表格面板&小组成员有1列共同点。称为user_id,例如,它是一个整数
桌面板:
————————————————– ——-user_id电子邮件…
UPDATE panelists SET email_active = '2' WHERE user_id in ( SELECT user_id FROM panel WHERE email LIKE '%dummy.com' );
这应该做的工作
您也可以像这样使用联接
UPDATE panelists JOIN panel ON panelists.user_id = panel.user_id SET panelists.email_active = '2' WHERE panel.email LIKE '%dummy.com';
您可以从1个查询中更改2个表,如下所示:
UPDATE panel t1, panelists t2 SET t1.email = "dummy", t2.email_active = 2 WHERE t1.user_id = t2.user_id AND t1.email like "%@dummy.com";
但是如果你的表很大(包含许多行),我宁愿建议你运行2个单独的查询,如下所示:
-- step 1 UPDATE panel SET email = "dummy" WHERE email like "%@dummy.com"; -- step 2 UPDATE panelists t1 JOIN panel t2 on t1.user_id = t2.user_id AND t2.email = "dummy" SET email_active = 2;
UPDATE panelists as a JOIN panel as s ON s.uid = a.uid SET a.pstatus = '2', s.u_name = 'dummy' WHERE s.u_email LIKE '%dummy.com';