MIN MAX和所有值以逗号分隔SQL Server

时间:2017-03-22 10:19:18

标签: sql-server pivot

我遇到以下问题:我有一张包含员工和电子邮件的表格。

  • employee1收到7封电子邮件
  • employee2收到4封电子邮件

现在我需要员工以逗号分隔的格式在一列中显示电子邮件的最长日期,最短日期和所有收到的邮件。

我有以下内容:

SELECT ... 
FROM
   (SELECT 
        MAX(Datum) as MDatum, MIN(Datum) as MinDatum,
        Sender, Betreff,
        CASE 
           WHEN MIN(Datum) = MAX(Datum) 
              THEN MIN(Datum) 
              ELSE dbo.explode(datum) <- my problem
        END AS VALUE,
        Nummer
    FROM 
        #NEWTABLE 
    WHERE 
        not nummer IS NULL
        AND Sender <> ''total'' 
    GROUP BY 
        Sender, Betreff, Nummer, datum <- second problem
         ) x 
PIVOT 
    (max(VALUE)
         FOR [Nummer] IN (' + @cols2 + ')
    ) p

@cols2是充满活力的员工

当我按基准分组时,我只获得数据中的最大值(值)

我怎么能解决这个问题?

最诚挚的问候,坦率地说

增加:

这是我的表(#NEWTABLE):enter image description here

Nummer |    Betreff |             Datum |
----   | ----       | ----              |
1       2.000 EUR Rente   06.03.2017
1       2.000 EUR Rente   NULL
1       2.000 Rente       04.03.2017
1       2.000 Rente       12.02.2017
1       2.000 Rente       12.03.2017
1       2.000 Rente       19.02.2017
1       2.000 Rente       19.03.2017
1       2.000 Rente       27.02.2017
1       2.000 Rente       NULL
1       2.000 Rente       27.02.2017
1       2.000 Rente       NULL
2       2.000 EUR Rente   06.03.2017
2       2.000 EUR Rente   NULL
2       2.000 Rente       04.03.2017
2       2.000 Rente       12.02.2017
2       2.000 Rente       12.03.2017
2       2.000 Rente       19.02.2017
2       2.000 Rente       19.03.2017
2       2.000 Rente       27.02.2017
2       2.000 Rente       NULL
2       2.000 Rente       27.02.2017
2       2.000 Rente       NULL

所以你有nummer,即employee,betreff和datum,收到的日期

预期输出应为:enter image description here

nummer  mindate     maxdate     alldates
1       12.02.2017  19.03.2017  12.02.2017,
                                19.02.2017,
                                27.02.2017,
                                27.02.2017,
                                04.03.2017,
                                06.03.2017,
                                12.03.2017,
                                19.03.2017
2       12.02.2017  19.03.2017  12.02.2017,
                                19.02.2017,
                                27.02.2017,
                                27.02.2017,
                                04.03.2017,
                                06.03.2017,
                                12.03.2017,
                                19.03.2017
最诚挚的问候,坦率地说

4 个答案:

答案 0 :(得分:2)

以下是FOR XML PATH('')的解决方案:

SELECT Nummer,min(Datum) miDat, max(Datum) maDat,
       STUFF( (SELECT ','+char(10) + CONVERT(char(10),Datum,104) FROM #tmp 
               WHERE NOT Datum is null AND Nummer=o.Nummer
               ORDER BY Datum
               FOR XML PATH('')), 1, 2, '' ) as dates 
FROM #tmp o GROUP BY Nummer

STUFF()函数只删除前两个字符。

可在此处找到工作演示: http://rextester.com/PUG64308

(我稍微改变了样本数据,以便得到两个数字略有不同的结果。)

修改:您可以通过以下方式修改查询来轻松实现更改后的结果:

SELECT min(Datum) miDat, max(Datum) maDat,
       STUFF( (SELECT ','+char(10) + CONVERT(char(10),Datum,104) FROM #tmp 
               WHERE NOT Datum is null AND Nummer=1
               ORDER BY Datum
               FOR XML PATH('')), 1, 2, '' ) as dates1,
       STUFF( (SELECT ','+char(10) + CONVERT(char(10),Datum,104) FROM #tmp 
               WHERE NOT Datum is null AND Nummer=2
               ORDER BY Datum
               FOR XML PATH('')), 1, 2, '' ) as dates2
FROM #tmp

