带有 Group By 子句的 SQL 逗号分隔行
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【中文标题】带有 Group By 子句的 SQL 逗号分隔行【英文标题】:SQL comma-separated row with Group By clause 【发布时间】:2011-11-18 21:49:11 【问题描述】:我有以下疑问:
SELECT
Account,
Unit,
SUM(state_fee),
Code
FROM tblMta
WHERE MTA.Id = '123'
GROUP BY Account,Unit
这当然会引发异常,因为代码不在group by
子句中。每个 state_fee 都有一个代码。如何让此代码显示在 1 条记录中(每个 state_fee 1 个代码,每个单元有多个 state_fee)作为逗号分隔的列表?我在这里查看了不同的解决方案,但找不到任何与 group by
一起使用的解决方案。
【问题讨论】:
【参考方案1】:你想使用FOR XML PATH
构造:
SELECT ACCOUNT,
unit,
SUM(state_fee),
Stuff((SELECT ', ' + code
FROM tblmta t2
WHERE t2.ACCOUNT = t1.ACCOUNT
AND t2.unit = t1.unit
AND t2.id = '123'
FOR XML PATH('')), 1, 2, '') [Codes]
FROM tblmta t1
WHERE t1.id = '123'
GROUP BY ACCOUNT,
unit
在此处查看其他示例:
SQL same unit between two tables needs order numbers in 1 cell SQL Query to get aggregated result in comma seperators along with group by column in SQL Server【讨论】:
【参考方案2】:没有用于连接的内置聚合函数,但本文讨论了几种替代解决方案,包括用户定义的连接聚合函数:
https://www.simple-talk.com/sql/t-sql-programming/concatenating-row-values-in-transact-sql/
【讨论】:
一开始我以为projectdms只是刮掉了没有署名的原始文章,但实际上该站点属于文章作者 链接已失效。这是另一种选择:simple-talk.com/sql/t-sql-programming/…【参考方案3】:这将显示表、索引名称、索引类型、索引列和包含的列:
with [indexes] (table_name, index_name, column_name, index_id, key_ordinal, object_id, type_desc)
as(
SELECT distinct
T.[name] AS [table_name], I.[name] AS [index_name],
AC.[name] AS [column_name],
I.[index_id], IC.[key_ordinal], T.[object_id], i.type_desc
FROM sys.[tables] AS T
INNER JOIN sys.[indexes] I ON T.[object_id] = I.[object_id]
INNER JOIN sys.[index_columns] IC ON I.[object_id] = IC.[object_id] and IC.index_id=I.index_id
LEFT OUTER JOIN sys.[all_columns] AC ON T.[object_id] = AC.[object_id] AND IC.[column_id] = AC.[column_id]
WHERE T.[is_ms_shipped] = 0 AND I.[type_desc] <> 'HEAP'
)
select
distinct
db_name() as dbname,
type_desc,
table_name,
index_name,
column_name,
STUFF((
select ', ' + column_name
from [indexes] t2
where t1.table_name=t2.table_name and t1.[index_name]=t2.[index_name] and t2.[key_ordinal] = 0
for xml path('')), 1, 2, '') inc_cols
from [indexes] t1
where t1.[key_ordinal] = 1
GROUP BY table_name, index_name, type_desc, column_name
【讨论】:
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