I want to send FirstName, LastName and Image through Ajax Call to PHP. I am able to send the data without image, and I am able to send the image without text using formdata but the problem is I am not able to do both at a time. Kindly check my code what I am doing wrong:
<script>
function createBasicInfo() {
// For Image Send
var formdata = new FormData();
var file = jQuery('#photo').prop('files')[0];
formdata.append('photo', file);
// For Text Send
var data = {
'firstname' : jQuery('#firstname').val(),
'lastname' : jQuery('#lastname').val(),
};
**//Here Is the Problem, I am not able to append the data with Text**
formdata.append(data);
//Ajax call Start Here
jQuery.ajax({
url: '/adminseller/parsers/sbasicinfo.php',
method: 'POST',
cache: false,
contentType: false,
processData: false,
data: formdata,
success: function(data) {
if (data != 'passed') {
jQuery('#modal_errors_1').html(data);
}
if (data == 'passed') {
jQuery('#modal_errors_1').html("");
location.reload();
}
},
error: function() {
alert("Something went wrong.");
},
});
}
</script>
In above code if I comment
//formdata.append(data); // Then only image will send to php
And If I use in Ajax Call
data: data, // Then only FirstName & LastName will send to php without Image
I am not able to send text data and image both at the same time to php file.
Any idea or suggestion would be welcome.
You could just do this insted of formdata.append(data):
formdata.firstname = jQuery('#firstname').val();
formdata.lastname = jQuery('#lastname').val();
Just append your firstname and lastname into formdata. Then send full formdata.
<script>
function createBasicInfo() {
var formdata = new FormData();
var file = jQuery('#photo').prop('files')[0];
formdata.append('photo', file);
formdata.append('firstname', jQuery('#firstname').val());
formdata.append('lastname', jQuery('#lastname').val());
//Ajax call Start Here
jQuery.ajax({
url: '/adminseller/parsers/sbasicinfo.php',
method: 'POST',
cache: false,
contentType: false,
processData: false,
data: formdata,
success: function(data) {
// ...
// ...
},
error: function() {
alert("Something went wrong.");
},
});
}
</script>
Related
I'm trying to send HTML form data using AJAX however I'm also trying to send other data along with the same AJAX POST call.
Is this possible?
$('#HTMLConForm').on('submit', function (e)
{
e.preventDefault();
$.ajax({
url: "***NewUserURL.com***",
type: "POST",
data:{
'otherinfo': otherinfo,
'form_data': new FormData(this),
},
processData: false,
contentType: false,
success: function (data)
{
alert('You Have Registered')
/*window.location = "index.html"; */
},
error: function (xhr, desc, err)
{
}
});
});
Any help with be appreciated!
Pass the FormData object itself to data, don't wrap it in a simple object.
Use the FormData object's append method to add additional data.
e.preventDefault();
const formdata = new FormData(this);
formdata.append("otherinfo", otherinfo);
$.ajax({
url: "***NewUserURL.com***",
type: "POST",
data: formdata,
Say I have an array which contains a few images:
var images = [image1, image2, image3]
How do I send these images to a php file using AJAX in a single request?
The following code did not work:
$.ajax({
url: 'PHP/posts.php',
type: 'post',
data: {data: images},
success: function(data) {
alert(data);
location.reload();
}
});
My HTML:
<div id="newPostFile">
<label for="newPostFiles"><i class="fa fa-file-text-o" id="newPostFileIcon" aria-hidden="true"></i></label>
<input type="file" name="newPostFiles" id="newPostFiles">
</div>
Endgoal:
Whenever a file is selected, the file is added to the array and when the submit button is clicked, all the files are uploaded at once.
You have to send files as formData
var images = [image1, image2, image3]
var data = new FormData();
images.forEach(function(image, i) {
data.append('image_' + i, image);
});
$.ajax({
url: 'PHP/posts.php',
type: 'post',
data: data,
contentType: false,
processData: false,
success: function(data) {
console.log(data);
location.reload();
}
});
But as you're reloading the page anyway, why use ajax at all ?
You can pass a JSON object to PHP, which would be the convenient solution here
var data ={'image1':image1,'image2':image2};
You can pass this object to the php code and then parse it:
Pass the object as a string:
AJAX call:
$.ajax({
type : 'POST',
url : 'PHP/posts.php',
data : {result:JSON.stringify(data)},
success : function(response) {
alert(response);
}
});
You can handle the data passed to the result.php as :
$data = $_POST["result"];
$data = json_decode("$data", true);
//just echo an item in the array
echo "image1 : ".$data["image1"];
Another option would be serializing the array before sending, or convert it into a comma separated string using array.join, then parse/split on the posts.php
array.join(",")
you can just use the list symbol after the input name [i]
for example:
let formData = new FormData();
let files = fileInput.prop('files');
for (let i = 0; i < TotalFiles; i++) {
formData.append('file['+i+']', files[i]);
}
Use FormData Object to send files using ajax.new FormData($('#your_form_id')[0]) automatically appends all form input in FormData object which we can be done by manually formData.append('file1',file1);
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<script>
let formData = new FormData($('#your_form_id')[0]);
$.ajax({
url: 'PHP/posts.php',
type: 'POST',
data: formData,
dataType:'json',
processData: false,
contentType: false,
success: function(response) {
console.log(response);
location.reload();
},
error: function() {
console.log('fail');
}
});
</script>
I want to make an ajax call that sends both JSON and file data to my PHP backend. This is my ajax call currently:
$.ajax({
type: 'POST',
dataType: 'json',
data: jsonData,
url: 'xxx.php',
cache: false,
success: function(data) {
//removed for example
}
});
The data(jsonData) is a JSON array that also holds the input from a file select as well(I am assuming this is wrong due to the type mismatch). I tried using contentType: false, and processData: false, but when I try to access $_POST data in PHP there is nothing there. The data I am passing does not come from a form and there is quite a bit of it so I do not want to use FormData and append it to that object. I am hoping i will not have to make two ajax calls to accomplish this.
