Question

基本上我的问题是，它适用于phpmyadmin而不是php本身。当我运行脚本时，它返回true，并且不会死。帮助！

这是查询：

DELETE FROM selections WHERE selectorID = '$userID' AND selectedID = '$ID'

这是php脚本：

<?php

include_once "connect_to_mysql.php";

$errorB = false;

$error = '';
$id = '';
$userID = '';
if (empty($_GET['ID']) || empty($_GET['userID'])) {
$errorB = true;

if (empty($_GET['ID'])) {
    $error = $error.'-No selected ID was sent';

}

if (empty($_GET['userID'])) {
    $error = $error.'-No selector ID was sent';

}


} else {
$id = $_GET['ID'];
$userID = $_GET['userID'];
echo $id.$userID;
$sqlInsert = mysql_query("DELETE FROM selections WHERE selectorID = '$userID' AND selectedID = '$ID'") or die (mysql_error());
echo 'shz';
}

if ($errorB) {
echo $error;

}



?>

我的数据库如下所示：

 id|selectorID|selectedID
 3  1          4
 4  1          5

Answer 1

你的第一个，但较小的麻烦是case sensitive nature of PHP variables：

下面

$sqlInsert = mysql_query("DELETE FROM selections WHERE selectorID = '$userID' AND selectedID = '$ID'") or die (mysql_error());

您引用的是$ID，但只有$id：

$id = $_GET['ID'];

正确：

$sqlInsert = mysql_query("DELETE FROM selections WHERE selectorID = '$userID' AND selectedID = '$id'") or die (mysql_error());

重要提示

虽然这很可能会奏效，但这很容易SQL injection！请改用PDO，或者至少使用escape您的值!!!

mysql DELETE适用于phpmyadmin但不适用于php脚本

1 个答案: