我有4个表:user
,education
,family
,work
表用户:
id_user | name | address |province|tlp|
---------------------------------------
1 |andi | xxx | xxy |123|
表教育:
id_education | id_user | year |school |
---------------------------------------
1 | 1 | 1990 | aaa |
2 | 1 | 1994 | bbb |
表系列:
id_family | id_user | name |status |
---------------------------------------
1 | 1 | ddd | wife |
2 | 1 | eee | first children |
3 | 1 | fff | second children |
表工作:
id_work | id_user | year |place |
---------------------------------------
1 | 1 | 2004 | ggg |
2 | 1 | 2010 | hhh |
我想获得具有sama id_user
public function show($id) {
$user = DB::table('user')
-> join('education', 'user.id_user', '=', 'education.id_user')
-> join('family', 'user.id_user', '=', 'family.id_user')
-> join('work', 'user.id_user', '=', 'work.id_user')
->where('id_user', '=', $id)
->first();
return View::make('user.show')->with('user', $user)->with('title', 'Show User');
}
但它给了我错误:
SQLSTATE[23000]: Integrity constraint violation: 1052 Column 'id_user' in where clause is ambiguous
请帮助我,
谢谢你:))
答案 0 :(得分:0)
尝试.. ->where('user.id_user', '=', $id) ..