我有下面的表格，它们像这样相互连接
Info_Table -> RoomGuests_Table -> ChildAge_Table
这些是表格
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 |
+-----------------------+
我想要这样的结果
如果 Info_Table 的 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]"}] |
+-----------------------------------------------------------------------------------------------+
我试过下面的代码，但没有用。我无法理解该怎么做
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;
或者有什么最好的方法来制作这样的数组请告诉我
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
(
)
)
)
)
最佳答案
尝试使用这个
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;
关于mysql - 如何在mysql中进行嵌套选择查询，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/39226567/