Question Re: Parse file name and path from full path in Technology Q&A

Parse file name and path from full path

Former Member — Mon, 18 Jan 2016 11:20:37 GMT

This is how I do it today, but I suspect there are better ways to do it?

BEGIN declare ls_path nvarchar(200); set ls_path = 'C:\\tempfolder\\test\\testfile.txt'; select left(ls_path, len(ls_path) -len(row_value) ) cc_path, row_value cc_filename from sa_split_list(ls_path, '\\') where row_value <> '' and line_num = (select max(line_num) from sa_split_list(ls_path, '\\') ); END

Re: Parse file name and path from full path

MarkCulp — Mon, 18 Jan 2016 11:44:15 GMT

I would think it is easier to use locate( ..., '\\\\', -1 ) to find the last backslash and then split the string at that location.

begin declare @path varchar(255) = 'c:\\\\my\\\\path\\\\name\\\\filename.jpg'; select locate( @path, '\\\\', -1 ) as len, left( @path, len-1 ) as path, substr( @path, len+1 ) as filename; end;

Note that backslashes should be doubled ... otherwise sequences like \\n will get interpreted as newline (similar for \\r and \\t and a few others)

HTH

Re: Parse file name and path from full path

VolkerBarth — Mon, 18 Jan 2016 11:46:18 GMT

Wow, was just about to post something comparable but had to lookup the third LOCATE() parameter, as usual:)