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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

Accepted Solution

by:
ozo earned 25 total points
ID: 17846196
awk -F: '{ print $6}'
0
 
LVL 23

Expert Comment

by:Mysidia
ID: 17846514
perl -e 'while(split(":",<>)) { print $_[5];}'
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LVL 23

Assisted Solution

by:Mysidia
Mysidia earned 25 total points
ID: 17846517
Err rather

perl -e 'while(split(":",<>)) { print $_[5]."\n";}'
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LVL 84

Expert Comment

by:ozo
ID: 17846565
perl -ne 'print ((split":")[5],"\n")'
perl -F: -lane 'print $F[5]'
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LVL 14

Assisted Solution

by:ygoutham
ygoutham earned 25 total points
ID: 17847048
cat /etc/passwd | cut -s -d':' -f6
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LVL 48

Assisted Solution

by:Tintin
Tintin earned 25 total points
ID: 17847340
awk -F: '/^user_A:/ {print $6}' /etc/passwd
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LVL 14

Expert Comment

by:ygoutham
ID: 17847688
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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LVL 84

Expert Comment

by:ozo
ID: 17847741
perl -F: -lane 'print $F[5] if /^user_A:/' /etc/passwd
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LVL 1

Assisted Solution

by:ltarc3
ltarc3 earned 25 total points
ID: 17858664
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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Assisted Solution

by:ctwaley
ctwaley earned 25 total points
ID: 17902752
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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LVL 48

Expert Comment

by:Tintin
ID: 17902951
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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