I'm trying to add a feature where I can attach an image in the form using Codeigniter, Ajax, jQuery. The form submits and when I check the database for the image file, it seems that it doesnt have anything on it. Usually they use FormData but, I already started using this method. I was wondering if there is anyway I could do it this way.
Here are my codes.
jQuery
$('#addForm').submit(function(event){
var emp_id = $("#agentNames").val();
var campaign = $("#addCampaign").val();
var k_type = $("#addKudosType").val();
var c_name = $("#addCustomerName").val();
var p_number = $("#addPhoneNumber").val();
var e_add = $("#addEmailAdd").val();
var comment = $("#addCustomerComment").val();
var supervisor = $("#addSupervisor").val();
var file = $("#addFile").val();
var p_reward = $("#addPrefReward").val();
var pfrd = $("#addProofreading").val();
var k_card = $("#addKudosCard").val();
var r_status = $("#addRewardStatus").val();
dataString = "emp_id="+emp_id+"&campaign="+campaign+"&k_type="+k_type+"&c_name="+c_name+"&p_number="+p_number+"&e_add="+e_add+"&comment="+comment+"&supervisor="+supervisor+"&file="+file+"&p_reward="+p_reward+"&pfrd="+pfrd+"&k_card="+k_card+"&r_status="+r_status;
$.ajax({
type: "POST",
url: "<?php echo base_url(); ?>index.php/Kudos/addKudos/",
data: dataString,
cache: false,
success: function(html)
{
alert("Succesfully Added!");
location.reload();
}
});
event.preventDefault();
});
Controller
public function addKudos()
{
$emp_id= $this->input->post('emp_id');
$campaign= $this->input->post('campaign');
$k_type= $this->input->post('k_type');
$c_name= $this->input->post('c_name');
$p_number= $this->input->post('p_number');
$e_add= $this->input->post('e_add');
$comment= $this->input->post('comment');
$supervisor= $this->input->post('supervisor');
$file= $this->input->post('file');
$config['upload_path'] = "uploads/images/";
$config['allowed_types'] = "jpg|png";
$config['file_name'] = $_FILES['addFile']['name'];
$this->load->library('upload', $config);
$this->load->initialize($config);
if($this->upload->do_upload('addFile')){
$uploadData = $this->upload->data();
$picture = $uploadData['file_name'];
} else {
$picture = '';
}
$p_reward= $this->input->post('p_reward');
$pfrd= $this->input->post('pfrd');
$k_card= $this->input->post('k_card');
$r_status= $this->input->post('r_status');
$this->KudosModel->add_kudos($emp_id,$campaign,$k_type,$c_name,$p_number,$e_add,$comment,$supervisor,$picture,$p_reward,$pfrd,$k_card,$r_status);
}
Model
function add_kudos($emp_id,$campaign,$k_type,$c_name,$p_number,$e_add,$comment,$supervisor,$picture,$p_reward,$pfrd,$k_card,$r_status)
{
$emp_id1 =mysqli_real_escape_string($this->db->conn_id,trim($emp_id));
$campaign1 =mysqli_real_escape_string($this->db->conn_id,trim($campaign));
$k_type1 =mysqli_real_escape_string($this->db->conn_id,trim($k_type));
$c_name1 =mysqli_real_escape_string($this->db->conn_id,trim($c_name));
$p_number1 =mysqli_real_escape_string($this->db->conn_id,trim($p_number));
$e_add1 =mysqli_real_escape_string($this->db->conn_id,trim($e_add));
$comment1 =mysqli_real_escape_string($this->db->conn_id,trim($comment));
$supervisor1 =mysqli_real_escape_string($this->db->conn_id,trim($supervisor));
$file1 =mysqli_real_escape_string($this->db->conn_id,trim($picture));
$p_reward1 =mysqli_real_escape_string($this->db->conn_id,trim($p_reward));
$pfrd1 =mysqli_real_escape_string($this->db->conn_id,trim($pfrd));
$k_card1 =mysqli_real_escape_string($this->db->conn_id,trim($k_card));
$r_status1 =mysqli_real_escape_string($this->db->conn_id,trim($r_status));
$query = $this->db->query("insert into tbl_kudos(emp_id,acc_id,kudos_type,client_name,phone_number,client_email,comment,uid,file,reward_type,proofreading,kudos_card,reward_status,is_given) values('$emp_id1','$campaign1','$k_type1','$c_name1','$p_number1','$e_add1','$comment1','$supervisor1','$file1','$p_reward1','$pfrd1','$k_card1','$r_status1',now())");
}
Thanks.
