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Php Jquery Ajax Post Request Example

In this example, i will share with you how to write simple ajax request example with jquery php. we will see jquery ajax post data example with php. you can simply form submit with pass ajax post data and get return all data with success.

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PHP
jQuery
Ajax
JSON

4.5/5.0
Last updated 08 September, 2022
By Admin

I will give you very simple example of ajax post request with php. you can also write server side validation using php logic. you can also pass form serialize ajax data to post method with php.

After this example you can easily write Ajax Get Request, Ajax Post Request, Ajax Put Request and Ajax Delete Request with jquery ajax and php.

You need to just follow bellow step to create ajax post request:

index.php

<!DOCTYPE html>
<html>
<head>
	<title>Php Ajax Form Validation Example</title>
	<script type="text/javascript" src="https://code.jquery.com/jquery-1.9.1.min.js"></script>
    <link rel="stylesheet" href="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.7/css/bootstrap.min.css">
</head>
<body>
<div class="container">
	<h1>Php Ajax Form Validation Example</h1>
	<form role="form" id="contactForm">
		<div class="alert alert-danger display-error" style="display: none">
		</div>
		<div class="form-group">
			<div class="controls">
				<input type="text" id="name" class="form-control" placeholder="Name">
			</div>
		</div>
		<div class="form-group">
			<div class="controls">
				<input type="email" class="email form-control" id="email" placeholder="Email" >
			</div>
		</div>
		<div class="form-group">
			<div class="controls">
				<input type="text" id="msg_subject" class="form-control" placeholder="Subject" >
			</div>
		</div>
		<div class="form-group">
			<div class="controls">
				<input type="text" id="message" class="form-control" placeholder="Massage" >
			</div>  
		</div>
		<button id="submit" class="btn btn-success"><i class="fa fa-check"></i> Send Message</button>
	</form>
</div>
</body>
<script type="text/javascript">
  $(document).ready(function() {
      $('#submit').click(function(e){
        e.preventDefault();
        var name = $("#name").val();
        var email = $("#email").val();
        var msg_subject = $("#msg_subject").val();
        var message = $("#message").val();
        $.ajax({
            type: "POST",
            url: "/formProcess.php",
            dataType: "json",
            data: {name:name, email:email, msg_subject:msg_subject, message:message},
            success : function(data){
                if (data.code == "200"){
                    alert("Success: " +data.msg);
                } else {
                    $(".display-error").html("<ul>"+data.msg+"</ul>");
                    $(".display-error").css("display","block");
                }
            }
        });
      });
  });
</script>
</html>

formProcess.php

<?php
$errorMSG = "";
/* NAME */
if (empty($_POST["name"])) {
    $errorMSG = "<li>Name is required</<li>";
} else {
    $name = $_POST["name"];
}
/* EMAIL */
if (empty($_POST["email"])) {
    $errorMSG .= "<li>Email is required</li>";
} else if(!filter_var($_POST["email"], FILTER_VALIDATE_EMAIL)) {
    $errorMSG .= "<li>Invalid email format</li>";
}else {
    $email = $_POST["email"];
}
/* MSG SUBJECT */
if (empty($_POST["msg_subject"])) {
    $errorMSG .= "<li>Subject is required</li>";
} else {
    $msg_subject = $_POST["msg_subject"];
}
/* MESSAGE */
if (empty($_POST["message"])) {
    $errorMSG .= "<li>Message is required</li>";
} else {
    $message = $_POST["message"];
}
if(empty($errorMSG)){
	$msg = "Name: ".$name.", Email: ".$email.", Subject: ".$msg_subject.", Message:".$message;
	echo json_encode(['code'=>200, 'msg'=>$msg]);
	exit;
}
echo json_encode(['code'=>404, 'msg'=>$errorMSG]);
?>

You can quick run our example by following command, so run bellow command for run PHP project.

php -S localhost:8000

Now, you can check from your url by bellow URL.

http://localhost:8000

I hope it can help you...

JSON

Php Jquery Ajax Post Request Example

index.php

formProcess.php

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