11-Apr-2023
.
Admin
Hello Friends,
I am going to explain you example of How to upload Multiple Image files with jQuery AJAX and PHP 8. I explained simply step by step PHP 8 Multiple Image Upload using Ajax Example Tutorial. This post will give you simple example of PHP 8 Ajax Multiple Image Upload Example. you'll learn Uploading both data and files in one form using Ajax PHP 8.
So, let's follow few step to create example of PHP 8 Multiple Files/Images Upload in MySQL Database.
This article will give you simple example of Upload Multiple Image Using Ajax in PHP 8 with Preview
You can use from how to upload multiple image in Database using PHP and MySQL. I will give you simple how to multiple image upload Jquery AJAX PHP MySQL.
So, let's see bellow solution:
index.php
<!DOCTYPE html>
<html lang="en">
<head>
<title>How to upload Multiple Image files with jQuery AJAX and PHP 8</title>
<meta charset="utf-8">
<meta name="viewport" content="width=device-width, initial-scale=1">
<link rel="stylesheet" href="https://maxcdn.bootstrapcdn.com/bootstrap/4.5.2/css/bootstrap.min.css">
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.5.1/jquery.min.js"></script>
</head>
<body>
<div class="container mt-5">
<div class="row">
<div class="col-md-12">
<div class="card m-auto w-75">
<div class="card-header text-center text-white" style="background: #0c06f3">
<h4>How to upload Multiple Image files with jQuery AJAX and PHP 8 - Nicesnippets.com</h4>
</div>
<div class="card-body">
<div class="alert alert-success alert-dismissible" id="success" style="display: none;">
<button type="button" class="close" data-dismiss="alert">×</button>
File uploaded successfully
</div>
<form id="submitForm">
<div id="preview"></div>
<div class="form-group">
<label><strong>Select Image : </strong><span class="text-danger"> *</span></label>
<input type="file" class="form-control" name="multipleFile[]" id="multipleFile" required="" multiple>
</div>
<div class="form-group d-flex justify-content-center">
<button type="submit" name="upload" class="btn btn-success">Upload</button>
</div>
</form>
</div>
</div>
</div>
</div>
</div>
<script type="text/javascript">
$(document).ready(function(){
function previewImages() {
var $preview = $('#preview').empty();
if (this.files) $.each(this.files, readAndPreview);
function readAndPreview(i, file) {
var reader = new FileReader();
$(reader).on("load", function() {
$preview.append($("<img/>", {src:this.result, height:100}));
});
reader.readAsDataURL(file);
}
}
$('#multipleFile').on("change", previewImages);
$("#submitForm").on("submit", function(e){
e.preventDefault();
$.ajax({
url :"store.php",
type :"POST",
cache:false,
contentType : false, // you can also use multipart/form-data replace of false
processData : false,
data: new FormData(this),
success:function(response){
$("#success").show();
$("#multipleFile").val("");
}
});
});
});
</script>
</body>
</html>
store.php
<?php
$sernamename = "localhost";
$username = "root";
$passoword = "";
$databasename= "store";
$con = mysqli_connect($sernamename, $username,$passoword,$databasename);
if ($con->connect_error) {
die("Connection failed". $con->connect_error);
}
if (!empty($_FILES['multipleFile']['name'])) {
$multiplefile = $_FILES['multipleFile']['name'];
foreach ($multiplefile as $name => $value) {
$allowImg = array('png','jpeg','jpg','');
$fileExnt = explode('.', $multiplefile[$name]);
if (in_array($fileExnt[1], $allowImg)) {
if ($_FILES['multipleFile']['size'][$name] > 0 && $_FILES['multipleFile']['error'][$name]== 0) {
$fileTmp = $_FILES['multipleFile']['tmp_name'][$name];
$newFile = rand(). '.'. $fileExnt[1];
$target_dir = 'c:/xampp7/htdocs/uploads/'.$newFile;
if (move_uploaded_file($fileTmp, $target_dir)) {
$query = "INSERT INTO user(image) VALUES('$newFile')";
mysqli_query($con, $query);
}
}
}
}
}
?>
Output:
It will help you...
#PHP 8