The database doesn't have any file because you aren't submitting any. You didn't show any HTML but you must have something like:
<input type="file" id="addfile">
But in jquery, you are retrieving the file using:
var file = $("#addFile").val();
Which will return the file name or an empty string because you won't get file contents (or image) using val(). You need to look for another method to upload form data with file contents using AJAX. Check this out: How can I upload files asynchronously?
Related
I want to upload image on the server on change event of jQuery but using codeigniter csrf I am able to upload image only one time. How can I upload images using ajax for multiple requests.Please keep in mind when I set this
config['csrf_protection'] = FALSE;
then I am able to send multiple request jQuery onchange event but when csrf_protection is going to be false then I think there is no advantage of csrf. so the question is how can I send multiple requests using ajax while csrf_protection is enable. My jquery code is following
$("#avatar").change(function(){
var link = $("#avatar").val();
$.ajax({
url : "<?php echo base_url('main/test'); ?>",
type: 'post',
data: {'<?php echo $this->security->get_csrf_token_name(); ?>':'<?php echo $this->security->get_csrf_hash(); ?>',"id":"hello","link":link},
success : function(data)
{
alert(data);
}
});
});
My controller:
public function test()
{
$config['upload_path'] = './uploads/';
$config['allowed_types'] = 'gif|jpg|png';
$config['max_size'] = 500;
$config['max_width'] = 260;
$config['max_height'] = 260;
$this->load->library('upload', $config);
if (!$this->upload->do_upload('link')) {
echo "error";
} else {
$data = array('upload_data' => $this->upload->data());
$image_name = $data['upload_data']['file_name'];
echo $image_name;
}
}
In my opinion you should try to recreate your csrf token each request
Try this code example...
For the js funcion
var csrfName = '<?php echo $this->security->get_csrf_token_name(); ?>',
csrfHash = '<?php echo $this->security->get_csrf_hash(); ?>';
("#avatar").change(function(){
var link = $("#avatar").val();
var dataJson = { [csrfName]: csrfHash, id: "hello", link: link };
$.ajax({
url : "<?php echo base_url('main/test'); ?>",
type: 'post',
data: dataJson,
success : function(data)
{
csrfName = data.csrfName;
csrfHash = data.csrfHash;
alert(data.message);
}
});
});
and for the controller
public function test() {
$config['upload_path'] = './uploads/';
$config['allowed_types'] = 'gif|jpg|png';
$config['max_size'] = 500;
$config['max_width'] = 260;
$config['max_height'] = 260;
$reponse = array(
'csrfName' => $this->security->get_csrf_token_name(),
'csrfHash' => $this->security->get_csrf_hash()
)
$this->load->library('upload', $config);
if (!$this->upload->do_upload('link')) {
$reponse['message'] = "error";
}
else {
$data = array('upload_data' => $this->upload->data());
$image_name = $data['upload_data']['file_name'];
$reponse['message'] = $image_name;
}
echo json_encode($reponse);
}
Let me know and good luck
Note: When someone ask you for posting more data to the question, don't post it as a comment or answer, it's better to edit the question itself and adding the stuff
You can set this in config.php
$config['csrf_regenerate'] = FALSE;
so the csrf protection is valid during all the session time it will solve your problem.
