topic Re: awk question in Operating System - HP-UX

awk question

Hunki — Thu, 28 Jun 2007 11:22:31 GMT

find . -type f

returns this :
./SECQUERYuninstall
./2007.02_uninstall
./2007.03_install
./2007.03_uninstall

and I want to remove the initial ./ from the output and tried this :

find . -type f |awk -F"./" '{print $NF}'

but it gave me the wrong output :

/SECQUERYuninstall
02_uninstall
03_install
03_uninstall

How shall i get the following output :
SECQUERYuninstall
2007.02_uninstall
2007.03_install
2007.03_uninstall

Re: awk question

Patrick Wallek — Thu, 28 Jun 2007 11:30:46 GMT

Just omit the '.' in your '-F "./"

# find . -type f | awk -F "/" '{print $NF}'

Re: awk question

A. Clay Stephenson — Thu, 28 Jun 2007 11:32:41 GMT

find . -type f | awk '{sub("^./","",$0); print $0}'

Re: awk question

Hunki — Thu, 28 Jun 2007 11:37:40 GMT

Thanks Patric , but a little change in requirement now :

find . -type f output is :

./shellscripts/A/SECQUERYuninstall
./shellscripts/A/B/2007.02_uninstall
./shellscripts/C/2007.03_install
./shellscripts/D/2007.03_uninstall

and I want :

shellscripts/A/SECQUERYuninstall
shellscripts/A/B/2007.02_uninstall
shellscripts/C/2007.03_install
shellscripts/D/2007.03_uninstall

Re: awk question

A. Clay Stephenson — Thu, 28 Jun 2007 11:41:28 GMT

find . -type f | awk -F "/" '{print $NF}'

This approach only works in a special case; you said that you want to remove the INITIAL ./ from the string; consider the case of directories below the CWD.

That is why sub("^./","",$0) is the better approach; it removes ONLY the initial ./ becase the search is anchored.

Re: awk question

James R. Ferguson — Thu, 28 Jun 2007 11:43:46 GMT

Hi Hunki:

# find . -type f | sed -e 's%^./%%'

Regards!

...JRF...

Re: awk question

Hunki — Thu, 28 Jun 2007 11:44:08 GMT

Thanks Clay but it gives me :

awk: syntax error near line 1
awk: illegal statement near line 1

Re: awk question

Patrick Wallek — Thu, 28 Jun 2007 11:48:00 GMT

I just cut-and-pasted Clay's solution into a terminal window and it worked fine for me.

Perhahps you forgot a " somewhere?

It would help to see the exact command line you used, not just the errors.

Re: awk question

A. Clay Stephenson — Thu, 28 Jun 2007 11:51:21 GMT

Then you have a personal problem because I just typed the command in and it produced exactly the desired output. Make sure that you do not confuse single and double quotes.

Re: awk question

Sandman! — Thu, 28 Jun 2007 11:57:23 GMT

Your original awk will work with a slight modification i.e.

# find . -type f |awk -F"^./" '{print $NF}'

~hope it helps

Re: awk question

Hunki — Thu, 28 Jun 2007 13:13:29 GMT

Thanks to Clay, JRF, Patrick.

Re: awk question

Dennis Handly — Thu, 28 Jun 2007 21:21:27 GMT

The correct solution was using sed to remove the leading "./".

>JRF: # find . -type f | sed -e 's%^./%%'

The "." needs to be quoted otherwise it matches "A/". But find probably won't return that.
# find . -type f | sed -e 's%^\./%%'

>Sandman: find . -type f |awk -F"^./" '{print $NF}'

This needs to quote the "." too.

Hmm, I had to do this to get it quoted if I left off the "^":
find . -type f | awk -F"\\\./" '{print $NF}'