Question

我有一个查询，结果如下：查询：

SELECT Tasks.TaskId, Comments.Comment, comments.timespent       
FROM   comments
INNER JOIN tasks ON comments.entityid = tasks.taskid                
WHERE  ( comments.entity = 1 ) 
GROUP  BY Tasks.TaskId, Comments.Comment, comments.timespent

结果：

TaskID  Comment  TimeSpent
__________________________
111754    C1        4
111754    C2        1
111754    C3       79

请告诉我如何编写查询以获得如下结果：

TaskID  Comment          TimeSpent
__________________________________
111754  ,C1,C2,C3           84

提前致谢。

Answer 1

这是工作的SQL小提琴：http://sqlfiddle.com/#!3/3597a/3

这是实际工作的SQL。

SELECT Tasks.TaskId, SUBSTRING(
(SELECT ',' + Comments.Comment
FROM Comments
INNER JOIN tasks ON comments.entityid = tasks.taskid
FOR XML PATH('')),2,200000) AS Comments
, SUM(comments.timespent) AS TimeSpent      
FROM   comments
INNER JOIN tasks ON comments.entityid = tasks.taskid                
WHERE  ( comments.entity = 1 ) 
GROUP  BY Tasks.TaskId

创建表格和填充数据

CREATE TABLE Tasks
(
  TaskID NVARCHAR(20) NOT NULL,
);

CREATE TABLE Comments
( 
  Entity  INT NOT NULL,
  EntityID NVARCHAR(20) NOT NULL,
  Comment NVARCHAR(50) NOT NULL,
  TimeSpent INT NOT NULL
);


INSERT INTO Tasks VALUES
( '111754' );

INSERT INTO Comments VALUES
(1,'111754', 'C1',4 ),
(1,'111754', 'C2',1 ),
(1,'111754', 'C3',79 );

执行SQL

SELECT Tasks.TaskId, SUBSTRING(
(SELECT ',' + Comments.Comment
FROM Comments
INNER JOIN tasks ON comments.entityid = tasks.taskid
FOR XML PATH('')),2,200000) AS Comments
, SUM(comments.timespent) AS TimeSpent     
FROM   comments
INNER JOIN tasks ON comments.entityid = tasks.taskid                
WHERE  comments.entity = 1 
GROUP  BY Tasks.TaskId

查看结果。

TASKID  COMMENTS    TIMESPENT
111754  C1,C2,C3    84

Answer 2

您应该查看FOR XML PATH。

Answer 3

您可以使用XML路径进行CROSS APPLY，例如：

   `
 Select *
   from table1 t1
     CROSS APPLY
      (
 SELECT
    [text()] = t.[Name] + '; '
    FROM table2 t2
    WHERE t1.[Id] = t2.[Id]
   ORDER BY t2.name
   FOR XML PATH('')
       ) a (Type)
`

Answer 4

好的，这有点复杂，但它没有使用xml，它可以用于除sql server之外的其他数据库：

    WITH orig
         AS (SELECT 1 AS f1, 'C11' AS f2
             UNION ALL
             SELECT 1 AS f1, 'C12' AS f2
             UNION ALL
             SELECT 1 AS f1, 'C13' AS f2
             UNION ALL
             SELECT 2 AS f1, 'C21' AS f2
             UNION ALL
             SELECT 2 AS f1, 'C22' AS f2
             UNION ALL
             SELECT 2 AS f1, 'C23' AS f2
             UNION ALL
             SELECT 3 AS f1, 'C31' AS f2)
       , orig2 AS (SELECT DISTINCT f1, f2 FROM orig)
       , orig3 AS (SELECT f1, f2, row_number() OVER(PARTITION BY f1 ORDER BY f2) AS RowNum FROM orig2)
       , orig4
         -- Use recursion to concatenate the fields
         AS (SELECT f1, CONVERT(VARCHAR(MAX), f2) AS val, rownum
               FROM orig3
              WHERE RowNum = 1
             UNION ALL
             SELECT orig4.f1, orig4.val + ', ' + orig3.f2 AS val, orig3.rownum
               FROM orig4
                    INNER JOIN orig3
                      ON orig4.RowNum + 1 = orig3.RowNum
                     AND orig4.f1 = orig3.f1)
    SELECT *
      FROM orig4
     -- select only the rows that match the maximum rownum
     WHERE NOT EXISTS
             (SELECT 1
                FROM orig4 o44
               WHERE o44.f1 = orig4.f1
                 AND o44.rownum > orig4.rownum)

另一种仅适用于sql server的方法是构建一个连接值的聚合CLR函数：http://msdn.microsoft.com/en-us/library/91e6taax%28v=vs.90%29.aspx。

如果你遇到过这篇文章但是你使用了oracle，你可以选择使用上面的查询或者在pl / sql（http://docs.oracle.com/cd/B28359_01/appdev.111/b28425/aggr_functions.htm）中定义自定义聚合函数。

Answer 5

您应该使用 STRING_AGG 函数。它是一个用于字符串的 sql 聚合器函数。我认为您的最终查询将是：

SELECT Tasks.TaskId, STRING_AGG(Comments.Comment, ', '), SUM(comments.timespent)       
FROM   comments
INNER JOIN tasks ON comments.entityid = tasks.taskid                
WHERE  ( comments.entity = 1 ) 
GROUP  BY Tasks.TaskId

如何在一个查询中连接一列中的多个行+内连接的字符串

5 个答案:

创建表格和填充数据

执行SQL

查看结果。