If you set
$config['csrf_regenerate'] = true; then CI generate new csrf token every request so your old csrf token not match with new generated csrf token
$config['csrf_regenerate'] = TRUE;
keep auto generate to true it will be more safer.
In similar case when csrf is expired in first request. What i have implemented
$(document).ajaxComplete(function (event, xhr, settings) {
let response = xhr.responseText,
let obj = JSON.parse(response),
let csrfData = obj.csrf;
document.querySelector('input[name="' + csrfData.name + '"]').value = csrfData.hash;
}); //Also here you can update any other non input element
In every ajax response we are passing csrf data in which latest csrf data will be replaced with current one
Sample response from request
{
csrf : {
name : 'csrf_name',
hash : 'qw76sd7s6f78sdfs8dfs9df8cx9'
}
}
I update csrf token in every ajax request. Also don't choose this method if you are working with multi tab environment
Each time you make a request, the csrf_token is being updated by CI. That's why the CSRF only work once. So everytime we make a request we need to update the csrf_token too. I solve this problem by doing this.
Conroller: get the updated csrf using this code.
public function update_csrf()
{
$data['csrf_hash'] = $this->security->get_csrf_hash();
echo json_encode($data);
}
AJAX: replace the old value of your csrf name="csrf_token_name"
var jqXHR = $.ajax({
url: $(this).attr('action'),
type: 'POST',
data: $(this).serialize(),
dataType: 'json',
})
jqXHR.done(function(response) {
$('input[name=csrf_token_name]').val(response.csrf_hash); //update the csrf to the form
})
jqXHR.fail(function(jqXHR, textStatus, errorThrown) {
console.log(jqXHR);
console.log(textStatus);
console.log(errorThrown);
});
Important!: use dataType: 'json'
So now each time you have a successful request, the csrf_token is updated too and you are now free from 403 (Forbidden) error.
All you need to do is reload the CSRF token in your AJAX response. It is that simple!.
Maybe you can try using jquery cookie
First you need to add this
<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery-cookie/1.4.1/jquery.cookie.js"></script>
then change your code to this
$("#avatar").change(function(){
var link = $("#avatar").val();
$.ajax({
url : "<?php echo base_url('main/test'); ?>",
type: 'post',
data: {csrf_test_name: $.cookie('csrf_cookie_name'),"id":"hello","link":link},
dataType : "JSON",
success : function(data)
{
alert(data);
}
});
Finally you can try to set your csrf_protection to true
[csrf_protection] = TRUE;
add this at a js file which is loaded every page (I put this at the end of jquery.js )
$.ajaxSetup({
beforeSend:function(jqXHR, Obj){
var value = "; " + document.cookie;
var parts = value.split("; csrf_cookie_name=");
if(parts.length == 2)
Obj.data += '&csrf_token='+parts.pop().split(";").shift();
}
});
(notice that in every ajax request you can not have empty data to send)
"csrf_cookie_name" at top defined in config.php
$config['csrf_cookie_name'] = 'csrf_cookie_name';
Please try like my code. its working my application
your view file
$token_name = $this->security->get_csrf_token_name();
$token_hash = $this->security->get_csrf_hash();
<input type="text" id="search-text" name="parent_name" placeholder="Search" value="" >
<input type="hidden" id="csrf" name="<?php echo $token_name; ?>" value="<?php echo $token_hash; ?>" />
after set jquery post method like below
// Get Keyup
jQuery( "#search-text").keyup(function() {
// Get Data
var val = jQuery("#search-text").val();
var hashValue = jQuery('#csrf').val();
// Get jquery post for ajax task
jQuery.post(
'<?php echo $base_controler; ?>',
{val:val,'<?php echo $this->security->get_csrf_token_name(); ?>':hashValue},
function(data)
{
// Get return data to decode
var obj = jQuery.parseJSON(data);
// Get csrf new hash value
var new_hash = obj.csrfHash;
// Set csrf new hash value update
jQuery('#csrf').val(new_hash);
}
);
});
please set your controller like below
$reponse = array(
'csrfName' => $this->security->get_csrf_token_name(),
'csrfHash' => $this->security->get_csrf_hash()
);
echo json_encode($reponse);
above all code recreate your csrf token each request.
