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Parameter to AWK

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# 8  
Old 07-23-2009
if you are using bash, just place your user_text inside of an array.
# 9  
Old 07-23-2009
Array in KSH

Originally Posted by RiSk
if you are using bash, just place your user_text inside of an array.

I tried with arrays in KSH.

Below code works fine
$ set -A x a b c
$ echo ${x[1]}
$ echo ${x[0]}

But when i try my input text i get errors as below

Try 1
$ set -A x -parm a=b -parm c=d -parm e=f -parm g=h \
ksh: -parm: 0403-010 A specified flag is not valid for this command.

Try 2
$  set -A x -parm a=b -parm c=d -parm e=f -parm g=h \\
ksh: -parm: 0403-010 A specified flag is not valid for this command.

Try 3
$ set -A x '-parm a=b -parm c=d -parm e=f -parm g=h \'
ksh: -parm a=b -parm c=d -parm e=f -parm g=h \: 0403-010 A specified flag is not valid for this command.

Any other suggestions?
# 10  
Old 07-23-2009
This is the input file I used from your request:

-parm a=b -parm c=d -parm e=f -parm g=h \
-parm a=b -parm c=d \
-parm a=b -parm c=d -parm e=f \

This is the code I used to do stuff based on this input:

while read -r line
  set -- $line
  while [ "$1" != "" ]
    case $1 in
      "a=b") echo "doing a=b stuff" ;;
      "c=d") echo "doing c=d stuff" ;;
      "e=f") echo "doing e=f stuff" ;;
      "g=h") echo "doing g=h stuff" ;;
done < /input_file

# 11  
Old 07-23-2009
It's working !!!

Thank you very much. It's working Smilie

Thanks for your suggestions and your valuable time. I appreciate it.
# 12  
Old 07-23-2009
Originally Posted by rakeshawasthi
Coming back to original question, how to achieve this by passing variable to awk.
user_text="first second third fourth"
echo $user_text | awk -v awk_var="$var" '{print $awk_var}'

---------- Post updated at 09:40 AM ---------- Previous update was at 09:39 AM ----------

$1 will be overwritten...
to avoid that, first store the passed argument and then use set.
in that case:
set $1 first second third fourth
echo $2
echo $5

# 13  
Old 07-23-2009
Need your help again...

I used set as below which is working fine.

$ set -- a b c d
$ echo $1
$ echo $2

But instead of using
$ echo $3

Is there a way of using it as below?
$ z=3
$ echo $z

Expected output is 'c' instead of '3'

The reason why I want to do this is like I mentioned before the number of arguements may vary in my input and one of my requirement is to get the last - 1 parameter.
For example
$ set -- -parm a=b -parm c=d -parm e=f \\
$ echo $1
$ echo $4
$ echo $7

This works fine when $4 or $1 is an arguement to echo.

But I need the 6th parameter of set which is e=f and this position varies based on my input line. Sometimes it might be 7th position as shown in above example and sometimes it may be 3rd.

Last edited by sacguy08; 07-23-2009 at 06:43 PM..
# 14  
Old 07-24-2009
I think there are two answers here. First, you seem to be asking how to access a variable-variable name. In other words, I want to define a variable name with a variable since I don't know something about the code before it runs. Try this:

set a b c d
eval echo '$'$field

Second, you seem to want this so you can grab the last field from the original input.

Original input from before:

-parm a=b -parm c=d -parm e=f -parm g=h \
-parm a=b -parm c=d \
-parm a=b -parm c=d -parm e=f \

This can be done simply with awk:

awk '{print $(NF-1)}' input_file

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