Parameter to AWK

sacguy08 · July 22, 2009, 7:40pm

Hi,

I need help with AWK. Here is the issue that i'm facing.

I have a variable in KSH script set as below.
user_text="first second third fourth"

My Requirement is to print 'first', 'second', 'third' or 'fourth' based on setting a variable

Example 1
user_text="first second third fourth"
var=1
echo $user_text | awk '{print $var}'

Ouput expected:
first

Example 2
user_text="first second third fourth"
var=3
echo $user_text | awk '{print $var}'

Ouput expected:
third

I'm new to AWK. I would be grateful if you could reply with correct AWK syntax. Your help is highly appreciated.

Thanks for your time.

vgersh99 · July 22, 2009, 7:58pm

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ryandegreat25 · July 22, 2009, 9:32pm

you need array not awk, please refer to your shell array assignment

-bash-3.2$ x=(first second third fourth)
-bash-3.2$ echo ${x[0]}
first
-bash-3.2$ echo ${x[1]}
second
-bash-3.2$ echo ${x[2]}
third
-bash-3.2$ echo ${x[3]}
fourth
-bash-3.2$

edidataguy · July 22, 2009, 9:42pm

 
set first second third fourth
echo $1 $4
Output==> first fourth

ryandegreat25 · July 22, 2009, 10:59pm

wow i didnt know that.. what happens if i use this to a script?
what happens to $1 if i have argumented script?

rakeshawasthi · July 23, 2009, 12:10am

Coming back to original question, how to achieve this by passing variable to awk.

user_text="first second third fourth"
var=1
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.

sacguy08 · July 23, 2009, 12:55pm

Hello Everyone,

Thanks for your replies. I tried set and array solutions but it didn't work.

I'm using ksh in AIX 5.3

Here is my input string
-parm a=b -parm c=d -parm e=f -parm g=h \

Note: At the end of each line in my input file there is a Back slash charater.

My input string can be of variable length. For example it can be
-parm a=b -parm c=d \
or
-parm a=b -parm c=d -parm e=f \

My requirement is to check for values for each -parm and do a particular action.

For example if my input string is
-parm a=b -parm c=d -parm e=f -parm g=h \

If the last -parm value is g=h then I need to set a variable X to 1.
If the 2nd -parm value is a=b then I need to set a variable Y to 1.
That is I need to read the value one at a time from my input string and perform a particular action.

I thought it would be possible in awk and hence my original question.

Your different approachs to this problem or solutions to this problem is highly appreciated.

Thanks for your time.

RiSk · July 23, 2009, 1:03pm

if you are using bash, just place your user_text inside of an array.

sacguy08 · July 23, 2009, 1:39pm

RiSk,

I tried with arrays in KSH.

Below code works fine

$ set -A x a b c
$ echo ${x[1]}
b
$ echo ${x[0]}
a

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?

peterro · July 23, 2009, 1:55pm

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:

#!/bin/ksh
while read -r line
do
  set -- $line
  while [ "$1" != "" ]
  do
    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" ;;
    esac
    shift
  done
done < /input_file

sacguy08 · July 23, 2009, 2:43pm

peterro,
Thank you very much. It's working

Others,
Thanks for your suggestions and your valuable time. I appreciate it.

edidataguy · July 23, 2009, 3:00pm

rakeshawasthi:

Coming back to original question, how to achieve this by passing variable to awk.
user_text="first second third fourth"
var=1
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
   first
echo $5
   fourth

sacguy08 · July 23, 2009, 5:20pm

I used set as below which is working fine.

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

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
-parm
$ echo $4
c=d
$ 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.

peterro · July 24, 2009, 11:11am

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
field=3
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

sacguy08 · July 24, 2009, 9:25pm

peterro,

Thanks in advance since I won't be able to have access to a UNIX machine until Monday. I'll try your solutions on Monday and let you know.

Thanks once again.

edidataguy · July 25, 2009, 12:43am

sacguy08:

I used set as below which is working fine.
$ set -- a b c d
$ echo $1
a
$ echo $2
b
$
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
-parm
$ echo $4
c=d
$ 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.

Try this:

set a b c d e f g h i
count=$#
lastone=`echo ${*:$count}`

echo $lastone
Output => i

---------- Post updated at 11:43 PM ---------- Previous update was at 10:17 PM ----------

user_text="first second third fourth"
echo ${user_text##* }

This is more simple

Did not realize you needed the last but one. Try this:

set aa bb cc dd ee ff gg hh ii
echo $* | sed 's/.* \([^ ]*\) [^ ]*$/\1/'

sacguy08 · July 27, 2009, 8:06pm

Thanks peterro and edidataguy.

It's working except for the below code given by edidataguy.

$ set a b c d e f g h i
$ count=$#
$ lastone=`echo ${*:$count}`
ksh: ${*:$count}: 0403-011 The specified substitution is not valid for this command.

Anyways my issues are solved. Thanks to everyone. I appreciate your help.

edidataguy · July 27, 2009, 8:20pm

sacguy08:

Thanks peterro and edidataguy.

It's working except for the below code given by edidataguy.
$ set a b c d e f g h i
$ count=$#
$ lastone=`echo ${*:$count}`
ksh: ${*:$count}: 0403-011 The specified substitution is not valid for this command.
Anyways my issues are solved. Thanks to everyone. I appreciate your help.

I think there is a version issue. It works fine for me.
Please post your final code for every one to see.