function details($id){
$query = $this->db->query("select t1.gradename, t2.* from grade as t1,student as t2 where studentid = '$id' and t2.gradeid = t1.gradeid ");
return $query->result();
}
有两个表,student表 和grade 表,调用方法传递过来id变量用于查询指定学号的学生信息,现在请教一下,我想查询sutdent的所有字段值和gradename的值。
由于student和grade是外键关联关系student表的gradeid和grade表的gradeid一样, gradename存在grade表里。
我写语句是错的,执行不了,请大家伙帮一下忙,谢谢。
回复讨论(解决方案)
报错信息是什么?sql没看出问题。。。
select s.*, g.gradename from student as s left join grade as g on s.gradeid = g.gradeid where s.studentid = $id
报错信息是什么?sql没看出问题。。。
#1052 - Column 'studentid' in where clause is ambiguous
wander_wind 写的正确,可我不明白我的为什么错了
Column 'studentid' in where clause is ambiguous
在where子句中的 studentid 列是含糊的。
在多表查询中,列名带上它的前缀是必须的,否则MySQL分不清楚。
select t1.gradename, t2.* from grade as t1,student as t2 where t1.studentid = '$id' and t2.gradeid = t1.gradeid
wander_wind 写的正确,可我不明白我的为什么错了
这个报错最常见的原因就是,关联的表中相同的字段导致的,像楼上说的那样加个表别名做前缀

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