Question

我正在尝试获取skill1，skill2和＆amp;的技能名称。来自table2的skill3使用Join。

当我试图单独skill1时，它工作正常。但是，当我尝试获取下一列的详细信息时， 1066不是唯一的表/别名错误。

表1（用户表）

======================================
ID  Name       skill   skill2   skill3
======================================
1   Ed           1       4       3    
--------------------------------------

表2（技能详情）

=========================
ID  Skill Name
=========================
1   php
2   html
3   css
4   mysql
-------------------------

这是我期望得到的：

[name]    => 'Ed'
[skill1]  => 'php'
[skill2]  => 'mysql'
[skill3]  => 'css'

这是我的代码，我正在使用laravel：

DB::table('table1')
   ->join('table2', function($join)
    {
         $join->on('table1.skill1', '=', 'table2.id');
    })
    ->join('table2', function($join)
    {
         $join->on('table1.skill2', '=', 'table2.id');
    })
    ->join('table2', function($join)
    {
         $join->on('table1.skill3', '=', 'table2.id');
    })
    ->get();

Answer 1

尝试此查询：

SELECT U.Name AS Name, S1.Skill Name AS Skill1, S2.Skill Name AS Skill2, S3.Skill Name AS Skill3
    FROM table1 U
    JOIN table2 S1 ON (S1.Id = U.skill1)
    JOIN table2 S2 ON (S2.Id = U.skill2)
    JOIN table2 S3 ON (S3.Id = U.skill3)

Answer 2

与@erickmcarvalho查询相同的结果

SELECT Table1.usrname,
(SELECT Table2.skillname FROM Table2 WHERE Table1.skill1 = Table2.Id) As skill1,
(SELECT Table2.skillname FROM Table2 WHERE Table1.skill2 = Table2.Id) As skill2,
(SELECT Table2.skillname FROM Table2 WHERE Table1.skill3 = Table2.Id) As skill3
FROM Table1

仍然会导致4次查询，最好重新构建表格

MYSQL从同一个表中加入多个列

2 个答案: