如何在mysql中进行嵌套选择查询
我有一个如下表，它彼此连接，就像如何在mysql中进行嵌套选择查询,mysql,sql,Mysql,Sql,我有一个如下表，它彼此连接，就像 Info_Table -> RoomGuests_Table -> ChildAge_Table 这些是表格 Info_Table +---------------------------+ | ID | Name | Rooms | +---------------------------+ | INFO1 | ABC | 2 | | INFO2 | DEF |
Info_Table -> RoomGuests_Table -> ChildAge_Table
Info_Table
+---------------------------+
| ID | Name | Rooms |
+---------------------------+
| INFO1 | ABC | 2 |
| INFO2 | DEF | 1 |
| INFO3 | GHI | 3 |
+---------------------------+
RoomGuests_Table
+-----------------------------------+
| ID | R_ID | Adult | Child |
+-----------------------------------+
| RG1 | INFO1 | 2 | 2 |
| RG2 | INFO1 | 3 | 0 |
| RG3 | INFO2 | 2 | 1 |
| RG4 | INFO3 | 2 | 1 |
| RG5 | INFO3 | 2 | 2 |
| RG6 | INFO3 | 2 | 1 |
+-----------------------------------+
ChildAge_Table
+-----------------------+
| ID | R_ID | Age |
+-----------------------+
| CA1 | RG1 | 4 |
| CA2 | RG1 | 5 |
| CA3 | RG3 | 2 |
| CA4 | RG4 | 7 |
| CA5 | RG5 | 1 |
| CA6 | RG5 | 5 |
| CA7 | RG6 | 3 |
+-----------------------+
ID='INFO1'
那么结果应该是这样显示的
Result
+-----------------------------------------------------------------------------------------------+
| ID | Name | Rooms | RoomGuests |
+-----------------------------------------------------------------------------------------------+
| INFO1 | ABC | 2 | [{"NoOfAdults" : "2", "NoOfChild" : "2", "ChildAge" : "[4,5]"}, |
| | | | {"NoOfAdults" : "3", "NoOfChild" : "", "ChildAge" : "[]"}] |
+-----------------------------------------------------------------------------------------------+
所有的结果都应该是这样的
Result
+-----------------------------------------------------------------------------------------------+
| ID | Name | Rooms | RoomGuests |
+-----------------------------------------------------------------------------------------------+
| INFO1 | ABC | 2 | [{"NoOfAdults" : "2", "NoOfChild" : "2", "ChildAge" : "[4,5]"}, |
| | | | {"NoOfAdults" : "3", "NoOfChild" : "", "ChildAge" : "[]"}] |
| | | | |
| INFO2 | DEF | 1 | [{"NoOfAdults" : "2", "NoOfChild" : "1", "ChildAge" : "[2]"}] |
| | | | |
| INFO3 | GHI | 3 | [{"NoOfAdults" : "2", "NoOfChild" : "1", "ChildAge" : "[7]"}, |
| | | | {"NoOfAdults" : "2", "NoOfChild" : "2", "ChildAge" : "[1,5]"}, |
| | | | {"NoOfAdults" : "2", "NoOfChild" : "1", "ChildAge" : "[3]"}] |
+-----------------------------------------------------------------------------------------------+
我尝试了下面的代码，但不起作用。我无法理解如何做
SELECT
S.`ID`, A.`Name`, A.`Rooms`,
CONCAT(
'[',
GROUP_CONCAT(
CONCAT(
'{
\"NoOfAdults\":\"', R.Adults,'\",
\"NoOfChild\":\"', R.Child,'\",
\"ChildAge\":
\"',
CONCAT(
'[',
GROUP_CONCAT(
CONCAT('{',C.Age,'}')
),
']'
),
,'\",
}'
)
),
']'
) AS RoomGuests,
FROM `Info_Table` AS I
LEFT JOIN `RoomGuests_Table` AS R ON R.`R_ID` = A.`ID`
LEFT JOIN `ChildAge_Table` AS C ON C.`R_ID` = R.`R_ID`
GROUP BY A.R_ID;
或者有没有最好的方法制作这样的阵列请告诉我
Array
(
[ID] => INFO1
[Name] => ABC
[Rooms] => 2
[RoomGuests] => Array
(
[0] => Array
(
[NoOfAdults] => 2
[NoOfChild] => 2
[ChildAge] => Array
(
[0] => 4
[1] => 5
)
)
[1] => Array
(
[NoOfAdults] => 3
[NoOfChild] => 0
[ChildAge] => Array
(
)
)
)
)
尽管我认为SGBD不应该参与数据的修饰，但您需要分两步来完成
首先，您需要执行一个group_concat来聚合子表数据。
然后，您将能够使用RoomGuests\u表和第一次查询的结果执行主组\u concat
如果您尝试只使用一个组进行一次查询，您将有重复的数据
例如：
SELECT ...
FROM info_table it
left outer join RoomGuests_Table rt on it.ID = rt.R_ID
left outer join
( SELECT ct.R_ID AS RG_ID, GROUP_CONCAT(ct.Age SEPARATOR ',') as concat_age
FROM ChildAge_Table as ct
GROUP BY ct.R_ID) tmp on tmp.RG_ID = rt.ID
GROUP BY ....
希望这有帮助
select i.name,i.rooms , group_concat(jsonres separator ',') from Info_Table i right join ( select r.r_id, CONCAT('[{"NoOfAdults":"', Adult,'","NoOfChild":"', Child,'","ChildAge":[',group_concat(C.Age separator ','),']"}') jsonres from test.RoomGuests_Table r left join test.ChildAge_Table c on r.ID=c.R_ID group by c.R_ID) as r on i.id= r.r_id group by i.name,i.rooms;
试着用这个
SELECT i.ID, i.name, i.rooms, RG.RoomGuests
FROM Info_Table i
LEFT JOIN (
SELECT
R.ID, R.R_ID AS RG_ID,
CONCAT(
'[',
GROUP_CONCAT(
CONCAT(
'{
\"NoOfAdults\":\"', Adult,'\",
\"NoOfChild\":\"', Child,'\",
\"ChildAge\":', IFNULL(CA.ChildAge, '[]'),'
}'
)
),
']'
) AS RoomGuests
FROM RoomGuests_Table R
LEFT JOIN (
SELECT
C.R_ID AS CA_ID,
CONCAT(
'[',
GROUP_CONCAT( Age SEPARATOR ','),
']'
) AS ChildAge
FROM ChildAge_Table C
GROUP BY CA_ID
) CA ON CA.CA_ID = R.ID
GROUP BY RG_ID
) RG ON RG.RG_ID = i.ID
WHERE i.ID = INFO1;
为什么你想用sql构建json格式？我已经编辑了帖子，请检查是否有其他方法忘记所有的CONCAT/GROUP\u CONCAT内容。只需返回一个有序数组并处理应用程序代码中的其余问题，因此您希望我为此使用一个循环是的，您是对的，我正在获取重复数据，因此如何做到这一点，请给我一个示例或一些链接，我仍在搜索我的答案，以提供最小的示例结构