Hi Gurus,
I need your help again...

I want to replace second col of each line of passwd file with word
TEST. I tried it in for loop but takes too long for a file with 10000
or more lines.

Here is my test file:

test1:x:Readres Test:/tmp:/bin/false

What i am trying is:

/usr/xpg4/bin/sed 's/:x:/:TEST:/g' passwd > $DEST_DIR/passwd.new

but there might me more instances of :x: so i want to use combination
of awk and sed but not in for loop.
Thanks

jürgen_kahrs

awk -F: '{$2="TEST"}1' /etc/passwd

chris f.a. johnson

awk -F: -v OFS=: '{$2 = "TEST"; print }' /etc/passwd --
Chris F.A. Johnson, author <http://cfaj.freeshell.org>
Shell Scripting Recipes: A Problem-Solution Approach (2005, Apress)
===== My code in this post, if any, assumes the POSIX locale
===== and is released under the GNU General Public Licence

Thanks a lot Chris. It worked. The only problem was root. I don't want
root user to be changed. How do i skip root for replacing this?

gazelle

!/^root/ { $2 = "TEST" }
19

Thanks but it completely removes root entry. I don't want to replace
root's second col. but still want to print as is.

jürgen_kahrs

awk -F: -v OFS=: '!/^root/{$2 = "TEST"}1' /etc/passwd

chris f.a. johnson

[ Please don't top post ]

awk -F: -v OFS=: '!/^root:/ {$2 = "TEST"}{ print }' /etc/passwd --
Chris F.A. Johnson, author <http://cfaj.freeshell.org>
Shell Scripting Recipes: A Problem-Solution Approach (2005, Apress)
===== My code in this post, if any, assumes the POSIX locale
===== and is released under the GNU General Public Licence

Thanks Chris It worked. Thanks a lot

gazelle

No, it doesn't. Look more carefully.

BTW, it should be:

!/^root:/ { $2 = "TEST" }
19