Edit the config:
$config['csrf_exclude_uris'] = ['controller/method'];
Array can include all whitelisted controllers/methods you want the csrf protection to be disabled for.
The array can also handle regular expressions such as:
$config['csrf_exclude_uris'] = array(
'api/record/[0-9]+',
'api/title/[a-z]+'
);
For more information visit Codeigniter Documentation - Security Class
I'm having a problem with sending a file from an app I'm developing with phonegap.
I'm new to phonegap, so I might be trying to solve this in an entirely wrong way, so let me describe the the end goal first.
I'm developing a car rental app, I need to make a contact form, so users can leave an order to rent a car.
The form requires user to put in some basic information, like name and phone number, and also attach a photo or a scan of driver's license.
I was able to figure out the basic info part. I'm using $.ajax dataType: 'jsonp', to send the data to the server and then simply e-mail it to my client's address.
But I can find a way to send the file to the server.
I'm using an input[type=file] field to let the user choose what file to upload.
I've tried uploading file using FileTransfer, but apparently input[type=file] gives you some fake file path, that can't be directly used by FileTransfer.upload()
Problem is, I can't understand how can I get a proper file path for FileTransfer.upload function.
I've tried doing it another way, by reading the file using FileReader.
I tried reading the file and setting an image src to the result, but it doesn't work (it show broken image icon instead of an image, the same code works on PC).
I also tried to output it as text, that does output some data (so why doesn't it work for image src?).
Because I did manage to output the data read from the file as text I thought I will send that to the server and save it.
So here is how the code would look like:
On input change I read the file into a global variable
$(".file1").change(function(e){
var caster = e.target;
var files = caster.files;
if(FileReader && files && files.length) {
var fr = new FileReader();
fr.onloadend = function(e) {
//$(".image").attr("src",e.target.result);
window.file1base64 = e.target.result;
}
fr.readAsDataURL(files[0]);
}
});
Then, when user presses a button, I run FileTransfer.upload and then check every 0.1 seconds, whether the file upload is complete
function uploadSuccess(r) {
$(".output").append(" Success ");
window.fileStatus = true;
}
function uploadError(error) {
$(".output").append(" Error "+error.code+" ");
window.fileStatus = true;
window.fileError = error.code;
}
function uploadFile() {
$(".output").append(" uploadFile ");
file = $('.file1').val().split('\\').pop();
$(".output").append(" File-"+file+" ");
if(file){
$(".output").append(" fileExists ");
var options = new FileUploadOptions();
options.fileKey = "file";
options.fileName = file;
options.mimeType = "image/jpeg";
options.chunkedMode = false;
options.headers = {
Connection: "close"
};
$(".output").append(" FileUploadOptions ");
window.fileStatus = false;
window.fileError = '';
//fileuri = $(".image").attr("src");
fileuri = window.file1base64;
$(".output").append(" fileuri ");
var ft = new FileTransfer();
ft.upload(fileuri, encodeURI("http://***.***/savefile.php"), uploadSuccess, uploadError, options);
$(".output").append(" upload ");
checkFile();
}
}
function checkFile() {
if(!window.fileStatus) {
$(".output").append(" check ");
setTimeout(checkFile, 100);
return;
}
}
After some checks, it prints out Error 3 and I can't figure out what that means or how to fix it.
