Question

我在sql中有数据（只需注意：SQL STudio是IDE），如：

data
a_10_b_c
a_1_b_c

我想获取前两个符号_之间的数据：

Output
10
1

Answer 1

这将是我的方法：

SELECT CAST('<x>' + REPLACE(data,'_','</x><x>') + '</x>' AS XML).value('/x[2]','int')
FROM YourTable

首先，将其转换为XML，然后选择第二个节点..

编辑：此方法有用的更多示例：

交叉申请：您可以使用此方法一次获得多个令牌

DECLARE @tbl TABLE(separated VARCHAR(100));
INSERT INTO @tbl VALUES('1_23:50_Look_this_is_a_test'),('2_12:00_that''s_one_more_test'),('3_13:30_great!_It_works!');

SELECT Converted.value('/x[1]','int') AS number
      ,Converted.value('/x[2]','time') AS time
      ,Converted.value('/x[3]','varchar(max)') AS text
FROM @tbl
CROSS APPLY(SELECT CAST('<x>' + REPLACE(separated,'_','</x><x>') + '</x>' AS XML) AS Converted) AS MySeparated
--type-safe and easy:
/*
number  time    text
1       23:50   Look
2       12:00   that's
3       13:30   great!
*/
GO

CTE：用作参数

DECLARE @Parameter VARCHAR(100)='1_12:30_SomeValue';
WITH MyParameters AS
(
    SELECT CAST('<x>' + REPLACE(@Parameter,'_','</x><x>') + '</x>' AS XML).value('/x[1]','int') AS IntParam
          ,CAST('<x>' + REPLACE(@Parameter,'_','</x><x>') + '</x>' AS XML).value('/x[2]','time') AS TimeParam
          ,CAST('<x>' + REPLACE(@Parameter,'_','</x><x>') + '</x>' AS XML).value('/x[3]','varchar(max)') AS TextParam
)
SELECT IntParam,TimeParam,TextParam
FROM MyParameters
/*
IntParam    TimeParam   TextParam
1           12:30:00    SomeValue
*/
GO

拆分字符串：转换为列表

DECLARE @MyIDs VARCHAR(100)='3,5,7';
SELECT A.B.value('.','int') TheIntValue
FROM(SELECT CAST('<x>' + REPLACE(@MyIDs,',','</x><x>') + '</x>' AS XML) AS MyListAsXML) AS x
CROSS APPLY MyListAsXML.nodes('/x') AS A(B)

/*
TheIntValue
3
5
7
*/
GO

动态IN语句

DECLARE @tbl TABLE(ID INT,Content VARCHAR(max));
INSERT INTO @tbl VALUES(1,'Value 1'),(2,'Value 2'),(3,'Value 3'),(4,'Value 4'),(5,'Value 5'),(6,'Value 6'),(7,'Value 7');

DECLARE @MyIDs VARCHAR(100)='3,5,7';
/*
This won't work (due to the fact, that @MyIDs is not a list of INTs but a text
SELECT * FROM @tbl WHERE ID IN(@MyIDs)
*/
WITH AsList AS
(
    SELECT A.B.value('.','int') TheIntValue
    FROM(SELECT CAST('<x>' + REPLACE(@MyIDs,',','</x><x>') + '</x>' AS XML) AS MyListAsXML) AS x
    CROSS APPLY MyListAsXML.nodes('/x') AS A(B)

)
SELECT * FROM @tbl WHERE ID IN(SELECT TheIntValue FROM AsList)

/*
ID  Content
3   Value 3
5   Value 5
7   Value 7
*/

Answer 2

您可以使用嵌套字符串函数执行此操作。通常，使用outer apply：

会更简单

select t3.output
from t outer apply
     (select stuff(t.col, 1, charindex('_', t.col), '') as col2
     ) t2 outer apply
     (select left(t2.col2, charindex('_', t2.col2)) as output
     ) t3;

获取SQL

2 个答案: