Question

也许它只是我是一个菜鸟，但对于我的生活，我无法阻止下面的代码恢复到以下错误。

Mysql错误：您的SQL语法中有错误;检查手册对应于您的MySQL服务器版本，以获得正确的语法在第4行'WHERE id = 113'附近使用

ID（在这种情况下为ID 113）确实与数据库匹配，但仍然没有乐趣。

非常感谢您的帮助，我真的很感激。

<?php

ob_start();

error_reporting(E_ALL & ~E_NOTICE);
ini_set('display_errors', TRUE);
ini_set('display_startup_errors', TRUE);

function isLoggedIn()
{
    if(isset($_SESSION['valid']) && $_SESSION['valid'])
        return true;
    return false;
}





session_start();
//if the user has not logged in
if(!isLoggedIn())
{
    header('Location: ../main');

    die();
}

    //! Checks that a restaurant is logged on
    if ($_SESSION['user_type'] !== 'restaurant')
    {


        echo('You need to be a restaurant user to do that!, If you are seeing this message in error, please contact the system administrator.');
        die();

    }


    //! Checks for direct access to page

    if (empty($_GET)) {
        header('location:../main/menu-manager.php?result=nothingentered');
        die();
    }


//! Get info from POST
$ID =                   $_GET['optionid'];
$rest_ID =              $_SESSION['rest_id'];



//! security real escape

$ID =                       mysql_real_escape_string($ID);

//! Connect to the database

require_once('../Connections/PropSuite.php');
mysql_select_db($database_Takeaway, $Takeaway);

//! Write the information to the database

$query = "UPDATE menu_cats
            SET rest_id = 'd.$rest_ID',

            WHERE id = $ID ";
mysql_query($query);



    if( mysql_errno() != 0){
     // mysql error
     // note: message like this should never appear to user, should be only stored in log
     echo "Mysql error: " . htmlspecialchars( mysql_error());
     die();
     }

else {

header('Location: ../main/menu-manager.php?result=success');

}


?>

Answer 1

在Where子句之前，您的SQL语句中不需要逗号。

Answer 2

您必须在where

之前删除逗号

$query = "UPDATE menu_cats
            SET rest_id = 'd.$rest_ID'
            WHERE id = $ID ";

Answer 3

您的查询中有一个迷路,。查询应该是这样的：

$query = "UPDATE menu_cats
        SET rest_id = 'd.$rest_ID'
        WHERE id = $ID ";

MySQL语法错误问题 - 无法解决问题

3 个答案: