Ahm*_*fiz 4 php mysql join laravel eloquent
我试图为获得技能名skill1
,skill2
,和skill3
从表2通过加入.
当我试图skill1
独自一人时,它工作正常.但是,当我尝试获取下一列的详细信息时,1066不会出现唯一的表/别名错误.
表1(用户表)
======================================
ID Name skill skill2 skill3
======================================
1 Ed 1 4 3
--------------------------------------
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表2(技能细节)
=========================
ID Skill Name
=========================
1 php
2 html
3 css
4 mysql
-------------------------
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这是我期望得到的:
[name] => 'Ed'
[skill1] => 'php'
[skill2] => 'mysql'
[skill3] => 'css'
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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();
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小智 14
试试这个查询:
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)
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