Server side code is simply this:
Get the file and save it
$dir_name = dirname(__FILE__)."/uploadedimages/";
move_uploaded_file($_FILES["file"]["tmp_name"], $dir_name."test.txt");
But no file is created on the server.
use the FormData object to get the form data (including input file) and submit it this way:
var data = new FormData($('#yourFormID')[0]);
$.ajax({
url: serverURL,
data: data,
cache:false,
contentType: false,
processData: false,
type: 'POST',
error: function(jqXHR,textStatus,errorThrown){
},
success: function(data){
}
});
You should set the FILEURL in some variable and image in some html image element and then use it to transfer the image.
like this:
function onPgCameraSuccess(imageData) {
fileEntry.file(
function(fileObj) {
var previewImage= document.getElementById('SomeImageElement');
fileName=imageData.substr(imageData.lastIndexOf('/')+1);
fileURL=imageData;
previewImage.src =imageData;
$('#SomeTextBox').val(fileName);
});
}
function SubmitPhoto(){
var uOptions = new FileUploadOptions();
var ft = new FileTransfer();
uOptions .fileKey = "keyofyourfileonserver";
uOptions .fileName = fileName;
uOptions .mimeType = "image/jpeg";
uOptions .httpMethod = "POST";
uOptions .params = params;
ft.upload(fileURL,
urlofsvc,
photoSuccess,
photoFail,
uOptions,
true
);}
I'm using laravel 4.2 and currently I don't how to save a csv file into public\csv\ directory using AJAX. I'm still finding some answers. Maybe someone can help me with this.
Here's my code:
In blade view:
{{Form::open(['route' => 'file_upload', 'files' => true, 'id' => 'upload_form', 'method' => 'POST'])}}
{{Form::file('csv_upload', ['id' => 'uploaded_file', 'accept' => 'text/csv'])}}
{{Form::submit('submit', ['class' => 'btn btn-primary btn-xs', 'id' => 'upload'])}}
{{Form::close()}}
Javascript Ajax:
var ajax_ready = 1
var token = {{Session::get('_token')}}
if($.type(originalOptions.data) === 'string') {
options.data = originalOptions.data+"&_token="+token;
}else if($.type(originalOptions.data) === 'object') {
//Here I got a new error
}else{
options.data = $.param(($.extend(originalOptions.data, {'_token':mmad_token})));
}
options.url = originalOptions.url.slice(0,originalOptions.url.indexOf("?_token="));
if (ajax_ready!=1){
jqXHR.abort();
}
ajax_ready = 0;
});
$('form#upload_form').on('submit', function(e){
e.preventDefault();
var uploadFile = $('#uploaded_file');
var ext = $("input#uploaded_file").val().split(".").pop().toLowerCase();
var file = $('input[name="csv_upload"]').val();
if($.inArray(ext, ["csv"]) === -1) {
alert("Please upload a .csv file!");
return false;
}
var csv = uploadFile[0].files;
var form = new FormData(this);
var csvFile = {lastModifed: csv[0].lastModified, fileName: csv[0].name, size: csv[0].size, fileType: csv[0].type};
$.post('{{ URL::route("file_upload") }}?_token={{Session::token()}}',{
data: form
}).done(function(response){
});
});
PHP:
public function upload_csv()
{
$inputs = Input::all();
$csvFile = $inputs['data']['fileName'];
$path = public_path().DIRECTORY_SEPARATOR.'csv'.DIRECTORY_SEPARATOR;
$path2 = public_path('csv/');
if(is_dir($path2))
{
#move_uploaded_file($csvFile, $path2.$csvFile); //This line can't move the uploaded files in my desired directory
}
return json_encode(['success' => 1, 'description' => 'Successfully Upload File']);
}
This code below does work when not using AJAX:
if(Input::hasFile('csv_upload'))
{
$file = Input::file('csv_upload');
$originalFilename = $file->getClientOriginalName();
$rules = ['csv_upload' => 'required|file:csv'];
$validate = Validator::make(['csv_upload' => $file], $rules);
if($validate->fails())
{
return json_encode(['error' => 1, 'description' => 'File must be in .csv format']);
}
$path = public_path('/csv/');
if(!file_exists($path))
{
mkdir($path);
}
}
Console.log of csv
You can not move file because when you submit form with ajax file is not being sent with ajax,For sending file you have to send file with FormData() javascript Object.
