Sending AJAX-Request to MySQL and getting JSON back via php/jQuery

Question

im trying to send a AJAX request to my database and get the information in JSON format back, then i want to replace my div with those json information. But my div wont get changed and i dont know why..

Here is my code and thank you very much for helping:

index.php

<form class="login-form" method="post" action="">
    <input type="text" name="lusername" placeholder="username"/>
    <input type="password" name="lpassword" placeholder="password"/>
    <input type="image" name="login" src="src/login.png" onClick="getLoginInfo()">
    <p class="message">Not registered? <a href="#" onClick="return false" onmousedown="changeContent('registerClick');"><span id="tab">Create an account</a></span></p>
    <div id="test">change me pls</div>
</form>

login.php

<?php
    require_once("connection.php");

    if(isset($_POST['login_x']) or isset($_POST['login_y'])) {
        $lusername = $_POST['lusername'];
        $lpassword = $_POST['lpassword'];
        $select = mysqli_query($connection, "SELECT Username, Password FROM Account WHERE Username='".$lusername."' AND Password='".$lpassword."'");

        if(mysqli_num_rows($select) == 0) {
            echo '<script>cantFindAccount()</script>';
        } else {
             $array = mysqli_fetch_row($select);
             echo json_encode($array);
        }
    }
?>

getLoginInfo.js

function getLoginInfo() {
    $.ajax({
        url: 'http://localhost/login.php',
        data: "",
        dataType: 'json',
        success: function(data) {
            var user = data[0];
            var pass = data[1];
            $('#test').html("<b>id: </b>"+user+"<b> name: </b>"+pass);
        }
    });
    return false;
}

If you `console.log(data);` in success what you do see in the debugger? — nerdlyist, Feb 17 '17 at 20:02
I thaught that the js function could get the data from this echo and i did but nothing appeared on my debugger log .. — Pachamachai, Feb 17 '17 at 20:08
Didn't see this the first time do not use `or` use `||` http://stackoverflow.com/questions/5998309/logical-operators-or-or also you are not making into your if because you are not passing any data. — nerdlyist, Feb 17 '17 at 20:10
@nerdlyist I make it through my if, when i pass wrong data the function cantFindAccount() starts. But when i pass correct data nothing happens, only echo json_encode($array) starts — Pachamachai, Feb 17 '17 at 20:12
Take a look at http://stackoverflow.com/questions/1960240/jquery-ajax-submit-form — nerdlyist, Feb 17 '17 at 20:13

score 0 · Answer 1 · answered Feb 17 '17 at 21:29

0

Try this, you have to post some data to your script so that it would process. Moreover you have to use method: 'POST' because you have used $_POST[]

 $.ajax({
    url: 'http://localhost/login.php',
    data: $(this).serialize(),
    method: 'POST',
    dataType: 'json',
    success: function(data) {
        var user = data[0];
        var pass = data[1];
        $('#test').html("<b>id: </b>"+user+"<b> name: </b>"+pass);
    }
});

answered Feb 17 '17 at 21:29

Ali Rasheed

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I tried it but unfortunately the same problem :/ Im hopeless – Pachamachai Feb 17 '17 at 22:26
Look at inspect element that when you submit form does it go with params and the returned response is in json format? – Ali Rasheed Feb 18 '17 at 08:59

Sending AJAX-Request to MySQL and getting JSON back via php/jQuery

1 Answers1