Tra*_*oki 6 sql postgresql concat concatenation subquery
我希望“user_names”成为用户列表,但我无法将多个子查询项分配给“user_names”。如果子查询只返回 1 个项目,则它起作用,但如果它返回多个则不起作用。
床桌
[id]
[1]
[2]
[3]
分配表:
[id, bed_id, user_id]
[1, 1, 1]
[2, 1, 2]
用户表:
[id, 'user_name']
[1, 'John Smith']
[2, 'Jane Doe']
sql = "SELECT
b.id,
(
SELECT
u.user_name
FROM
assignments AS a
INNER JOIN
users as u
ON
a.user_id = u.id
WHERE a.bed_id = b.id
) AS user_names
FROM beds AS b"
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期望的结果是:
[1, 'John Smith, Jane Doe']
[2, '']
[3, '']
我尝试对床位 ID 进行硬编码并运行此段以获取名称列表。它没有用:
sql = """
(
SELECT
array_agg(user_name)
FROM
roomchoice_assignment AS a
INNER JOIN
roomchoice_customuser as u
ON
a.user_id = u.id
WHERE
a.bed_id = 1
GROUP BY user_name
)"""
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它返回以下内容:
[
[
[
"John Smith"
]
],
[
[
"Jane Doe"
]
]
]
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我希望这样:
['John Smith, Jane Doe']
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您所拥有的查询的一个问题是您正在按应用 array_agg 的列进行分组。如果您删除组,您会得到"{"John Smith","Jane Doe"}",但您仍然会缺少床位 id 列,并且如果您想要所有床位的列表,即使没有分配,您也应该使用左连接而不是子查询(这也应该更好性能和可读性)。
您可以string_agg按照重复问题的指示使用。
这个查询:
SELECT b.id, string_agg(u.user_name, ', ') users
FROM beds AS b
LEFT JOIN assignment AS a ON a.bed_id = b.id
LEFT JOIN users as u ON a.user_id = u.id
GROUP by b.id;
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会给你一个结果:
1;"John Smith, Jane Doe"
2;""
3;""
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