If you check in upload_csv controller by putting print_r($_FILES); you will get empty array.
So use FormData on client side for appending file, then try agian.
You aren't getting error beacuse you have used php Error Control Operators likes#move_uploaded_file($csvFile, $path2.$csvFile);.
if you need working example then tell me i will give it to you.
Code For Your Help:
1. In blade view:
<script type="text/javascript">
$('form#upload_form').on('submit', function(e){
e.preventDefault();
var uploadFile = $('#uploaded_file');
var ext = $("input#uploaded_file").val().split(".").pop().toLowerCase();
var file = $('input[name="mmad_csv_upload"]').val();
if($.inArray(ext, ["csv"]) === -1) {
alert("Please upload a .csv file!");
return false;
}
var csv = uploadFile[0].files;
var formData = new FormData($(this)[0]);
formData.append('uploaded_file', $("#uploaded_file")[0].files[0]);
formData.append('lastModifed', csv[0].lastModified);
formData.append('fileName', csv[0].name);
console.log(formData);
$.ajax({
url: '{{ URL::route("file_upload") }}',
type: 'POST',
data: formData,
async: true,
cache: false,
contentType: false,
processData: false,
success: function (returndata) { //alert(returndata); return false;
}
});
});
</script>
2.Controller
public function file_upload(Request $request)
{
$inputs = Input::all();
$csvFile = $inputs['fileName'];
$path = public_path().DIRECTORY_SEPARATOR.'csv'.DIRECTORY_SEPARATOR;
$path2 = public_path('/csv/');
if(is_dir($path2))
{
$success = $request->file('uploaded_file')->move($path2, $csvFile);
}
return json_encode(['success' => 1, 'description' => 'Successfully Upload File']);
}
To move the uploaded file to a new location, you should use the move method. This method will move the file from its temporary upload location (as determined by your PHP configuration) to a more permanent destination of your choosing:
Input::file('fileName')->move($destinationPath, $fileName);
If you need additional validations, you can check it at http://laravel.com/docs/5.1/requests#files
Default AJAX POST does not support file uploads. Use jQuery Form to upload files successfully. Full documentation of file upload at http://malsup.com/jquery/form/#file-upload
Below my example of a recentlty build script... My Controller uploads the files to S3, but is easy to be implemented with local storage.
var progress = function(event, position, total, percent) {
$(".progress-bar").width(percent + '%');
$(".progress-bar").html(percent + '%');
if(percent > 50) {
$(".progress-bar").css('color','#fff');
}
if(percent == 100) {
setTimeout(function(){
$(".progress").html('<span class="processing-msg">Processing... Please be patient!</span>');
$(".processing-msg").fadeIn('slow');
}, 1000);
}
}
var success = function(data) {
var obj = $.parseJSON(data);
$("#"+obj.hidden, parent.document).val(obj.filename);
var src = 'https://s3.amazonaws.com/spincms'+obj.path+'thumb_'+obj.filename;
$("#uploaded-"+obj.hidden, parent.document).html('<img class="img-circle uploaded-img" src="' + src + '">');
$(".progress").html('<span class="processing-msg-next">File has been uploaded and processed. Do not forget to submit the form!</span>');
}
var options = {
target: '#output',
uploadProgress: progress,
success: success,
resetForm: true
};
$(document).on('click', "#upload-now", function(e) {
$(".progress").html('<div class="progress-bar progress-bar-success" role="progressbar" aria-valuenow="60" aria-valuemin="0" aria-valuemax="100"></div>');
if($("#upload-form input[type=file]")[0].files.length == 0) {
