Shell Programming and Scripting

I've got a file that contains a large list of links in this type of style:

Code:

'home_dir\2009\09\01\file.html'

I'd like to cut off all of the characters left of 'file.html'. I tried:

Code:

 
cat file.txt | cut -d\ -f4

but it told me that I had an invalid delimiter.

So I tried:

Code:

cat file.txt | cut -d "\" -f4

and it presented me with a line prompt

Next I tried

Code:

 
cat file.txt | cut -d "/\" -f4

but I got the same thing.

Can anyone tell me what I'm doing wrong? Thanks

including or excluding quotes?
Did you try basename?

Code:

basename 'home_dir\2009\09\01\file.html'

---------- Post updated at 02:18 PM ---------- Previous update was at 02:00 PM ----------

Is basename working for you with backslash?

Or:

Code:

awk -F'\' '{print $5}' filename

cut -d\ -f4: In this case, the shell's parser eats the backslash. The error message you see is the result of cut not receiving an argument to the -d option. It is equivalent to: cut -d -f4

cut -d "\" -f4: You opened a double quoted string, within which \" is an escape sequence representing a literal double quote. The shell is prompting you for further input because you have an open string pending. If you closed it (with another double quote), it would call cut with the -d option being passed a long string as an argument (definitely not what you want, resulting in another 'bad delimiter' error).

cut -d "/\" -f4: Same problem as above, with a forward slash that does not affect the semantics explained in the previous case.

cut -d\\ -f4 or cut -d'\' is what you'd want to do to pass \ as the delimiter.

If the above explanations don't make much sense to you, review sh quoting rules in the man page. It may clear things up further.

Regards,
alister

yes it works, (I have cygwin on my home pc)

Code:

/home->basename 'home_dir\2009\09\01\file.html'
file.html
/home->