How to extract values from jquery `formData` to insert into Mysql? - php

Question

My code is about submitting a multipart form, through $.ajax, which is successfully doing it & upon json_encode in submit.php , its giving me this :
{"success":"Image was submitted","formData":{"fb_link":"https:\/\/www.google.mv\/",
"show_fb":"1",
"filenames":[".\/uploads\/homepage-header-social-icons\/27.jpg"]}}
Can someone explain me, in submit.php, how can i extract values from formData, to store in mysql table ? I have tried, so many things including :
$fb_link = formData['fb_link'];
$show_fb = formData['show_fb'];
and
$arr = json_encode($data);
$fb_link=$arr['fb_link'];
and
$fb_link = REQUEST['fb_link'];
$show_fb = REQUEST['show_fb'];
but, nothing seems to work ? Any Guesses ?
Thanks
update :
JS code on parent-page is :
$(function()
{
// Variable to store your files
var files;
// Add events
$('input[type=file]').on('change', prepareUpload);
$('form#upload_form').on('submit', uploadFiles);
// Grab the files and set them to our variable
function prepareUpload(event)
{
files = event.target.files;
}
// Catch the form submit and upload the files
function uploadFiles(event)
{
event.stopPropagation(); // Stop stuff happening
event.preventDefault(); // Totally stop stuff happening
// START A LOADING SPINNER HERE
// Create a formdata object and add the files
var data = new FormData();
$.each(files, function(key, value)
{
data.append(key, value);
});
//var data = new FormData($(this)[0]);
$.ajax({
url: 'jquery_upload_form_submit.php?files=files',
type: 'POST',
data: data,
//data: {data, var1:"fb_link" , var2:"show_fb"},
cache: false,
dataType: 'json',
processData: false, // Don't process the files
contentType: false, // Set content type to false as jQuery will tell the server its a query string request
success: function(data, textStatus, jqXHR)
{
if(typeof data.error === 'undefined')
{
// Success so call function to process the form
submitForm(event, data);
}
else
{
// Handle errors here
console.log('ERRORS: ' + data.error);
}
},
error: function(jqXHR, textStatus, errorThrown)
{
// Handle errors here
console.log('ERRORS: ' + textStatus);
// STOP LOADING SPINNER
}
});
}
function submitForm(event, data)
{
// Create a jQuery object from the form
$form = $(event.target);
// Serialize the form data
var formData = $form.serialize();
// You should sterilise the file names
$.each(data.files, function(key, value)
{
formData = formData + '&filenames[]=' + value;
});
$.ajax({
url: 'jquery_upload_form_submit.php',
type: 'POST',
data: formData,
cache: false,
dataType: 'json',
success: function(data, textStatus, jqXHR)
{
if(typeof data.error === 'undefined')
{
// Success so call function to process the form
console.log('SUCCESS: ' + data.success);
}
else
{
// Handle errors here
console.log('ERRORS: ' + data.error);
}
},
error: function(jqXHR, textStatus, errorThrown)
{
// Handle errors here
console.log('ERRORS: ' + textStatus);
},
complete: function()
{
// STOP LOADING SPINNER
}
});
}
});
UPDATE CODE OF - submit.php :
<?php
// Here we add server side validation and better error handling :)
$data = array();
if(isset($_GET['files'])) {
$error = false;
$files = array();
$uploaddir = './uploads/homepage-header-social-icons/';
foreach ($_FILES as $file) {
if (move_uploaded_file($file['tmp_name'], $uploaddir . basename($file['name']))) {
$files[] = $uploaddir . $file['name'];
//$files = $uploaddir . $file['name'];
}
else {
$error = true;
}
}
$data = ($error) ? array('error' => 'There was an error uploading your image-files') : array('files' => $files);
}
else {
$data = array(
'success' => 'Image was submitted',
'formData' => $_POST
);
}
echo json_encode($data);
?>

Answer 1

If you're using POST method in ajax, then you can access those data in PHP.
print_r($_POST);
Form submit using ajax.
//Program a custom submit function for the form
$("form#data").submit(function(event){
//disable the default form submission
event.preventDefault();
//grab all form data
var formData = new FormData($(this)[0]);
$.ajax({
url: 'formprocessing.php',
type: 'POST',
data: formData,
async: false,
cache: false,
contentType: false,
processData: false,
success: function (returndata) {
alert(returndata);
}
});
return false;
});
You can access the data on PHP
$json = '{"countryId":"84","productId":"1","status":"0","opId":"134"}';
$json = json_decode($json, true);
echo $json['countryId'];
echo $json['productId'];
echo $json['status'];
echo $json['opId'];
If you want to access file object, you need to use $_FILES
$profileImg = $_FILES['profileImg'];
$displayImg = $_FILES['displayImg'];

Answer 2

This problem is different in terms that :
1. Its sending MULTIPLE files, to upload
2. Contains a SELECT
3. Contains an INPUT
So, images are serialized & sent via GET . And, rest FORM data is being sent via POST. So Solution is :
var data = new FormData($(this)[0]);
inside : function uploadFiles(event){}
And in submit.php :
print_r($_POST);
print_r($_GET);
die();
This will give you both values. Then comment these & as per the Code, make these changes :
$filename_wpath = serialize($files);
$fb_link = $_POST['fb_link'];
$show_fb = $_POST['show_fb'];
$sql = "INSERT INTO `tblbasicheader`(fldFB_image, fldFB_link, fldHideShow)
VALUES('$filename_wpath','$fb_link','$show_fb')";
mysqli_query($db_conx, $sql);
Works like Charm ! :)