$(".progress").html('<span class="processing-msg-next">No file selected!</span>');
return false;
} else {
var name = $("#upload-form input[name='name']").val();
var token = $("#upload-form input[name='_token']").val();
var file_name = $("#upload-form input[type=file]")[0].files[0].name;
$("#upload-form").ajaxSubmit(options);
}
}
});
Since you are using jQuery you can use the form plugin as it will make things much more easier for you to work with for example , this is the jquery part that you will use :
$(document).ready(function() {
// bind 'myForm' and provide a simple callback function
$('#upload_form').ajaxForm(function() {
alert("Your file has been uploaded, thanks");
});
});
and in your controller you can code it like :
pubilc function postUpload()
{
$success = false;
if(Request::ajax())
{
if(Input::hasFile('csv_upload'))
{
$file = Input::file('csv_upload');
if(!File::isDirectory(storage_path('csv'))) {
File::createDirectory(storage_path('csv'));
}
$file->move(storage_path('csv'), $file->getClientOriginalName());
// now your file is on app/storage/csv folder
$filePath = storage_path('csv/'.$file->getClientOriginalName());
$success = true;
}
}
return Response::json(['success'=>$success]);
}
Admittedly, there are similar questions lying around on Stack Overflow, but it seems none quite meet my requirements.
Here is what I'm looking to do:
Upload an entire form of data, one piece of which is a single file
Work with Codeigniter's file upload library
Up until here, all is well. The data gets in my database as I need it. But I'd also like to submit my form via an AJAX post:
Using the native HTML5 File API, not flash or an iframe solution
Preferably interfacing with the low-level .ajax() jQuery method
I think I could imagine how to do this by auto-uploading the file when the field's value changes using pure javascript, but I'd rather do it all in one fell swoop on for submit in jQuery. I'm thinking it's not possible to do via query strings as I need to pass the entire file object, but I'm a little lost on what to do at this point.
Can this be achieved?
It's not too hard. Firstly, take a look at FileReader Interface.
So, when the form is submitted, catch the submission process and
var file = document.getElementById('fileBox').files[0]; //Files[0] = 1st file
var reader = new FileReader();
reader.readAsText(file, 'UTF-8');
reader.onload = shipOff;
//reader.onloadstart = ...
//reader.onprogress = ... <-- Allows you to update a progress bar.
//reader.onabort = ...
//reader.onerror = ...
//reader.onloadend = ...
function shipOff(event) {
var result = event.target.result;
var fileName = document.getElementById('fileBox').files[0].name; //Should be 'picture.jpg'
$.post('/myscript.php', { data: result, name: fileName }, continueSubmission);
}
Then, on the server side (i.e. myscript.php):
$data = $_POST['data'];
$fileName = $_POST['name'];
$serverFile = time().$fileName;
$fp = fopen('/uploads/'.$serverFile,'w'); //Prepends timestamp to prevent overwriting
fwrite($fp, $data);
fclose($fp);
$returnData = array( "serverFile" => $serverFile );
echo json_encode($returnData);
Or something like it. I may be mistaken (and if I am, please, correct me), but this should store the file as something like 1287916771myPicture.jpg in /uploads/ on your server, and respond with a JSON variable (to a continueSubmission() function) containing the fileName on the server.
Check out fwrite() and jQuery.post().
On the above page it details how to use readAsBinaryString(), readAsDataUrl(), and readAsArrayBuffer() for your other needs (e.g. images, videos, etc).
