cannot export result from awk into a variable in a bash script

Emmanuel_uk · 03-04-2005, 01:57 AM

Hi

I want to export the result from a awk command into a variable in a bash
script. How do I do that?
PS: I do not want to have to create a file and then read from it to
load the variable. I tried redirection, but no luck, not enough experience.

(I have read the awk manual and fiddled quite a bit.
I also looked at some existing scripts, no inspiration so far.
Will have to learn by example am afraid. thanks)

I call my script with ./myscript.sh --*.txt
(In practice I want to get a substr of the argument passed to the script)
There must be zillion way of doing it... but fow now I am trying with awk.

Code:

 
#!/bin/bash
myvar=$1
echo "myvar is $myvar"

myshortvar=""
gawk 'BEGIN { print substr("'$myvar'",3) }'
#this above works fine. Now how do I return the result to myshortvar?

#ls -l $myvar
#mydir=`pwd`
#df -hl $mydir

jkobrien · 03-04-2005, 04:14 AM

Hi,

Backticks should do it...

myshortvar=`gawk 'BEGIN { print substr("'$myvar'",3) }'`

John

theYinYeti · 03-04-2005, 05:40 AM

It is usually better to use the $() notation, rather than backticks, because parenthesis can be used several times; For example: RPMforX=$(rpm -qf $(which X)).

So you would write something like that:
myVar=$(awk command)

But be carefull! Be it with backticks or with parenthesis, if you want to keep the original amounts of spaces, as well as line breaks, you'll have to use quotes, too:
myVar="$(awk command)"

Note that (at least using parenthesis) you can use quotes in the command. So this would work for example:
pidList="$(ps -e | grep -v "PID" | awk "{print \"${USER}: \" \$1}")"

Yves.

Emmanuel_uk · 03-04-2005, 06:55 AM

Thanks/Merci,

I tried backquoting but maybe with too many spaces, or wrong nesting,
and it was also before I could get the syntax right for the inner variable in substr.
I let you know once I have tried your suggestions

Thanks for the help

Emmanuel_uk · 03-07-2005, 01:54 AM

Thanks guys, it now works

I have put this home-made newbee function in my bashrc.
Zillions did it before, I am sure, in a more elegant way

dirr ()
{
# This function is equivalent to a dir / ls
# with added at the last line the space left in the current drive / partition
#
# call the function with dirr [--arg1]
# where -- must preceed the argument (otherwise *.txt as an arugment for example will
# not work because of a global file auto-completion)
#
argu1=$1 #retrieve fist argument

#Retrieve -- and trap errors
shortargu0=`gawk 'BEGIN { print substr("'$argu1'",0,2) }'`
if [ "$shortargu0" == "--" ] || [ "$shortargu0" == "" ]; then
shortargu0="valid argument at start"
else
echo "Argument must be preceeded by -- (minus-minus)"
return 0
fi

# gawk 'BEGIN { print substr("'$argu1'",3) }'
#extract the substring that ignores the 2 first characters (--)
shortargu1=`gawk 'BEGIN { print substr("'$argu1'",3) }'`

#echo "myvar is now $shortargu1"
ls -l -F --color=auto $shortargu1 #call a long ls
mydir=`pwd` #find where we hare, more to the point we are interested in the drive / partition itself
df -hl $mydir #display space left
return 0
}