awk question

custangro · 08-01-2008, 06:17 PM

Hello,

How can I get awk to print all fields except the first one...

For example...

I want this

Code:

/path/to/the/file

To output like this...

Code:

/to/the/file

Ive gotten this far

Code:

awk -F'/'

Sorry but I'm a newbie with awk and sed...I should also mention that I am using Sun Solaris (so I do not have the fancy functionality of gsed and gawk

)

Any help would be great...thanks

-C

jschiwal · 08-01-2008, 06:37 PM

You could either loop through 2 to NF printing out the indexed field, or delete $1.

Here is an example:

Code:

 find Documents/LinuxGazette/ | awk -F/ '{ for ( i=3; i<=NF; i++ ) \
{ printf "%s/", $i } print
}' | head
/Documents/LinuxGazette/
issue61/Documents/LinuxGazette/issue61
issue61/okopnik.html/Documents/LinuxGazette/issue61/okopnik.html
issue61/lg_tips61.html/Documents/LinuxGazette/issue61/lg_tips61.html
issue61/andreiana.html/Documents/LinuxGazette/issue61/andreiana.html
issue61/dellomodarme.html/Documents/LinuxGazette/issue61/dellomodarme.html
issue61/index.html/Documents/LinuxGazette/issue61/index.html
issue61/collinge.html/Documents/LinuxGazette/issue61/collinge.html
issue61/ward.html/Documents/LinuxGazette/issue61/ward.html
issue61/lg_backpage61.html/Documents/LinuxGazette/issue61/lg_backpage61.html

I hope this works with a generic awk.

Here I looped from the second to the last field printing out each field. I needed to add the "/" character because the -F/ option is using it as a field separator. After the loop I use a "print" for the new line character.

p.s. And if I used the "-type f" option to the find command, the input to awk would have matched the expected pattern.

custangro · 08-01-2008, 06:56 PM

Quote:

Originally Posted by jschiwal

You could either loop through 2 to NF printing out the indexed field, or delete $1.

Here is an example:

Code:

 find Documents/LinuxGazette/ | awk -F/ '{ for ( i=3; i<=NF; i++ ) \
{ printf "%s/", $i } print
}' | head
/Documents/LinuxGazette/
issue61/Documents/LinuxGazette/issue61
issue61/okopnik.html/Documents/LinuxGazette/issue61/okopnik.html
issue61/lg_tips61.html/Documents/LinuxGazette/issue61/lg_tips61.html
issue61/andreiana.html/Documents/LinuxGazette/issue61/andreiana.html
issue61/dellomodarme.html/Documents/LinuxGazette/issue61/dellomodarme.html
issue61/index.html/Documents/LinuxGazette/issue61/index.html
issue61/collinge.html/Documents/LinuxGazette/issue61/collinge.html
issue61/ward.html/Documents/LinuxGazette/issue61/ward.html
issue61/lg_backpage61.html/Documents/LinuxGazette/issue61/lg_backpage61.html

I hope this works with a generic awk.

Here I looped from the second to the last field printing out each field. I needed to add the "/" character because the -F/ option is using it as a field separator. After the loop I use a "print" for the new line character.

p.s. And if I used the "-type f" option to the find command, the input to awk would have matched the expected pattern.

That worked perfectly!

So basically you are looping from printing from field 2 to NF and adding / in the process right?

...I think I get it

-C

jschiwal · 08-01-2008, 07:06 PM

Here is the second method. The results are a little different. Each line starts with "/" but maybe that is what you want. One thing that needs to be done here is to restore "/" between fields with "OFS="/". Then printing $0 prints "/" between fields:

Code:

find Documents/LinuxGazette/ -type f | awk -F/ 'BEGIN {OFS="/"} { $1=""; print $0}'

The first record is still printed, it's just empty. So this only works if you want remove only the first record. A preceding / on each line will cause different results. In that case, you may need to use the first method. The first record $1 will be empty when it's read in, and in my first example, I would have needed to start with field 3 instead.

Looking back at my first example, while it works, I realize now that I should have expected each line to end with a "/" but using "print" instead of 'printf "\n"' swallowed up that trailing "/" character. I admit it, I got lucky!
---
To answer your question,
Yes, that's just what the first method does.

jschiwal · 08-01-2008, 07:28 PM

Here is a sed solution that removes the first part.

Code:

find ./Documents/LinuxGazette/ -type f | sed '/^\//s#^/[^/]*/##;/^\/!/s#^[^/]*/##;/^\.\//s#\.\/[^/]*/##'

Like most sed commands, it's easier to write than understand. Since "/" is normal separator in a sed command, I used "#" instead but was stuck with using "/" for the selection tests. I tested for 3 types of matches:
Documents/LinuxGazette/
/Documents/LinuxGazette/
./Documents/LinuxGazette/

The oneliner contains 3 sed commands, separated by a semi-colon. I don't know if this is a gnu extension or not.
/^\//s#^/[^/]*/##
/^\/!/s#^[^/]*/##
/^\.\//s#\.\/[^/]*/##

ghostdog74 · 08-01-2008, 08:39 PM

another way

Code:

# echo "/path/to/the/file" | awk '{match($0,"/[a-zA-Z0-9]+/");print substr($0,RLENGTH)}'
/to/the/file

chakka.lokesh · 08-03-2008, 11:51 PM

dear custangro

how about the following?

Code:

 sed s/^[^\/]*// inputfile