现在我做了一个非常简单的版本(这没有多大意义,但显示了我的错误) UPDATE a_test SET categoryID = '2956' WHERE id IN ( SELECT)
) 这将以以下错误结束: #1093 - Table 's_articles_categories' is specified twice, both as a target for 'UPDATE这是我想使用的完整查询,但有相同的错误: UPDATE s_art
我正疯狂地尝试着用MySQL中的CTE进行更新。这是我第一次天真的尝试,它说我在UPDATE附近有一个错误。嗯哼。PARTITION BY `Date`, `Name`, Statement_s) AS Trans, FROM sa_general_journalUPDATE sa_general_journal gj, cteWHE
UPDATE mapping SET FILTER = 0 WHERE id IN( CONCAT(Field1, Field2, F3, F4, F5, F6, F7) NOT LIKE "%searchTerm%"这对我的本地开发人员和MySQL来说就像一种魅力。但是,我的主机使用MariaDB,我收到一条错误消息:
#1064 -您的SQL语法有一个错误;请检查与您的MariaDB服务器版本相对应的手册,以便在第4行使用“WHE
mysql> SELECT Ext, Pass, Name, Context FROM temp_Users WHERE temp_Users.Pass NOT IN (SELECT Pass FROM| Hello | WebPone | DLPN_Admin |1 row in set (0.00 sec)
-> SET (Pass, Name, Context) =
$user, $pass) or die(mysql_error());if ($_SERVERid, username, email, ip FROM users WHERE username='". mysql_real_escape_string($username) . "'");= $arr['email'];
$
update share set holder = 22 where SHARE_ID IN (select SHARE_ID from SHARE WHERE holder=1 LIMIT 10)#1235 - This version of MySQL doesn't yet support 'LIMIT & IN/ALL/ANY/SOME subquery'