES['file']['name']); $i++){
$target_path = "uploads/";
$ext = explode('.', basename( $_FILES['file']['name'][$i]));
$target_path = $target_path . md5(uniqid()) . "." . $ext[count($ext)-1];
if(move_uploaded_file($_FILES['file']['tmp_name'][$i], $target_path)) {
echo "The file has been uploaded successfully <br />";
} else{
echo "There was an error uploading the file, please try again! <br />";
}
}
Any suggestions on how I should write my '.submit()' function will be really helpful.
回答1:
Finally I have found the solution by using the following code:
$('body').on('click', '#upload', function(e){
e.preventDefault();
var formData = new FormData($(this).parents('form')[0]);
$.ajax({
url: 'upload.php',
type: 'POST',
xhr: function() {
var myXhr = $.ajaxSettings.xhr();
return myXhr;
},
success: function (data) {
alert("Data Uploaded: "+data);
},
data: formData,
cache: false,
contentType: false,
processData: false
});
return false;
});
回答2:
HTML
<form enctype="multipart/form-data" action="upload.php" method="post">
<input name="file[]" type="file" />
<button class="add_more">Add More Files</button>
<input type="button" value="Upload File" id="upload"/>
</form>
Javascript
$(document).ready(function(){
$('.add_more').click(function(e){
e.preventDefault();
$(this).before("<input name='file[]' type='file'/>");
});
});
for ajax upload
$('#upload').click(function() {
var filedata = document.getElementsByName("file"),
formdata = false;
if (window.FormData) {
formdata = new FormData();
}
var i = 0, len = filedata.files.length, img, reader, file;
for (; i < len; i++) {
file = filedata.files[i];
if (window.FileReader) {
reader = new FileReader();
reader.onloadend = function(e) {
showUploadedItem(e.target.result, file.fileName);
};
reader.readAsDataURL(file);
}
if (formdata) {
formdata.append("file", file);
}
}
if (formdata) {
$.ajax({
url: "/path to upload/",
type: "POST",
data: formdata,
processData: false,
contentType: false,
success: function(res) {
},
error: function(res) {
}
});
}
});
PHP
for($i=0; $i<count($_FILES['file']['name']); $i++){
$target_path = "uploads/";
$ext = explode('.', basename( $_FILES['file']['name'][$i]));
$target_path = $target_path . md5(uniqid()) . "." . $ext[count($ext)-1];
if(move_uploaded_file($_FILES['file']['tmp_name'][$i], $target_path)) {
echo "The file has been uploaded successfully <br />";
} else{
echo "There was an error uploading the file, please try again! <br />";
}
}
/**
Edit: $target_path variable need to be reinitialized and should
be inside for loop to avoid appending previous file name to new one.
*/
Please use the script above script for ajax upload. It will work
回答3:
My solution
- Assuming that form id = "my_form_id"
- It detects the form method and form action from HTML
jQuery code
$('#my_form_id').on('submit', function(e) {
e.preventDefault();
var formData = new FormData($(this)[0]);
var msg_error = 'An error has occured. Please try again later.';
var msg_timeout = 'The server is not responding';
var message = '';
var form = $('#my_form_id');
$.ajax({
data: formData,
async: false,
cache: false,
processData: false,
contentType: false,
url: form.attr('action'),
type: form.attr('method'),
error: function(xhr, status, error) {
if (status==="timeout") {
alert(msg_timeout);
} else {
alert(msg_error);
}
},
success: function(response) {
alert(response);
},
timeout: 7000
});
});
回答4:
Using this source code you can upload multiple file like google one by
one through ajax. Also you can see the uploading progress
HTML
<input type="file" id="multiupload" name="uploadFiledd[]" multiple >
<button type="button" id="upcvr" class="btn btn-primary">Start Upload</button>
<div id="uploadsts"></div>
Javascript
<script>
function uploadajax(ttl,cl){
var fileList = $('#multiupload').prop("files");
$('#prog'+cl).removeClass('loading-prep').addClass('upload-image');
var form_data = "";
form_data = new FormData();
form_data.append("upload_image", fileList[cl]);
var request = $.ajax({
url: "upload.php",
cache: false,
contentType: false,
processData: false,
async: true,
data: form_data,
type: 'POST',
xhr: function() {
var xhr = $.ajaxSettings.xhr();
if(xhr.upload){
xhr.upload.addEventListener('progress', function(event){
var percent = 0;
if (event.lengthComputable) {
percent = Math.ceil(event.loaded / event.total * 100);
}
$('#prog'+cl).text(percent+'%')
}, false);
}
return xhr;
}
})
.success(function(res,status) {
if(status == 'success'){
percent = 0;
$('#prog'+cl).text('');
$('#prog'+cl).text('--Success: ');
if(cl < ttl){
uploadajax(ttl,cl+1);
}else{
alert('Done ');
}
}
})
.fail(function(res) {
alert('Failed');
});
}
$('#upcvr').click(function(){
var fileList = $('#multiupload').prop("files");
$('#uploadsts').html('');
var i
for ( i = 0; i < fileList.length; i++) {
$('#uploadsts').append('<p class="upload-page">'+fileList[i].name+'<span class="loading-prep" id="prog'+i+'"></span></p>');
if(i == fileList.length-1){
uploadajax(fileList.length-1,0);
}
}
});
</script>
PHP
upload.php
move_uploaded_file($_FILES["upload_image"]["tmp_name"],$_FILES["upload_image"]["name"]);