With jQuery (and without FormData API) you can use something like this:
function readFile(file){
var loader = new FileReader();
var def = $.Deferred(), promise = def.promise();
//--- provide classic deferred interface
loader.onload = function (e) { def.resolve(e.target.result); };
loader.onprogress = loader.onloadstart = function (e) { def.notify(e); };
loader.onerror = loader.onabort = function (e) { def.reject(e); };
promise.abort = function () { return loader.abort.apply(loader, arguments); };
loader.readAsBinaryString(file);
return promise;
}
function upload(url, data){
var def = $.Deferred(), promise = def.promise();
var mul = buildMultipart(data);
var req = $.ajax({
url: url,
data: mul.data,
processData: false,
type: "post",
async: true,
contentType: "multipart/form-data; boundary="+mul.bound,
xhr: function() {
var xhr = jQuery.ajaxSettings.xhr();
if (xhr.upload) {
xhr.upload.addEventListener('progress', function(event) {
var percent = 0;
var position = event.loaded || event.position; /*event.position is deprecated*/
var total = event.total;
if (event.lengthComputable) {
percent = Math.ceil(position / total * 100);
def.notify(percent);
}
}, false);
}
return xhr;
}
});
req.done(function(){ def.resolve.apply(def, arguments); })
.fail(function(){ def.reject.apply(def, arguments); });
promise.abort = function(){ return req.abort.apply(req, arguments); }
return promise;
}
var buildMultipart = function(data){
var key, crunks = [], bound = false;
while (!bound) {
bound = $.md5 ? $.md5(new Date().valueOf()) : (new Date().valueOf());
for (key in data) if (~data[key].indexOf(bound)) { bound = false; continue; }
}
for (var key = 0, l = data.length; key < l; key++){
if (typeof(data[key].value) !== "string") {
crunks.push("--"+bound+"\r\n"+
"Content-Disposition: form-data; name=\""+data[key].name+"\"; filename=\""+data[key].value[1]+"\"\r\n"+
"Content-Type: application/octet-stream\r\n"+
"Content-Transfer-Encoding: binary\r\n\r\n"+
data[key].value[0]);
}else{
crunks.push("--"+bound+"\r\n"+
"Content-Disposition: form-data; name=\""+data[key].name+"\"\r\n\r\n"+
data[key].value);
}
}
return {
bound: bound,
data: crunks.join("\r\n")+"\r\n--"+bound+"--"
};
};
//----------
//---------- On submit form:
var form = $("form");
var $file = form.find("#file");
readFile($file[0].files[0]).done(function(fileData){
var formData = form.find(":input:not('#file')").serializeArray();
formData.file = [fileData, $file[0].files[0].name];
upload(form.attr("action"), formData).done(function(){ alert("successfully uploaded!"); });
});
With FormData API you just have to add all fields of your form to FormData object and send it via $.ajax({ url: url, data: formData, processData: false, contentType: false, type:"POST"})
I'm trying to code my own php uploader so I can learn and understand how it all works. Trying not to use any libraries/plugins/API's. So basically I have the javascript stuff doin what it's supposed to be (i hope..). Now I'm stuck on the PHP. I want it to upload and save it as img/1.png. (not using any variables just to test it...). Here are my scripts. Keeping it very simple without any error checks just to get basic uploading working. Would this work for multiple files? May somebody guide me on what my PHP file should look like just for very basic (multi)file upload? Thanks!
album_create.js
$(document).ready(function() {
$('#album_create').submit(function() {
var file = document.getElementById('album-files').files[0]; //Files[0] = 1st file
var reader = new FileReader();
reader.onload = function(f) {shipOff(f);};
reader.onerror = function(event) {
console.error("File could not be read! Code " + event.target.error.code);
};
reader.readAsText(file, 'UTF-8');
return false;
});
});
function shipOff(event) {
var result = event.target.result;
var fileName = document.getElementById('album-files').files[0].name; //Should be 'picture.jpg'
console.log(fileName);
$.post('ajax/album_create.php', { data: result, name: fileName }, function(data){console.log(data);});
return false;
}
ajax/album_create.php
$data = $_POST['data'];
move_uploaded_file($data[0], 'img/1.png');
you are posting:
data: result, name: fileName
but album_create.php expecting:
$data = $_POST['data'];
$fileName = $_POST['fileName'];
$fileName = $_POST['fileName']; should be: $fileName = $_POST['name'];