Question

我有一个字符串，如：

Games/Maps/MapsLevel1/Level 1.swf
Games/AnimalWorld/Animal1.1/Level 1.1.swf
Games/patterns and spatial understanding/Level 13.5/Level 13.5.swf

我想只获得没有扩展名（String After last Slash and before Last dot）的文件名，即Level 1和Level 1.1以及Level 13.5，甚至我想删除所有空格，最后的字符串应该是小写的，即最终的输出应该是

level1
level1.1
level13.5 and so on..

我尝试了以下查询，但我得到了Level 1.swf，我如何更改此查询？

SELECT SUBSTRING(vchServerPath, LEN(vchServerPath) - CHARINDEX('/', REVERSE(vchServerPath)) + 2, LEN(vchServerPath)) FROM Games

Answer 1

SELECT (left((Path), LEN(Path) - charindex('.', reverse(Path))))
FROM 
(
    SELECT SUBSTRING(vchServerPath, 
                     LEN(vchServerPath) - CHARINDEX('/', REVERSE(vchServerPath)) + 2, 
                     LEN(vchServerPath)) Path
    FROM Games
) A

这样可以工作，我保留了你的内部子串，让你分道扬and，我添加了点的剥离。

我已经包含了一个sql小提琴链接，供您查看动作sql fiddle

<强>编辑：以下将删除空格并返回小写...

SELECT REPLACE(LOWER((left((Path), LEN(Path) - charindex('.', reverse(Path))))), ' ', '')
FROM 
(
    SELECT SUBSTRING(vchServerPath, 
                     LEN(vchServerPath) - CHARINDEX('/', REVERSE(vchServerPath)) + 2, 
                     LEN(vchServerPath)) Path
    FROM Games
) A

Answer 2

试试这个：

select
 case 
 when vchServerPath is not null
 then reverse(replace(substring(reverse(vchServerPath),charindex('.',reverse(vchServerPath))+1, charindex('/',reverse(vchServerPath))-(charindex('.',reverse(vchServerPath))+1)),' ',''))
 else ''
end

Answer 3

这应该可以正常工作;删除了扩展名。

select 
REVERSE(
SUBSTRING(
reverse('Games/patterns and spatial understanding/Level 13.5/Level 13.5.swf'),
5,
(charindex('/',
reverse('Games/patterns and spatial understanding/Level 13.5/Level 13.5.swf')) - 5)
))

在SQL中获取两个字符之间的子字符串并删除所有空格

3 个答案: