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extracting a path from a string

Posted on 2006-10-31
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Last Modified: 2010-04-20

Hello,

I want to extract a portion of a string. The string is a path which beloings to an entry in /etc/passwd file

Let's assume the line is:

user_A:x:100:0:ABC User Account:/Dir1/Dir2/Dir3:/bin/sh

I did:

      cat /etc/passed | grep user_A  | awk '{ print $3}'  

this prints:  Account:/Dir1/Dir2/Dir3:/bin/sh

but I need to extract "/Dir1/Dir2/Dir3/" only, neither before it nor after it.

Any suggestions?
Thanks
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Question by:akohan
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14 Comments
 
LVL 84

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by:
ozo earned 25 total points
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awk -F: '{ print $6}'
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by:Mysidia
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perl -e 'while(split(":",<>)) { print $_[5];}'
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by:Mysidia
Mysidia earned 25 total points
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Err rather

perl -e 'while(split(":",<>)) { print $_[5]."\n";}'
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by:ozo
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perl -ne 'print ((split":")[5],"\n")'
perl -F: -lane 'print $F[5]'
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by:ygoutham
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cat /etc/passwd | cut -s -d':' -f6
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by:Tintin
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awk -F: '/^user_A:/ {print $6}' /etc/passwd
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by:ygoutham
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cat /etc/passwd | grep 'user_A' | cut -s -d':' -f6

the earlier would give all the home directories of all users.  this one gives only specific user. can be put in a script and a parameter can also be given as $1 for any specific user
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by:ozo
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perl -F: -lane 'print $F[5] if /^user_A:/' /etc/passwd
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by:ltarc3
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Look at all these solutions!  Perl is overkill it seems to me, awk is your solution.  But generally you use what you are most comfortable with.  You almost had, just needed help with the syntax.  Ozo had it right with the first comment and gave you a nice Perl alternative, as well. Also, note that you want the 6th field for the user directory, not the third.

You can grep a file directly, you do not need to cat and then pipe to it:

grep ^user_A /etc/passwd | awk -F: {print $6}

The -F: tells awk that the fields are separated by colons.

And just as you do not need to cat and then pipe to grep, you do not even need to grep and then pipe to awk.  Awk can work directly on the file and do the grepping itself:

awk -F: '/^user_A/ {print $6}' /etc/passwd

This tells awk that the field separator is a colon, match lines beginning with user_A, print the 6th field, operate on /etc/passwd.  One command.   Nice, clean, concise.
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by:ctwaley
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One more to the mix:

cut -d: -f6 <(grep 'user_A' /etc/passwd)

(basically the same as ygoutham's post above, w/o the 'cat' command)
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by:Tintin
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If you are using grep/cut, the more usual way is

grep 'user_A' /etc/passwd|cut -d: -f6

But note that will match *any field in /etc/passwd that *contain* user_A, eg:

Myuser_ABC

The awk solution is the easiest and most sensible.
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