如何使用 STRING_AGG 将一些值组合成一行?

How to use STRING_AGG to group some value into one line?

当所有列都相同但只有一列时,我想将几​​列合并为一列。这是一个例子:

ID  |  Name  | CP  | Job
-----------------------------
1   |  Muse  | 13  | Job1
1   |  Muse  | 13  | Job2
1   |  Muse  | 13  | Job3
2   |  Tort  | 51  | Job4
2   |  Tort  | 51  | Job5

我想要:

ID  |  Name  | CP  | Job
-----------------------------
1   |  Muse  | 13  | Job1, Job2, Job3
2   |  Tort  | 51  | Job4, Job5

我试过这样使用 STRING_AGG :

SELECT ID, Name, CP, STRING_AGG(Job, ',')
FROM myTable
GROUP BY ID, Name, CP, Job
ORDER BY ID

我读过这个:https://database.guide/how-to-return-query-results-as-a-comma-separated-list-in-sql-server/

感谢您的帮助

实际上,我只是自己发现的....我不需要按工作分组:

SELECT ID, Name, CP, STRING_AGG(Job, ',')
FROM myTable
GROUP BY ID, Name, CP
ORDER BY ID

只是一个有趣的观察和一些关于使用 STRING_AGG 的知识。如果您需要以任何特定顺序返回新工作列中的项目,您可以利用 WITHIN GROUP

-- Sample data with the inserts rearranged in random order
DECLARE @myTable TABLE 
(
  ID     INT,
  [Name] VARCHAR(100),
  CP     INT,
  Job    VARCHAR(100)
);

INSERT @myTable
VALUES
(1,'Muse',13,'Job3'),
(1,'Muse',13,'Job1'),
(1,'Muse',13,'Job2'),
(2,'Tort',51,'Job5'),
(2,'Tort',51,'Job4');

-- STRING_AGG ordering NOT guaranteed
SELECT ID, Name, CP, Job = STRING_AGG(Job, ',')
FROM @myTable
GROUP BY ID, Name, CP
ORDER BY ID;

-- STRING_AGG ordering IS guaranteed    
SELECT ID, Name, CP, job = STRING_AGG(Job, ',') WITHIN GROUP (ORDER BY Job)
FROM @myTable
GROUP BY ID, Name, CP
ORDER BY ID;

未排序 STRING_AGG 结果:

ID          Name        CP          Job
----------- ----------- ----------- ------------------
1           Muse        13          Job3,Job1,Job2
2           Tort        51          Job5,Job4

未排序 STRING_AGG 结果:

ID          Name        CP          job
----------- ----------- ----------- ------------------
1           Muse        13          Job1,Job2,Job3
2           Tort        51          Job4,Job5