asked on

finding text after /

Want to extract last part of string following last forward slash

eg
Have a txt file with say the following
/fahah/rjyjke/23454/test.rpt
/agddddddddd/ff/ff/ff/ff/ff/ff/fred.rpt
/a/test4.rpt
etc etc
I want to return ..
test.rpt
fred.rpt
test4.rpt

Regards

farzanj

Try:

line = "/fahah/rjyjke/23454/test.rpt"
re = /.*\/([^/]+)$/

match1 = line.match re

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PHIL Sawyer

ASKER

Ruby does not like this - it throws error
syntax error, unexpected ')'
re = /.*\/([^]/+)$/

Regards

Gertone (Geert Bormans)

ln = "/fahah/rjyjke/23454/test.rpt"
regex = /.*\/([^\/]+)$/
puts ln.gsub(regex, '\1')

Gertone (Geert Bormans)

hi Phil,
there is a little typo in farzanj's post
the / inside the [] needs escaping

SOLUTION

farzanj

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PHIL Sawyer

ASKER

OK - Yes I saw the typo and changed my logic to ..

list.each do |z|
z=z.scan(/.*\/([^\/]+)$/)
puts z
end

Thanks Guys

Hi Gertone

farzanj

@Gertone

That typo is not in my post. It is in Author's post.

Gertone (Geert Bormans)

and the line.match re will return the whole match, including the part matched by .*\/
what you would need to do then is this

puts line.match(re)[1]

PHIL Sawyer

ASKER

Gertone
Is "puts ln.gsub(regex, '\1') " bringing back last match - is that what the '\1' is doing?

Phil

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Gertone (Geert Bormans)

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farzanj

Hi Gertone:

See my second post. I never used Ruby. I do regex only.

Gertone (Geert Bormans)

Is "puts ln.gsub(regex, '\1') " bringing back last match - is that what the '\1' is doing?

it is substituting the line with the part that is in the first pair of brackets
there is only one pair of brackets, so it is returning that

ln.match(regex)[1] is returning the same
ln.match regex
equals
ln.match(regex)[0], returns the whole match

PHIL Sawyer

ASKER

Gertone
Just missed you post earlier which explains my last post

Phil

PHIL Sawyer

ASKER

Thanks

Gertone (Geert Bormans)

welcome