Question

我有一个示例数据

     DECLARE @Table1 table ([name] varchar(62));

     INSERT INTO @Table1
            ([name])
        VALUES
            ('2018-08-08 23:02:57,731 INFO  [AllRequestInterceptor] CRTST020'),
            ('2018-08-08 23:03:11,687 INFO  [SOAPLoggingHandler] CRTST020'),
            ('2018-08-08 23:03:02,028 ERROR [AJAXController] CRTST003');

我正在尝试根据提供的空格创建4列。

     SELECT 
         Reverse(ParseName(Replace(Reverse([name]), ' ', '.'), 1)) As [M1]
         ,Reverse(ParseName(Replace(Reverse([name]), ' ', '.'), 2)) As [M2]
         ,Reverse(ParseName(Replace(Reverse([name]), ' ', '.'), 3)) As [M3]
         ,Reverse(ParseName(Replace(Reverse([name]), ' ', '.'), 4)) As [M4]

      FROM  (Select [name] from @Table1
      ) As [x]

预期输出：

Date                    Name    Req                     Code
8/8/2018 23:02:57,731   INFO    [AllRequestInterceptor] CRTST020
8/8/2018 23:03:11,687   INFO    [SOAPLoggingHandler]    CRTST020
8/8/2018 23:03:02,028   ERROR   [AJAXController]        CRTST003

Answer 1

这不是理想的选择，但是假设，唯一的空格将出现在date值中（而不是作为分隔符），您可以使用字符串分隔符和然后将数据旋转回去。使用delimitedsplit8k_LEAD，因为顺序位置很重要：

WITH CTE AS(
    SELECT T1.name,
           DS.ItemNumber,
           DS.Item,
           ROW_NUMBER() OVER (PARTITION BY T1.[name] ORDER BY DS.ItemNumber ASC) AS RN
    FROM @Table1 T1
         CROSS APPLY dbo.delimitedsplit8k_LEAD(T1.[name],' ') DS
    WHERE DS.Item <> '')
SELECT MAX(CASE WHEN RN = 1 THEN Item END) + ' ' + MAX(CASE WHEN RN = 2 THEN Item END) AS [Date],
       MAX(CASE WHEN RN = 3 THEN Item END) AS [Name],
       MAX(CASE WHEN RN = 4 THEN Item END) AS Req,
       MAX(CASE WHEN RN = 5 THEN Item END) AS Code
FROM CTE
GROUP BY [Name];

db<>fiddle

大声，如果您不能创建功能，则可以做错事（但很糟糕）：

DECLARE @Table1 table ([name] varchar(62));

INSERT INTO @Table1 ([name])
VALUES ('2018-08-08 23:02:57,731 INFO  [AllRequestInterceptor] CRTST020'),
       ('2018-08-08 23:03:11,687 INFO  [SOAPLoggingHandler] CRTST020'),
       ('2018-08-08 23:03:02,028 ERROR [AJAXController] CRTST003');

DECLARE @Delimiter char(1) = ' ';

WITH E1  (N) AS
    (SELECT 1
     UNION ALL
     SELECT 1
     UNION ALL
     SELECT 1
     UNION ALL
     SELECT 1
     UNION ALL
     SELECT 1
     UNION ALL
     SELECT 1
     UNION ALL
     SELECT 1
     UNION ALL
     SELECT 1
     UNION ALL
     SELECT 1
     UNION ALL
     SELECT 1),     --10E+1 or 10 rows
E2  (N) AS
    (SELECT 1
     FROM E1 AS a,
          E1 AS b), --10E+2 or 100 rows
E4  (N) AS
    (SELECT 1
     FROM E2 AS a,
          E2 AS b), --10E+4 or 10,000 rows max
cteTally (N) AS
    ( --==== This provides the "zero base" and limits the number of rows right up front
    -- for both a performance gain and prevention of accidental "overruns"
    SELECT 0
    UNION ALL
    SELECT TOP 62
           ROW_NUMBER() OVER (ORDER BY (SELECT NULL))
    FROM E4),
cteStart (N1, [name]) AS
    ( --==== This returns N+1 (starting position of each "element" just once for each delimiter)
    SELECT t.N + 1,
           T1.[name]
    FROM cteTally AS t
         CROSS JOIN @Table1 AS T1
    WHERE (SUBSTRING(T1.[name], t.N, 1) = @Delimiter
        OR t.N = 0)),
Splits AS
    (SELECT s.[name],
            ROW_NUMBER() OVER (ORDER BY s.N1) AS ItemNumber,
            SUBSTRING(s.[name], s.N1, ISNULL(NULLIF((LEAD(s.N1, 1, 1) OVER (PARTITION BY s.[name] ORDER BY s.N1) - 1), 0) - s.N1, 8000)) AS item
     FROM cteStart AS s),
CTE AS
    (SELECT name,
            ItemNumber,
            item,
            ROW_NUMBER() OVER (PARTITION BY [name] ORDER BY ItemNumber ASC) AS RN
     FROM Splits
     WHERE item <> '')
SELECT MAX(CASE WHEN RN = 1 THEN item END) + ' ' + MAX(CASE WHEN RN = 2 THEN item END) AS [Date],
       MAX(CASE WHEN RN = 3 THEN item END) AS [Name],
       MAX(CASE WHEN RN = 4 THEN item END) AS Req,
       MAX(CASE WHEN RN = 5 THEN item END) AS Code
FROM CTE
GROUP BY [name];

Answer 2

尝试一下

Date                     Name   Req                     Code
----------------------------------------------------------------
2018-08-08 23:02:57,731  INFO   [AllRequestInterceptor] CRTST020
2018-08-08 23:03:02,028  ERROR  [AJAXController]        CRTST003
2018-08-08 23:03:11,687  INFO   [SOAPLoggingHandler]    CRTST020

结果

document

Answer 3

您的数据几乎是固定长度的格式。对于示例数据，可以使用字符串函数：

select left(name, 23) as date,
       trim(substring(name, 25, 6)) as name,
       trim(substring(name, 31, len(name) - 39)) as req,
       right(name, 8) as code
from @table1 t1;

Here是db <>小提琴。

Answer 4

您可以为此使用多个apply：

select substring(name, 0, n) as [Date], substring(d, 0, d1) as Name, substring(d, d1, d2-d1+1) as Req, substring(d, d2+1, len(name)) as Code
from @Table1 t1 cross apply
     ( values (patindex('%[a-z]%', name)) 
     ) tt(n) cross apply
     ( values (substring(name, n, len(name)))
     ) ttt(d) cross apply
     ( values (charindex('[', d), charindex(']', d)) 
     ) tttt(d1, d2);

如何根据空格分割字符串？

4 个答案: