我的数据库中有一个有点疯狂的结构,其中用户和组织在同一个表中(下面的示例)
他们都关联了PARENTID和MEMBERID,我需要得到的是:
获取所有“人”的 name
,他们的 MEMBERID
以及他们 parent 的 name
结果应该是这样的:
UserName1 | DHAD781 | OrgName1
表格示例:
这里是数据库的例子:
-------*/*------------*/*------------*/*------------*/*------------*/*------------
MEMBERID level name kind status PARENTID
-------*/*------------*/*------------*/*------------*/*----------*/*------------
EMD123F | 2 | OrgName1 | Org | | rootID
---------------------------------------------------------------------------------
DHAD781 | 3 | UserName1 | Person | active | EMD123F
---------------------------------------------------------------------------------
7AJIZU7 | 3 | UserName2 | Person | active | EMD123F
---------------------------------------------------------------------------------
DME123F | 2 | OrgName2 | Org | | rootID
---------------------------------------------------------------------------------
TT5451AL| 3 | UserName3 | Person | active | DME123F
---------------------------------------------------------------------------------
RRMI7481| 2 | OrgName3 | Org | | rootID
---------------------------------------------------------------------------------
PPUNSAD9| 2 | OrgName4 | Org | | rootID
---------------------------------------------------------------------------------
GJASDNZB| 3 | UserName4 | Person | inactive | PPUNSAD9
---------------------------------------------------------------------------------
KJNSCZM7| 2 | OrgName5 | Org | | rootID
---------------------------------------------------------------------------------
1UZGOPAS| 3 | UserName5 | Person | deleted | KJNSCZM7
---------------------------------------------------------------------------------
我做了什么尝试:
SELECT t1.NAME, t1.MEMBERID
FROM roles t1
inner join roles t2 on t2.PARENTID = t1.MEMBERID
WHERE t1.kind= 'Person'
我从这个查询中得到了什么: 找到 0 个结果
最佳答案
似乎是相反的:
SELECT persons.NAME, persons.MEMBERID, orgs.NAME, orgs.MEMBERID
FROM roles persons
INNER JOIN roles orgs on persons.PARENTID = orgs.MEMBERID
WHERE persons.kind = 'Person'
此外,选择明确的别名使查询更具可读性(并且更不容易出错 :))
- 参见 this fiddle
关于MySQL SELECT 里面有另一个 SELECT,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/32952480/