Question

列名主题

Value = Something TEST001 something

我需要从“Something TEST001 Something”获得TEST001

如果在TEST001的1之后有空格或任何特殊字符，它们将被删除。

我只有这个

SELECT   
    Subject
    , REPLACE(SUBSTRING(MailSubject, CHARINDEX('TEST', MailSubject), LEN(MailSubject)),'', '') AS Assingment
FROM    
    AssingmentEmail

来自TEST001的数字可以更多，但如果有任何空格或非数字，它将被删除。

Answer 1

基于评论，这里是编辑的解决方案：

DECLARE @t TABLE(v VARCHAR(100))
INSERT INTO @t VALUES
('Something TEST001 something'),
('Something [TEST0001] something'),
('Something TEST001 something 123 something'),
('Something {TEST0001} something 123 something')

;WITH cte AS(SELECT SUBSTRING(v, CHARINDEX('TEST', v), LEN(v)) AS v FROM @t)
SELECT SUBSTRING(v, 1, PATINDEX('%[^TEST0-9]%', v) - 1) AS v FROM cte

输出：

v
TEST001
TEST0001
TEST001
TEST0001

说明：

在cte中，您正在选择子字符串：

TEST001 something
TEST0001] something
TEST001 something 123 something
TEST0001} something 123 something

然后，您正在搜索第一次出现的带有T E S T 0 1 2 3...9 %[^TEST0-9]%的符号，并获得子串直到该符号。

Answer 2

我会使用子查询来完成此操作。处理中间结果更容易。首先，在＆＃34; TEST＆＃34;之前摆脱该部分。然后，将结果放在第一个第一个空格中：

select left(ms1, charindex(' ', ms1)) as TestStuff
from (select stuff(MailSubject, 1, charindex(' TEST', MailSubject), '') as ms1
      from assignment email
     ) ae

Answer 3

结合使用SUBSTRING，CHARINDEX＆amp; LEN。无论TESTXXXX中有多少位数，它都可以工作。

SELECT MailSubject,
LEFT(SUBSTRING([MailSubject], CHARINDEX('TEST', [MailSubject]), LEN([MailSubject])), CHARINDEX(' ', SUBSTRING([MailSubject], 
CHARINDEX('TEST', [MailSubject]), LEN([MailSubject]))) - 1) AS [MailSubject]
FROM AssingmentEmail

INPUT：

MailSubject
something TEST001 something 
something TEST201 something
something TEST20122 something
something TEST601 something

输出：

MailSubject                      MailSubject
something TEST001 something      TEST001
something TEST201 something      TEST201
something TEST20122 something    TEST20122
something TEST601 something      TEST601

SQL FIDDLE：http://sqlfiddle.com/#!3/93660d/1/0

如何从sql中的列中获取某个字符串

3 个答案: