How to extract all characters after “/ ” from a column consisting varchar strings?-CodePudding

 Ex: Value= 'abc/efg'  
 Result=efg

"To be performed in MS SQL"

CodePudding user response：

there exist numerous approches to get your result such as

select  PARSENAME(REPLACE('abc/efg'  ,'/','.'),1)

select right('abc/efg', charindex('/', reverse('abc/efg')) - 1)

SELECT LTRIM(RIGHT('abc/efg', CHARINDEX('/', 'abc/efg') - 1))

SELECT SUBSTRING('abc/efg',CHARINDEX('/','abc/efg') 1,LEN('abc/efg'))

CodePudding user response：

Please see if this helps. CHARINDEX will get the position of specific characters. LEN will get variable length and RIGHT will get specified number of strings from the Right.

delcare @testvalue = 'abc/efg'
SELECT RIGHT(@testvalue, LEN(@testvalue) - CHARINDEX('/', @testvalue))

CodePudding user response：

If you have only one / you could use:

select  case when  CHARINDEX('/',column1 ) = 0 then column1 
            else right(column1, charindex('/', column1) - 1) end 
from table_name;

Demo

It takes the right string after the / even if you have more than one / , still it will take right string after the latest /.

CodePudding user response：

To make sure to pick up the last '/' in the string you could use the REVERSE function and then locate the starting point by subtracting from the LEN

select substring(column1, len(column1)-charindex('/', reverse(column1)) 2, charindex('/', reverse(column1)))
from table_name;

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