[英]ajax with php don't do anything
我在這里匆忙寫了代碼,我知道我可以進行 sql-injection,但是在我將發布的版本中我會修復所有內容,現在我只需要了解它為什么不起作用
我有這個名為 user.php 的 php 頁面
<?php
$to = mysqli_query($con,"SELECT COUNT(username) FROM coins_logs WHERE id= '5';");
while($totaleusertt = mysqli_fetch_array($to))
{
$totaleuser = $totaleusertt[0] - 1;
}
$contatore = 0;
$query = mysqli_query($con,"SELECT username FROM coins_logs WHERE id= '5';");
while($row = mysqli_fetch_array($query, MYSQLI_NUM))
{
$utentiU[$contatore] = $row[0];
$contatore = $contatore + 1;
}
$prova = '<div class="table">
<div class="table-header">
<div class="header__item"><a id="name" class="filter__link" href="#">bids history</a>. </div>
</div>
<div class="table-content">
<div class="table-row">
<div class="table-data">'. $utentiU[$totaleuser-1].'</div>
</div>
<div class="table-row">
<div class="table-data">'.$utentiU[$totaleuser-2].'</div>
</div>
<div class="table-row">
<div class="table-data">'. $utentiU[$totaleuser-3] .'</div>
</div>
<div class="table-row">
<div class="table-data">'.$utentiU[$totaleuser-4].' </div>
</div>
<div class="table-row">
<div class="table-data">'.$utentiU[$totaleuser-5].'</div>
</div>
</div>
</div>';
echo $prova;
?>
我有這個 JS 的 html 頁面
<head>
<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.3/jquery.min.js"></script>
</head>
<div id="mytablepage">
</div>
<script type="text/javascript">
$(document).ready(function(){
function(){
$.ajax({
url: '/user.php',
method:'POST',
success:function(msg){
$('#mytablepage').html(msg);
}
});
}
});
</script>
但什么也沒發生,我嘗試為 ajax 代碼添加一個按鈕,但我得到了相同的結果:什么都沒有。 我能做些什么來解決這個問題?
您需要致電 function
$(document).ready(function() { (function() { $.ajax({ url: '/user.php', method: 'POST', success: function(msg) { $('#mytablepage').html(msg); } }); }()) });
聲明:本站的技術帖子網頁,遵循CC BY-SA 4.0協議,如果您需要轉載,請注明本站網址或者原文地址。任何問題請咨詢:yoyou2525@163.com.