Search and replace in unix(Replace colon with space in /etc/passwd)

[Sammy_2]

How do you replace :(colon) with a space in the example below , using cut or awk, etc?
Should look like:
sam 111
dave 232

==========unix command below=======
# cut -d: -f1,3 /etc/passwd > output
sam:111
dave:232

==============
Not modifying /etc/passwd . Just sending the output to another file

[Geoff Wild · Accepted solution]

cut -d: -f1,3 /etc/passwd |sed 's/:/ /' > output



Rgds...Geoff

[Chris Watkins_1]

And the awk way:

# cut -d: -f1,3 /etc/passwd |awk -F: '{print $1 " " $2}' > output

[Rajeev Shukla]

Hi,
Use
sed 's/:/ /g' /etc/passwd > output

[Chris Watkins_1]

Shouldn't have said "the" awk way...
there are always lots of ways to do things :-)

awk -F: '{print $1 " " $3}' < /etc/passwd > output

[Dave Johnson_1]

how about
awk -F: '{print $1 " " $2}' output

-Dave

[Dave Johnson_1]

Chris,
I guess greate minds think alike.
You just type faster than I.

-Dave

[Bill Hassell]

And yet another way...

cut -d: -f1,3 /etc/passwd | tr -s ":" " "

[Chris Hulihan]

if you have perl, you could do
perl -pi -e 's|\:| |g' output

to update the "output" file.

[Shahul]

Hi,

Please find below command, this helps. But don't do it on original passwd file.

#sed 's/:/ /g' passwd > passwd.new

Best of luck
Shahul

[H.Merijn Brand (procura]

# perl -aF: -nle'print"@F[0,2]"' /etc/passwd

or shorter, but more cryptic

# perl -aF: -ple'$_="@F[0,2]"' /etc/passwd

Enjoy, Have FUN! H.Merijn [ Who thinks -i is extremely dangerous on /etc/passwd ]

[Jordan Bean]

Even the posix shell can do it:

#!/usr/bin/sh
while IFS=: read login pw uid gid etc
do
echo $login $uid
done < /etc/passwd > output

[Sammy_2]

Thanks everyone. All of them work. AWK , SED , Bill's Tr was different and Procur and Chris added perl to it.