[SOLVED] how can I use shell variable in side awk

[brijith]

Hai,


How can I use A shell variable in side an awk when I use AWK in side a shell script. Is it possible? If it is can you give a simple example that shows how it can be..



Thanks

[geirha]

You generally don't want to do that. Perhaps if you give some example code, someone will find a better solution for it.

I'll try to explain why your variables don't work as expected though. Bash treats the quotes '' and "" differently. The single-quotes '' are called hardquotes, bash treats everything inside it as text, and will not replace variables with their values. The double-quotes "" are called soft quotes. Bash treats some parts of it as code, including variables. That means, that whenever you use a symbol in your awk, that bash may interpret instead, you need to escape it with a \.


$ word=world
$ echo hello | awk '{ print $1,$word; }' # Both are treated as awk variables, so it
# won't work

$ echo hello | awk "{ print $1,$word; }" # Both are treated as bash variables, so $1
# will not be hello, and it will fail if $1
# is empty or unset.

$ echo hello | awk "{ print \$1,$word; }" # Since the $ in \$1 is escaped, bash
# will change it to $1 and leave it alone
# for awk to interpret. $word is replaced by
# bash, with $word's value.


Try doing it a different way and you might save yourself from a lot of headaches.

[ghostdog74]

sure. you can pass your shell variable into awk using -v option


myvar="something"
awk -v var="$myvar" '{ print var }' file

[geirha]

Right, and if I had just waited for ghostdog74 to come around with an elegant solution, I could've saved me the trouble of explaining why my bad solution is bad. Ignore my previous post ;)

[ghostdog74]

actually, without -v option


# var="help"
# awk 'BEGIN{print "'$var'" }'
help

just a bit extra quoting.

[brijith]

Hi,


Another Doubt, How can I use cut on a shell variable.


str="mac:maya"

cat -d: -f1 $str # does not works
`echo "$str" | cat -d: -f1` # does not works


Please help!!

Thanks

[ghostdog74]

first , get your spelling correct and secondly, you don't need the backticks


str="mac:maya"
echo "$str" | <get_your_spelling_correct> -d: -f1

lastly, you can also use awk


echo "$str" | awk -F":" '{print $1}'

[dtmilano]

This also works and you don't have to explicitily define every variable on the command line

$ export A=A
$ awk END {print ENVIRON["A"]} < /dev/null

[brijith]

#!/bin/bash
usernames="brijith"
lst=`awk -v user="$usernames" -F: 'BEGIN {OFS=":"};user==$1 {print $3,$4}' /etc/passwd`
uid=echo "$str" | cut -d: -f1
gid=echo "$str" | cut -d: -f2
echo "$usernames -- $uid $gid"

Above Scripts Does Not Work Can you Correct it? What I need is the uid and gid of the given user. In two shell variable.

[geirha]

uid=`grep ^$username: /etc/passwd | cut -d: -f3`
gid=`grep ^$username: /etc/passwd | cut -d: -f4`
uid=`id -u $username`
gid=`id -g $username`

[ghostdog74]

#!/bin/bash
usernames="brijith"
lst=`awk -v user="$usernames" -F: 'BEGIN {OFS=":"};user==$1 {print $3,$4}' /etc/passwd`
uid=echo "$str" | cut -d: -f1
gid=echo "$str" | cut -d: -f2
echo "$usernames -- $uid $gid"

Above Scripts Does Not Work Can you Correct it? What I need is the uid and gid of the given user. In two shell variable.
firstly, where does "$str" comes from. anyway, unless your requirement is more complicated, just do it in awk.


awk 'BEGIN { user="user2"; FS=":"}
$1==user { print "uid: " $3, "gid: "$4}
' /etc/passwd

[brijith]

Hi,
Me again with another issue :)


for files in `find <some_folder_name> -name "*"`; do
echo "File Name $files"
done


The above code works Perfectly if there is no space in file name and directory names in the searching folder -<some_folder_name>-. But it fails if there is space in the file name or folder names.

I hope you understood what I said. How to solve this issue ?

Thanks

[ghostdog74]

Hi,
Me again with another issue :)


for files in `find <some_folder_name> -name "*"`; do
echo "File Name $files"
done


The above code works Perfectly if there is no space in file name and directory names in the searching folder -<some_folder_name>-. But it fails if there is space in the file name or folder names.

I hope you understood what I said. How to solve this issue ?

Thanks

use a while loop and quote your variables!


find <some_folder_name> -name "*" | while read FILE
do
echo "$FILE"
done