我有這個奇怪的問題,發生在我的PHP腳本,頁面加載AJAX腳本運行,也是第二次後,AJAX腳本運行它的工作原理和發送數據到PHP,但我似乎不明白爲什麼PHP腳本在第二次輸入POST請求時沒有處理傳入的POST請求,當我清理輸入文本框並再次輸入時,我得到一個空白的響應。我的代碼更詳細。PHP運行AJAX腳本只能運行一次
的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
}
}
?>
請正確格式化你的代碼。這不好讀! –
'當我警告我沒有任何東西'PHP沒有警報功能,那也是JS也在哪裏調用這個js函數? 'searchmedia()'通信總是從客戶端到服務器。 – ArtisticPhoenix
@ArtisticPhoenix它在第一個輸入字段中從'onkeyup =「searchmedia(this)」調用。 – Barmar