运行AJAX脚本的PHP只能运行一次

Question

I have this strange issue, happening to my PHP script, On page load the AJAX script runs and also after the second time the AJAX script runs it works and sends data to PHP, but i seem to not understand why the PHP script doesn't process the incoming POST request the second time it is sent in when i clean the input text box and type again, i get a blank response.My code for more expatiation.

index.php :

<input type="text" onkeyup="searchmedia(this)" placeholder="Search for seller with UNIQUE ID or Name.">

<div id="resut" style="margin-top:-24px!important;">
    //where the ajax result is returned
</div>
<div style="margin-top:-24px!important;" id="normal">
    //bla bla data here
</div>
<div id="hui" style="display:none;"><img src="../ajax5.gif">
</div>

<script>
    function searchmedia(e) {
        var tuq = $(e).val();
        if (tuq == "") {
            $('#resut').hide();
            $('#normal').show();
            $('#hui').hide();
        } else {
            $('#normal').hide();
            $('#hui').show();
            $.ajax({
                type: 'POST',
                url: 'sellersmessageajax.php',
                data: {tuq: tuq},
                timeout: 5000,
                cache: false,
                success: function (r) {
//console.log(r); 
                    $('#resut').html(r);
                    $('#normal').hide();
                    $('#hui').hide();
                },
                error: function () {
                    alert("Could not search, reload the page and try again.");
                    $('#normal').show();
                    $('#hui').hide();
                }
            });
        }
    }
</script>

sellersmessageajax.php :

<?php include('../connect.php'); ?>


<?php
if (isset($_POST['tuq']))
{

    $term = $_POST['tuq'];

    $term = mysqli_real_escape_string($con,
        $term); //WHEN I ALERT HERE THE SECOND TIME I SEE THE INPUT TEXT DATA THAT CAME IN BUT PLEASE CHECK AFTER THE **FOREACH**


    $condition = '';
    $query     = explode(" ", $term);
    foreach ($query as $text)
    {
        $condition .= "name LIKE '%" . mysqli_real_escape_string($con,
                $text) . "%' OR reign_uniqeer LIKE '%" . mysqli_real_escape_string($con, $text) . "%' OR ";
    }

//WHEN I ALERT HERE I GET NOTHING

    $condition = substr($condition, 0, -4);
    $zobo      = "ORDER BY name";
    $sql_query = "SELECT * FROM sellers_login WHERE " . $condition . $zobo;
    $result    = mysqli_query($con, $sql_query);
    if (mysqli_num_rows($result) > 0)
    {
        while ($row = mysqli_fetch_array($result))
        {
            $v_ida            = $row['id'];
            $v_namea          = $row['name'];
            $v_reign_uniqeera = $row['reign_uniqeer'];
            ?>

            <div style="border-bottom:0.1px solid #eee;padding-bottom:20px;margin-top:20px;">
                <a class="zuka" title="<?php echo $v_ida ?>" id="<?php echo $v_ida ?>"
                   style="color:#666;text-decoration:none;outline:none!important;cursor:pointer;">
                    <b style="color:blue;"><?php echo $v_namea ?></b>
                    <br/>
                    <div style="height:auto;max-height:30px;">
                        <b>UNIQUE ID :</b> <b style="color:red;"><?php echo $v_reign_uniqeera ?></b>
                    </div>
                </a>
            </div>

            <?php
        }
    }
    else
    {
        ?>
        <h1 class="zuka" style="text-align:center;margin-top:20%;"> No result found.</h1>
        <?php
    }
}
?>

Answer 1

you are sending the var tuq wrongfully. Try this:

 data :  {"tuq": tuq}