删除部分

Question

SELECT 
    right(name,7), 
    substring(params, charindex('|-|', params)+3,LEN(params)) as 'List Name', 
    convert(varchar,dateadd(hh,-8,created_date), 101) as Date, 
    convert(char, dateadd(hh,-8,created_date), 108) as Time 
FROM 
    [meldb].[dbo].[mr_message] 
WHERE 
    name in ('CL_LIST_STARTED', 'CL_LIST_STOPPED') 
    AND dateadd(hh,-8,created_date) > '7/1/2014' 
ORDER BY 
    created_date ASC 

列表名称将返回类似：

firstname.lastname-|LISTNAME|-|PARENTLISTNAME 

我想LISTNAME和PARENTLISTNAME隔离在单独的列，但由于它们可以在煤焦大小不等我可以”牛逼只需指定向左或向右

顺便说一句我没有创建使用它

任何想法，这个表我只是坚持？

Answer 1

你试过了吗？
好......高兴周五:)

declare @str varchar(100); 
set @str = 'jim.smith|-|firstItem|-|secondItem'; 



--- for your query, change @str to the column name, obviously --- 
select 
    substring(
        @str 
        , charindex('|-|', @str) + 3 
        , ((charindex('|-|', @str, charindex('|-|', @str) + 3)) - (charindex('|-|', @str) + 3)) 
       ) 
    ,substring(
        @str 
        , charindex('|-|', @str, charindex('|-|', @str) + 3) + 3 
        , len(@str) -- guaranteed to be past the end, to catch all 
       ) 

Answer 2

是否要将params分为三列？请检查以下查询。

SELECT 
    SUBSTRING(params, 1, CHARINDEX('-', params)-1) AS FullName, 
    SUBSTRING(STUFF(params, CHARINDEX('|-|', params), LEN(params), ''), CHARINDEX('-', params) + 2, LEN(params)) AS 'List Name', 
    SUBSTRING(params, CHARINDEX('|-|', params) + 3, LEN(params)) AS 'Parent List Name', 
    CONVERT(VARCHAR,DATEADD(hh,-8,created_date), 101) AS DATE, 
    CONVERT(CHAR, DATEADD(hh,-8,created_date), 108) AS TIME 
FROM 
    [meldb].[dbo].[mr_message] 
WHERE 
    name IN ('CL_LIST_STARTED', 'CL_LIST_STOPPED') 
    AND DATEADD(hh,-8,created_date) > '7/1/2014' 
ORDER BY 
    created_date ASC 

Answer 3

这里是我想出了使用PAT索引的格式和你提到的分隔符：

SELECT name, 
substring(name, 0, charindex('.', name)) as 'FirstName', 
substring(name, charindex('.', name) + 1, patindex('%|-|%', name) - patindex('%-|%', name) -2) as 'LastName', 
substring(name,patindex('%-|%', name)+2, patindex('%|-|%', name) - patindex('%-|%', name)-2) as 'ListName', 
substring(name, patindex('%|-|%', name)+3, len(name) - patindex('%|-|%', name)) as 'ParentListName' 
from FancyNames 

链接到SQL小提琴：http://sqlfiddle.com/#!6/03c2c/38