If you want to send data along with any file than you can use FormData object.
Send your jsonData as like that:
var jsonData = new FormData(document.getElementById("yourFormID"));
Than in PHP, you can check your data and file as:
<?php
print_r($_POST); // will return all data
print_r($_FILES); // will return your file
?>
Try Using formdata instead of normal serialized json
Here's an example:
var formData = new FormData();
formData.append("KEY", "VALUE");
formData.append("file", document.getElementById("fileinputID").files[0]);
then in your ajax
$.ajax({
type: 'POST',
url: "YOUR URL",
data: formData,
contentType: false,
processData: false,
dataType: 'json',
success: function (response) {
CallBack(response, ExtraData);
},
error: function () {
alert("Error Posting Data");
}
});
You can try like this also
You can visit this answer also
https://stackoverflow.com/a/35086265/2798643
HTML
<input id="fuDocument" type="file" accept="image/*" multiple="multiple" />
JS
var fd = new FormData();
var files = $("#fuDocument").get(0).files; // this is my file input in which We can select multiple files.
fd.append("kay", "value"); //As the same way you can append more fields
for (var i = 0; i < files.length; i++) {
fd.append("UploadedImage" + i, files[i]);
}
$.ajax({
type: "POST",
url: 'Url',
contentType: false,
processData: false,
data: fd,
success: function (e) {
alert("success");
}
})
I have read in other post on SO, similar to what is happening to me, but I still can not get the solution
$('#edit-continue').on('click', function(e) {
e.preventDefault();
var photo = new FormData(); <-----
jQuery.each(jQuery('#photo')[0].files, function(i, file) {<----- SO suggest for file upload
photo.append('file-' + i, file); <-----
});
$.ajax({
type: "POST",
url: "/templates/staycation/common/edit-profile.php",
data: {
id: $('#id').val(),
email: $('#email').val(),
birthday: $('#birthday').val(),
gender: $("input[name='gender']:checked").val(),
photo: photo,
},
success: function(data) {
console.log('pass');
console.log(data);
},
error: function(data) {
console.log('not pass');
console.log(data);
},
cache: false,
contentType: false,
processData: false, <------ i think my error is here
});
if i leave processData: false, , the post arrives empty, by making echo of my array, all data is empty, in the other case, if I comment, the console throws Uncaught TypeError: Illegal invocation
I need to do is send the values entered by the user such as "email", "gender", ... and profile picture, to make an update to the database and to save the image to a folder
this is what working perfectly for me
// formData will wrap all files and content of form
var formData=new FormData($('#formId')[0]);
// you can add more data ot formData after this to
$.ajax({
url : 'upload.php',
type : 'post',
data : formData,
processData:false,
contentType:false,
success : function(e)
{
alert('uploaded successfully');
},
error : function()
{
alert('hello from here');
}
});
When I try to retrieve the File path it's shows me the result like this: "C:\fakepath\amine.jpeg" so the upload in the server is not working as a result of that problem.
$('input[type=file]').change(function () {
var filePath=$('#file-input').val();
$.ajax({
url : "{{path('upload_file')}}",
type : 'POST',
data: {
filePath : filePath,
method: 'post',
params: {
action: "uploadFile"
}
},
success : function(data, textStatus, jqXHR) {
alert(data);
}
});
});
You are doing this all wrong.
You have to create a form object and send it via $.ajax.
And I assume you have written the correct serverside code to save the image.
var f = new FormData();
f.append('img',document.getElementById("file-input").files[0]);
var url= "{{Your URL}}";
$.ajax({
url: url,
type:"post",
data: f,
dataType:"JSON",
processData: false,
contentType: false,
success: function (data, status)
{
console.log(data);
},
error: function (data)
{
if (data.status === 422) {
console.log("upload failed");
} else {
console.log("upload success");
}
});
Your file by default upload to a temporary folder. You need to move it to an actual location to get access to it like in php:
move_uploaded_file($_FILES["fileToUpload"]["tmp_name"], $target_file)
For reference, see http://www.w3schools.com/php/php_file_upload.asp