这里是确切的错误:笨数据库错误编号1052
Error Number: 1052
Column 'id' in where clause is ambiguous
SELECT * FROM (`membership_personal`) JOIN `membership_account` ON `membership_account`.`memberid` = `membership_personal`.`id` JOIN `swapaccount` ON `swapaccount`.`memberid` = `membership_account`.`memberid` WHERE `id` = '5'
Filename: /home/phpgod/public_html/johnnyarias/ci_website/models/generalfeaturesmodel.php
Line Number: 57
我知道,它与id字段存在于的连接多个表的事情。所以我的问题是如何确保where子句中的id字段仅应用于member_personal表中的id字段?
它说:未知列“member_personal.id = 5”在“where子句” – user1549397 2012-08-06 06:06:37
究竟如何将ü把上面的其中CodeIgniter的数据库方法的条款? – user1549397 2012-08-06 06:07:10