虽然在我看来,“Sender”和Betreff这两个专栏“在这样的结果场景中不再有意义。”

可在此处找到一个工作示例:http://rextester.com/EATE35782

答案 1 :(得分:2)

希望这有帮助

;WITH cte_Table (Nummer,Betreff,Datum) AS
(
SELECT 1,'2.000 EUR Rente',CAST('2017-03-07' AS DATE) UNION ALL
SELECT 1,'2.000 EUR Rente',NULL UNION ALL
SELECT 1,'2.000 Rente','2017-03-04' UNION ALL
SELECT 1,'2.000 Rente','2017-02-12' UNION ALL
SELECT 1,'2.000 Rente','2017-03-12' UNION ALL
SELECT 1,'2.000 Rente','2017-02-19' UNION ALL
SELECT 1,'2.000 Rente','2017-03-19' UNION ALL
SELECT 1,'2.000 Rente','2017-02-27' UNION ALL
SELECT 1,'2.000 Rente',NULL UNION ALL
SELECT 1,'2.000 Rente','2017-02-27' UNION ALL
SELECT 1,'2.000 Rente',NULL UNION ALL
SELECT 2,'2.000 EUR Rente','2017-03-06' UNION ALL
SELECT 2,'2.000 EUR Rente',NULL UNION ALL
SELECT 2,'2.000 Rente','2017-03-04' UNION ALL
SELECT 2,'2.000 Rente','2017-02-12' UNION ALL
SELECT 2,'2.000 Rente','2017-03-12' UNION ALL
SELECT 2,'2.000 Rente','2017-02-19' UNION ALL
SELECT 2,'2.000 Rente','2017-03-19' UNION ALL
SELECT 2,'2.000 Rente','2017-02-27' UNION ALL
SELECT 2,'2.000 Rente',NULL UNION ALL
SELECT 2,'2.000 Rente','2017-02-27' UNION ALL
SELECT 2,'2.000 Rente',NULL
)
,cte_MinMax(nummer,mindate,maxdate) AS
(
SELECT Nummer,MIN(Datum),MAX(Datum)
FROM cte_Table
GROUP BY Nummer
)
,cte_CSV AS
(
SELECT DISTINCT a.Nummer, STUFF((    SELECT ', ' + CONVERT(VARCHAR(10),b.Datum,104) AS [text()]
                        FROM cte_Table b
                        WHERE a.Nummer = b.Nummer
                        FOR XML PATH('')), 1, 1, '' ) AS alldates
FROM cte_Table a
)
SELECT b.nummer,b.mindate,b.maxdate,a.alldates 
FROM cte_CSV a
JOIN cte_MinMax b
ON a.Nummer = b.nummer

答案 2 :(得分:1)

它就像一个魅力...... 但我犯了一个错误,对不起,预期的输出应该是这样的(带有支点) enter image description here

sender | betreff | mindate      | maxdate     | 1             | 2
blubb  | blah    | 12.02.2017   | 19.03.2017  | 12.02.2017,   | 12.02.2017,
                                                19.02.2017,     19.02.2017,
                                                27.02.2017,     27.02.2017,
                                                27.02.2017,     27.02.2017,
                                                04.03.2017,     04.03.2017,
                                                06.03.2017,     06.03.2017,
                                                12.03.2017,     12.03.2017,
                                                19.03.2017      19.03.2017
最诚挚的问候,坦率地说

答案 3 :(得分:0)

我明白了:

SELECT ... 
FROM
   (SELECT sender,Nummer,min(Datum) MinDatum, max(Datum) MDatum,betreff,
                    STUFF( (SELECT '',''+char(10) + CONVERT(char(10),Datum,104) FROM #NEWTABLE 
                    WHERE NOT Datum is null AND Nummer=o.Nummer and sender = o.sender and betreff = o.betreff
                    group by Datum,sender,Nummer,betreff
                    ORDER BY Datum
                    FOR XML PATH('''')), 1, 2, '''' ) as VALUE 
                    FROM #NEWTABLE o where sender <> ''total'' GROUP BY sender,Nummer,betreff
         ) x 
PIVOT 
    (max(VALUE)
         FOR [Nummer] IN (' + @cols2 + ')
    ) p
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