Question

我想检查一些事情......但我无法......（查看下面的内容）。

Okey所以这就是它的工作方式，但我想添加更多的支票。

mysql_query('UPDATE characters SET voted=1 where account_name like \''.$row['login'].'\' and online=1;') ;

无论如何我要添加的是在线= 1后检查哪里 MIN(lastAccess)

我尝试了一些东西，但我失败了...... 像：

mysql_query('UPDATE characters SET voted=1 where account_name like \''.$row['login'].'\' and online=1 having min(lastaccess);') ;

Answer 1

您甚至可以根据需要进一步加入表格，

UPDATE  characters a
        INNER JOIN
        (
            SELECT  account_name, MIN(lastaccess) min_date
            FROM    characters
            GROUP   BY account_name
        ) b ON  a.Account_Name = b.Account_name AND
                a.lastAccess = b.min_date
SET     a.voted = 1
WHERE   a.Account_Name = 'nameHere' AND
        a.online = 1

更新1

尝试使用double quotes，例如

$userName = $row['login'];
$result = mysql_query(" UPDATE  characters a
                                INNER JOIN
                                (
                                    SELECT  account_name, MIN(lastaccess) min_date
                                    FROM    characters
                                    GROUP   BY account_name
                                ) b ON  a.Account_Name = b.Account_name AND
                                        a.lastAccess = b.min_date
                        SET     a.voted = 1
                        WHERE   a.Account_Name = '$userName' AND
                                a.online = 1");

作为旁注，如果变量的值（ s ）来自外部，则查询易受SQL Injection攻击。请查看下面的文章，了解如何防止它。通过使用PreparedStatements，您可以摆脱在值周围使用单引号。

How to prevent SQL injection in PHP?

尝试检查时出错

1 个答案: