我想創建一個通過按下按鈕來更新資料庫的程序。但是，當您像下面的代碼一樣撰寫它時，當您單擊按鈕時，它會轉到該admin_cp.php頁面。頁面凍結，資料庫的內容保持不變。我的代碼有什么問題？
管理費用.php
<form action="admin_cp.php" method="POST" onsubmit="return confirm('Charging handle?');">
<input type="submit" neme="charge"/>
</form>
admin_cp.php
include('Aconfig.php');
if (isset($_POST["charge"])){
$query = "UPDATE userinfo
INNER JOIN chargeINFO ON (userinfo.u_id = chargeINFO.u_id)
SET userinfo.u_charged = chargeINFO.u_chargewait;";
$result = mysqli_query($conn, $query);
if ($result) {
echo "<script>alert('Charged complete.')</script>";
echo "<meta http-equiv='refresh' content='0;url=admincharge.php'>";
} else {
echo "<script>alert('fail')</script>";
echo "<meta http-equiv='refresh' content='0;url=admincharge.php'>";
}
}
uj5u.com熱心網友回復：
你檢查過這部分嗎：
$query = "UPDATE userinfo
INNER JOIN chargeINFO ON (userinfo.u_id = chargeINFO.u_id)
SET userinfo.u_charged = chargeINFO.u_chargewait;";
在 .unchargewait;"; delete ; 的末尾，看起來像這樣：
$query = "UPDATE userinfo
INNER JOIN chargeINFO ON (userinfo.u_id = chargeINFO.u_id)
SET userinfo.u_charged = chargeINFO.u_chargewait";
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