sql-server - 如何在 VARCHAR(MAX) 中的 SQL Server 中的两个已知字符串之间选择可变长度字符串，其中某些列没有适用的字符串

Question

使用 SQL Server 2012，我需要从这个示例中获取

ColumnName    
--------------------------------
 Enroll to: Carol Goals are many and varied  
 characters that don't include desired results
 Enroll to: Jan Levinson Goals will be discussed at first encounter
 Enroll to: Stephon-Anderson   Goals none
 NULL
 Enroll to:   David   Goals  --Note uneven spaces, Need David 

提取列看起来像：

Name    
-----------
Carol
NULL   
Jan Levinson   
Stephon-Anderson 
NULL 
David 

这段代码让我非常接近我正在寻找的结果，但有时会错误地修剪名称。

Select 
CASE WHEN AssignedTo like '%Enroll To:%' THEN  SUBSTRING(AssignedTo, CHARINDEX('%Enroll To:%', AssignedTo) + LEN('%Enroll To:%') 
            ,CHARINDEX('Goals', AssignedTo) - CHARINDEX('%Enroll To:%', AssignedTo) + LEN('Goals'))
          ELSE 'None'
          END AS 'Name'
FROM
(

  Select 
CASE WHEN ColumnName like '%Enroll To:%' THEN SUBSTRING    (ColumnName, CHARINDEX('Enroll To:', ColumnName), 40) 


ELSE 'None'
END AS 'AssignedTo'

FROM TABLE ) A

我不知道该怎么感谢你才足够！

Answer 1

您可以使用apply和字符串函数：

select left(v.s1, charindex(' ', s1) - 1)
from t cross apply
     (values (stuff(t.col, 1, 11, '')) v(s1) 

Answer 2

这是戈登答案的替代方案：

SELECT
    SUBSTRING(ColumnName,
              CHARINDEX(':', ColumnName) + 2,
              CHARINDEX(' ', ColumnName, CHARINDEX(':', ColumnName) + 2) -
                  CHARINDEX(':', ColumnName) - 2) AS Name
FROM yourTable;

演示

Answer 3

这产生了预期的结果，并且似乎处理了目标字符串的可变长度。希望它可以帮助某人。

DECLARE @pretext as NVARCHAR(100) = 'Enroll to:' 
DECLARE @posttext as NVARCHAR(100) = 'Goals'

Select 
,CASE When CHARINDEX(@posttext, ColumnName) - (CHARINDEX(@pretext, ColumnName) + len(@pretext)) < 0 THEN NULL
    Else
    SUBSTRING(ColumnName, CHARINDEX(@pretext, ColumnName) + len(@pretext)
    ,CHARINDEX(@posttext, ColumnName) - (CHARINDEX(@pretext, ColumnName) + len(@pretext)) )    
    END as betweentext 

FROM TABLE 

Answer 4

这是您要以表格形式测试的数据：

declare @goals table (string nvarchar(255));
insert @goals values 
    ('Enroll to: Carol Goals are many and varied  characters that don''t include desired results'),
    ('Enroll to: Jan Levinson Goals will be discussed at first encounter'),
    ('Enroll to: Stephon-Anderson   Goals none'),
    (NULL),
    ('Enroll to:   David   Goals  '), --Note uneven spaces, Need David 
    (' '); -- I (psw) added this

以下代码似乎可以毫无错误地执行您想要的操作。但它假设你的名字后面的句子总是以“目标”开头。

select      *,
            result = 
                case 
                when isValid = 1 then 
                    ltrim(rtrim(
                        substring(string, colonPos + 1, goalsPos - colonPos - 1)
                    ))
                end

from        @goals
cross apply (select 
                colonPos = charindex(':', string),
                goalsPos = patIndex('%goals%', string)
            ) positions
cross apply (select
                isValid = 
                    case 
                    when colonPos = 0 or goalsPos = 0 or colonPos > goalsPos then 0
                    else 1
                    end
            ) validity