I am trying to send some data via POST method to a PHP file without using form in HTML. This is the code I have. Why doesn't it do anything?
<input type="file" name="fileToUpload" id="fileToUpload">
<input type="hidden" value="<?php echo $row['Gallery_Id']; ?>" name="gid" id="gid">
<input type="hidden" value="User" name="user" id="user">
<button onclick="myFormData()">Upload Image</button>
<script>
$('#fileToUpload').on('change', function() {
var myFormData = new FormData();
var file = document.getElementById('fileToUpload').value;
var gid = document.getElementById('gid').value;
var user = document.getElementById('user').value;
myFormData.append('file', file);
myFormData.append('gid', gid);
myFormData.append('user', user);
});
$.ajax({
url: 'imgupload.php',
type: 'POST',
processData: false, // important
contentType: false, // important
dataType : 'json',
data: myFormData
});
</script>
On imgupload.php I get the POST data like this
$gid = $_POST['gid'];
$user = $_POST['user'];
It worked when I used the HTML form method. What's wrong here?
FormData.append() takes key-value pairs, so this is wrong:
myFormData.append(file,gid,user);
You need something like:
myFormData.append('file', file);
myFormData.append('gid', gid);
myFormData.append('user', user);
Appart from that you need to put this code inside an event handler so that it triggers when you need it to.
For example:
$('#fileToUpload').on('change', function() {
// your javascript code
});
And you should probably also put it inside a document.ready block.
Related
I am trying to upload multiple images from one input field and want to send form-data with images and action via ajax. Below I am sharing my code
jQuery(document).ready(function() {
var ajax_url = 'admin-ajax.php';
// When the Upload button is clicked...
jQuery(document).on('click', '#upload_multiImages', function(e){
e.preventDefault;
var fd = new FormData();
var files_data = jQuery('.files-data'); // The <input type="file" /> field
// Loop through each data and create an array file[] containing our files data.
jQuery.each($(files_data), function(i, obj) {
jQuery.each(obj.files,function(j,file){
fd.append('files[' + j + ']', file);
})
});
// our AJAX identifier
fd.append('action', 'cvf_upload_files');
jQuery.ajax({
type: 'POST',
url: ajax_url,
data: fd,
contentType: false,
processData: false,
success: function(response){
console.log(response);
jQuery('.upload-response').html(response); // Append Server Response
}
});
});
});
<input name="my_file_upload[]" id="my_file_upload" type="file" multiple="multiple" accept = "image/*" class="files-data form-control multi with-preview" value="Drag and Drop" />
<input name="all_vendor_file" type="hidden" value="<?php //echo implode(',', $vendor_images);?>">
<button type="submit" name="upload" id="upload_multiImages">Upload</button>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.2.0/jquery.min.js"></script>
But in response I am only getting action files are not appending, due to which I am not able to get files data over ajax function.
I am trying to make a form where there will be user data(name,dob etc) and an image. When user submits the form a pdf will be generated with the user given data and the image. I can successfully serialize the data but failed to get image in my pdf. I am using simple ajax post method to post data. Below is my code.
HTML code
<form onsubmit="submitMe(event)" method="POST" id="cform">
<input type="text" name="name" placeholder="Your Name" required>
<input type="file" name="pic" id="pic" accept="image/*" onchange="ValidateInput(this);" required>
<input type="submit" value="Preview"/>
</form>
Jquery code
function submitMe(event) {
event.preventDefault();
jQuery(function($)
{
var query = $('#cform').serialize();
var url = 'ajax_form.php';
$.post(url, query, function () {
$('#ifr').attr('src',"http://docs.google.com/gview?url=http://someurl/temp.pdf&embedded=true");
});
});
}
PHP code
<?php
$name=$_POST['name'];
$image1=$_FILES['pic']['name'];
?>
Here I am not getting image1 value. I want to get the url of the image.
You need FormData to achieve it.
SOURCE
Additionally, you need to change some stuff inside ajax call(explained in link above)
contentType: false
cache: false
processData:false
So the full call would be:
$(document).on('change','.pic-upload',uploadProfilePic);
#.pic-upload is input type=file
function uploadProfilePic(e){
var newpic = e.target.files;
var actual = new FormData();
actual.append('file', newpic[0]);
var newpic = e.target.files;
var actual = new FormData();
actual.append('file', newpic[0]);
$.ajax({
type:"POST",
url:"uploadpic.php",
data: actual,
contentType: false,
cache: false,
processData:false,
dataType:"json",
success: function (response){
#Maybe return link to new image on successful call
}
});
}
Then in PHP you handle it like this:
$_FILES['file']['name']
since you named it 'file' here:
actual.append('file', newpic[0]);
I am making a simple page where user can upload a image without refreshing the whole page. But if(isset($_post[oneimgtxt])) is not working..
here is my serverSide Code that upload image :
<?php
$maxmum_size = 3145728; //3mb
$image_type_allowed = array(IMAGETYPE_GIF, IMAGETYPE_JPEG, IMAGETYPE_PNG, IMAGETYPE_BMP);
if ($_SERVER["REQUEST_METHOD"] == "POST") {
if(isset($_POST["oneimgtxt"])){//<!------------------ this line is not working
if((!empty($_FILES[$_FILES['upimage']['tmp_name']])) && ($_FILES["upimage"]['error'] == 0)){
$file=$_FILES['upimage']['tmp_name'];
$image_count = count($_FILES['upimage']['tmp_name']);
if($image_count == 1){
$image_name = $_FILES["upimage"]["name"];
$image_type = $_FILES["upimage"]["type"];
$image_size = $_FILES["upimage"]["size"];
$image_error = $_FILES["upimage"]["error"];
if(file_exists($file)){//if file is uploaded on server in tmp folder (xampp) depends !!
$filetype =exif_imagetype($file); // 1st method to check if it is image, this read first binary data of image..
if (in_array($filetype, $image_type_allowed)) {
// second method to check valid image
if(verifyImage($filename)){// verifyImage is function created in fucrenzione file.. using getimagesize
if($ImageSizes < $maxmum_size){//3mb
$usr_dir = "folder/". $image_name;
move_uploaded_file($file, $usr_dir);
}else{
$error_container["1006"]=true;
}
}else{
$error_container["1005"]=true;
}
}else{
$error_container["1004"]=true;
}
}else{
$error_container["1003"]=true;
}
}else{
$error_container["1002"]=true;
}
}else{
$error_container["1007"]=true;
}
}else{//this else of image issset isset($_POST["oneimgtxt"])
$error_container["1001"]=true;//"Error during uploading image";
}
echo json_encode($error_container);
}
?>
in chrome inspect element i got this..
image
and this is my js code with ajax...
$(".sndbtn").click( function(e){
var form = $("#f12le")[0];
var formdata = new FormData(form)
$.ajax({
type:'POST',
//method:'post',
url: "pstrum/onphotx.php",
cache:false,
data: {oneimgtxt : formdata},
processData: false,
contentType: false,
success:function (e){console.log(e);}
});
});
Here is html code:
<form method="post" id="f12le" enctype="multipart/form-data">
<input type="file" name="upimage"/>
<label for="imgr">Choose an Image..</label>
<textarea placeholder="Write something about photo"></textarea>
<input type="button" name="addimagedata" value="Post" class="sndbtn"/>
</form>
Thanks for any help.
You should send your FormData as a whole data object not a part of another data object. So, it should be like this -
$(".sndbtn").click( function(e){
var form = $("#f12le")[0];
var formdata = new FormData(form)
$.ajax({
type:'POST',
//method:'post',
url: "pstrum/onphotx.php",
cache:false,
data: formdata,
processData: false,
contentType: false,
success:function (e){console.log(e);}
});
});
Now, you should be able to access the form as it is. For example if you have any input with name inputxt inside the form, you should be able to access it with $_POST['inputxt']. And if you have any input type="file" with the name upimage, you need to access through $_FILES['upimage']. So, if you want to do isset() for that. You can do like this :
if(isset($_FILES['upimage'])){
add enctype on form any time using file inputs
<form enctype="multipart/form-data" >
<input type=file />
...
</form>
and make sure it's always a POST request.
Good luck...!
I had headaches for this thing! you should use $_FILES['name_of_dom_element']; in your php code.
So, I have the following form and js/php:
php
<form enctype="multipart/form-data">
<input id="fileupload" type="file" name="files[]" class="files " onChange="UploadImage" accept='image/*'/>
<input type="button" class="submit_form" value="submit">
</form>
<?php
add_action( 'wp_ajax_UploadImage', 'UploadImage' );
function UploadImage()
{
$upload_dir = wp_upload_dir();
$files = $_FILES['files'];
//Some function
}
?>
JS
function UploadImage(e)
{
jQuery('#fileupload').fileupload({
url: upload_image.ajax_url,
});
if(jQuery('#fileupload')) {
var form = document.forms.namedItem("upload_video");
var formdata = new FormData(form);
formdata.append('action', 'UploadImage');
jQuery.ajax({
success : function(data){
alert('sddsf');
}
})
}
};
As you can see from here, when an image is selected using Blueimp jQuery File upload (which the js is not properly written), I want the image file to be handled by the php function.
In other words, the js is not correctly written and I am not sure how to initiate the plugin then when the image is selected, it is processed by the php function via ajax (meaning, how do I parse the file info to php function via ajax?)
Don't use $.ajax directly. The plugin already does that behind the scenes.
Here's a working example, based on your code, but adapted to run on JSFiddle:
$(document).ready(function(){
var url = '/echo/json/';
var formdata = {json: JSON.stringify({field1: 'value1'})};
jQuery('#fileupload').fileupload({
url: url,
formData : formdata,
dataType: 'json',
type: "POST",
contentType:false,
processData:false,
success : function(data){
alert('success...');
console.dir(data);
}
});
});
Demo: http://jsfiddle.net/pottersky/8usb1sn3/3/
function chkFile(file1) {
var file = file1.files[0];
var formData = new FormData();
formData.append('formData', file);
$.ajax({
type: "POST",
url: "chkFileType.php",
contentType: false,
processData: false,
data: formData,
success: function (data) {
alert(data);
}
});
}
<form action="" method="post" name="myForm" id="myForm" enctype="multipart/form-data">
<input type="hidden" name="MAX_FILE_SIZE" value="30000" />
Upload Files
<input type="file" name="uploadFile" id="uploadFile" onChange="chkFile(this)"/>
<input type="submit" name="submitbutt" value="Checkout">
chkFileType.php
<?php
print_r($_FILE)
?>
I want to create a form that when the user uploads a file, it will do a check on the uploaded file before submitting the whole form. I use onChange when a file is uploaded and then pass the formData value to Ajax to call my chkFileType.php to do the checks and alert back the response.
The function is running without any errors, but no response from alert(data);
I know I am doing something wrong, but have no idea which direction to go from. Am I doing the right way?
Everything looks fine. You have done in right way. But to get any response from an ajax call, you have to print the required stuff in chkFileType.php.
Like,
if($ext =="jpg" || $ext == "png"){
echo "Image"; // data in alert will alert as Image
} else if(check for txt file){
echo "Text File"; // data in alert will alert as Text File
} else if(chck for pdf) {
echo "Pdf";// data in alert will alert as Pdf
}
EDIT
change this
var formData = new FormData( $("#formID")[0] );
Hope you understand what i meant to say.
I think the problem is in that you have written "type" instead of "method", "POST" is a method not a type.