[英]'id' in where clause is ambiguous
I have a table with some fields and some relations with others. 我有一张桌子,里面有一些字段,还有一些与其他人的关系
I get the list view with no problem, but when I try to filter the results (using the search feature in flexigrid table) I get: 我得到列表视图没有问题,但当我尝试过滤结果(使用flexigrid表中的搜索功能)时,我得到:
A Database Error Occurred
发生数据库错误
Error Number: 1052
错误号码:1052
Column 'id' in where clause is ambiguous
where子句中的列'id'是不明确的
SELECT
gee_job_boards
.*, j32e2cb0f.dp_name AS s32e2cb0f FROMgee_job_boards
LEFT JOINgee_distribution_partner
asj32e2cb0f
ONj32e2cb0f
.SELECT
gee_job_boards
。*,j32e2cb0f.dp_name AS s32e2cb0f FROMgee_job_boards
LEFT JOINgee_distribution_partner
asj32e2cb0f
ONj32e2cb0f
。id
=gee_job_boards
.id
=gee_job_boards
。dp_id
WHEREid
LIKE '%27%' ESCAPE '!'dp_id
WHEREid
LIKE'%27%'ESCAPE'!' LIMIT 25限制25
Filename: models/Grocery_crud_model.php
文件名:models / Grocery_crud_model.php
Line Number: 87
行号:87
Error Number: 1052
错误号码:1052
Column 'id' in where clause is ambiguous
where子句中的列'id'是不明确的
how could i solve this ? 我怎么能解决这个问题?
thanks 谢谢
您应该在where子句中限定您的id,例如这样
j32e2cb0f.id LIKE ...
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.