Question

我有一个表（＆＃39;名称＆＃39;），其中包含与依赖于ID的其他表中的其他数据相关的数据。例如：

*名称表

id |     name     | predecessor | successor | house | birthplace 
-----------------------------------------------------------------
10    Bayezid II         9           11         4         NULL
11    Selim I            10          12         4         5
12    Suleiman           11          13         4         61

*房屋表

id |  house
--------------
4    House of Osman

*地方表

id |  place
--------------
5    Amasya
61   Trabzon

我想要完成的是构建一个查询，导致根据id返回整个信息，如：

{"result":[{
    "id":"11",
    "name":"Selim I",
    "predecessor": "Bayezid II",
    "successor": "Suleiman",
    "house":"House of Osman",
    "birthplace":"Amasya"
}]}

所以，房子和出生地的名称来自其他表格（＆＃39;房屋＆＃39;地点＆＃39;），而前任和后继者来自同一张桌子。我需要帮助构建此查询。谢谢。

Answer 1

只需自我加入几次，一次获得前一行（下面有别名n0），再一次为继任者（n2）：< / p>

SELECT n1.id, n1.name, n0.name AS predecessor, n2.name AS successor
FROM names n1
LEFT JOIN names n0 ON n1.predecessor = n0.id
LEFT JOIN names n2 ON n1.successor = n2.id

SQL Fiddle demo

加入房屋和出生地是留给读者的练习。

Answer 2

试试这个：

select n.id,
n.name,
n1.name as predecessor,
n2.name as successor,
h.house,
p.place
from names n
inner join names n1 on n.id = n1.predecessor
inner join names n2 on n.id = n2.successor
left join Houses h on n.house = h.id
left join Place p on n.birthplace = p.id
where n.id = 11

MySql选择结果从同一表和其他表组合多个查询

2 个答案: