如何从 SQL 选择中获取 JSON 数组

Question

我想在我的 SQL 选择中获取一个对象数组。

SELECT id, a.name antibodyName, c.name colorName, c.location colorLocation 
FROM Antibodies a 
JOIN AssignedColors c ON id = antiId 
WHERE colorId IS NOT NULL

我得到了这样的回应：

[
  { id: 1, antibodyName: 'antibody 1', colorName: 'red', colorLocation: 'A5/C3' },
  { id: 2, antibodyName: 'antibody 2', colorName: 'red', colorLocation: 'C3/A1' },
  { id: 2, antibodyName: 'antibody 2', colorName: 'yellow', colorLocation: 'E4/F2' }
]

有可能得到这样的东西吗？

[
  { id: 1, antibodyName: 'antibody 1', colors: [{name: 'red, location: 'A5/C3'}] },
  { id: 2, antibodyName: 'antibody 2', colors: [{name: 'red, location: 'C3/A1'}, {name: 'yellow', location: 'E4/F2'}] },
]

Answer 1

使用 json_build_object 和 array_agg。

WITH foobar AS ( 
    SELECT id, a.name antibodyName, c.name colorName, c.location colorLocation 
    FROM Antibodies a JOIN AssignedColors c
    ON id = antiId WHERE colorId IS NOT NULL
    )
SELECT 
     json_build_object('data',array_agg(json_build_object('id', id, 'antibodyName', antibodyName, 'colorName', colorName, 'colorLocation', colorLocation))
FROM 
    foobar

Answer 2

有点生疏，但试试这个：

SELECT
a.id AS 'Id',
a.Name AS 'antibodyName',
c.Name AS 'name',
c.location AS 'location'
FROM Antibodies a
LEFT JOIN AssignedColors c ON c.id = a.Id
FOR JSON AUTO;

更新：以上内容不适用于 SQLite。对于 SQLite，您应该在此处查看 json1 扩展：https://www.sqlite.org/json1.html

另一个更新： 它可以在没有 JSON 扩展的情况下完成...

SELECT
a.id AS 'id',
a.Name AS 'antibodyName',
(SELECT '[' || GROUP_CONCAT('{name:"' || Name || '", location:"' || location || '"}') ||']'
 FROM AssignedColors
 WHERE c.id=Id) AS colors
 FROM Antibodies a
 LEFT JOIN AssignedColors c ON c.id = a.id
 GROUP BY a.id

这里有一个小问题：颜色列/节点被视为字符串而不是数组...这是正常的，因为 sqlite 不支持数组类型的列。需要进行一些解析才能将字符串转换为数组...