php - 表单不发送值或 PHP 代码不更新

Question

形式

<form method="post" action="menu_duzenle_islem.php">
Şuanki İsim : <input name="menu_isim" value="<?php echo $_POST['menu_isim'] ?>"readonly>
Yeni İsim : <input name="yeni_menu_isim" placeholder="isim giriniz.">
Şuanki URL : <input name="menu_url" value="<?php echo $_POST['menu_url'] ?>"readonly>
Yeni URL : <input name="yeni_menu_url" placeholder="url giriniz">
<input type="submit" value="Düzenle">
</form>

menu_duzenle_islem.php

<?php
$con=mysqli_connect("localhost","root","","mmogezgini");
mysqli_query($con,"UPDATE menuler SET isim='$_POST[yeni_menu_isim]' and url='$_POST[yeni_menu_url]'
WHERE isim='$_POST[menu_isim]' AND url='$_POST[menu_url]'");
mysqli_close($con);
Header("Location:menuler.php");
?>

当我更新“menu_isim”和“menu_url”时 menu_isim 显示“0”并且 menu_url 没有改变。

Answer 1

您的 SQL 查询不正确，请参阅以下内容：

mysqli_query($con,"UPDATE menuler SET isim='". $_POST['yeni_menu_isim']."' ,url='".$_POST['yeni_menu_url'] ."'
WHERE isim='".$_POST['menu_isim']."' AND url='".$_POST['menu_url']."'");