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SSIS Expression FileName

I’ve created a variable called strFileName as below with the following expression

substring(@[User::strFileName], 1, FINDSTRING(@[User::strFileName], "Test", 1)-2)
####_#########_#######_###_######_Test_18-1-2014.csv

Which works fine and returns the file name but the problem is the file name changes tomorrow it could be

####_#########_#######_###_######_Test_19-1-2014.csv or
####_#########_#######_###_######_Test_18-10-2014.csv

So what I need to do is extract the last part of the filename and add it to my expression.
If I get the total length of the filename, then from the left get the first 34 characters.
0
aneilg
Asked:
aneilg
1 Solution
 
fiboCommented:
- You have the filename in string
- You want to extract the part that starts at the 35th character (position 34 if starting at 0)

Is that correct?

If so, then I presume that
substring(@[User::strFileName], 35,length(@[User::strFileName]))
should do the trick (note that "extracting more characters" than there are in the string is OK and will return gracefully the expected result)
0
 
aneilgAuthor Commented:
thanks. i'll give it a go.
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