Ajax form validation with php WITHOUT jQuery

Question

I am learning JavaScript without jQuery.

Right now I am trying to pass some data from an input field to php and than pass a $variable from php to javascript. In jQuery this is easy with $.ajax.

But how do I do this with ONLY JavaScript? Here is my attempt. Right now I only want to pass the $_POST content from the inputfield. I didn't do any validation at this moment.

My plan is to make a validation with php and then pass an error message or more than one. Or in case of success an success message.

But out of my console log I am only getting NULL.

window.onload = function () {
    var Input = document.querySelector('input#Input');
    var InputButton = document.querySelector('button.formBtn');
    InputButton.onclick = function () {

        var InputRequest = new XMLHttpRequest();

        InputRequest.open("POST", "ajax.php", true);
        InputRequest.send();
        InputRequest.onreadystatechange = function () {
            if (this.readyState == 4 && this.status == 200) {
                 var obj = JSON.parse(InputRequest.response)
                 console.log(obj);
            }
        }
        return false;
    }
}

<!DOCTYPE html>
<html lang="en">

<head>
    <meta charset="UTF-8">
    <title>Ajax Example</title>
    <style>
        #Input {
            width: 200px;
            height: 15px;
            padding: 10px 0;
            text-indent: 5px;
        }
        #Input:focus {
            outline: none;
            border: 1px solid lightblue;
        }
    </style>
</head>

<body>
    <form name="form" action="ajax.php" method="post">
        <input type="text" id="Input" name="inputTest">
        <button type="submit" class="formBtn">Absenden</button>
    </form>
    <script src="ajax.js"></script>
</body>

</html>

<?php
$inputResponse = $_POST["inputTest"];
echo json_encode($inputResponse)
?>

Rasclatt · Accepted Answer · 2017-08-09 20:02:59Z

1

You are missing a line and need to modify your send() line for sending POST content:

// You need to send the type
InputRequest.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
// Send the post values in the send
InputRequest.send('key=value&key2=value2');

In the case of the send() you have to turn the keys and values to a query string. I think this is why so many use jQuery, it's done all this for you.

answered Aug 9, 2017 at 20:02

Rasclatt

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