new to awk!

[bluefrog]

could anyone help out with this?

The following command line works in korn shell
awk '/SCC_DAILY_UPDATE/ {print $2}' service_stream_mapping.dat

but this doesn't

export JOB_STREAM="SCC_DAILY_UPDATE"
awk '/$JOB_STREAM/ {print $2}' service_stream_mapping.dat

I need to be able to pass the string to search as a parameter.
Any ideas ?

[gerard4143]

Try this

export JOB_STREAM="SCC_DAILY_UPDATE"
awk '{printf("%s\n", ENVIRON["JOB_STREAM"])}' testfile.dat

It works with bash...

[bluefrog]

That doesn't seem to work in ksh,

but I figured it out eventually using this:

$ awk '/'^${JOB_STREAM}'/ {print$2}' service_stream_mapping.dat
DOLNSGC3

I have now encountered a different problem, I cannot assign the result to a variable:

e.g.
export GET_RESULT
GET_RESULT=`awk '/'^${JOB_STREAM}'/ {print$2}' service_stream_mapping.dat`
echo $GET_RESULT

doesn't print anything!
Any further help would be much appreciated!

[drl]

Hi.

Getting shell variables into awk has been improved with modern versions of awk. The method that appeals to me is:

       -v var=val
       --assign var=val
              Assign  the  value  val to the variable var, before execution of
              the program begins.  Such variable values are available  to  the
              BEGIN block of an AWK program.

-- excerpt from man awk

For example:

#!/usr/bin/env ksh

# @(#) s1	Demonstrate method of getting shell variables into awk.

echo
export LC_ALL=C
echo "Environment: LC_ALL = $LC_ALL"
echo "(Versions displayed with local utility \"version\")"
version >/dev/null 2>&1 && version "=o" $(_eat $0 $1) awk
echo

echo "Results:"
greeting=" Hello, world."
awk -v hi="$greeting" '
BEGIN	{ print hi ; exit }
'

echo
echo "Results, assignment:"
GET_RESULT=`awk -v hi="$greeting" '
BEGIN	{ print hi ; exit }
'`
echo " GET_RESULT is \"$GET_RESULT\""

echo
echo "Results, assignment, modern syntax:"
GET_RESULT=$( awk -v hi="$greeting" '
BEGIN	{ print hi ; exit }
' )
echo " GET_RESULT is \"$GET_RESULT\""

exit 0

producing:

% ./s1

Environment: LC_ALL = C
(Versions displayed with local utility "version")
OS, ker|rel, machine: Linux, 2.6.26-2-amd64, x86_64
Distribution        : Debian GNU/Linux 5.0 
ksh 93s+
GNU Awk 3.1.5

Results:
 Hello, world.

Results, assignment:
 GET_RESULT is " Hello, world."

Results, assignment, modern syntax:
 GET_RESULT is " Hello, world."

I had no trouble with the backquote-execution. However, I prefer the syntax that uses

$( ... )

for a number of reasons.

Best wishes ... cheers, drl

[bluefrog]

thx drl

I have tried the following:

export $JOB_STREAM="DAILY_UPDATE"
$SERVICE_NAME="$(awk -v VAR=$JOB_STREAM' '/VAR/ {print $2}' service_stream_mapping.dat)"

and the following syntax error appears:

awk: syntax error near line 1
awk: bailing out near line 1
ksh[6]: =: not found


An error does not appear when I try the following;
$SERVICE_NAME=$(awk '/$x/ {print $2}' x='${JOB_STREAM}' service_stream_mapping.dat)

but I get the following run time message:
ksh[6]: =: not found

[drl]

Hi.

Do not use:

export $JOB_STREAM="DAILY_UPDATE"

but rather:

export JOB_STREAM="DAILY_UPDATE"

that is, remove the "$" on the assignments; and, if you use the "-v" syntax, you will not need the export, just the assignment ... cheers, drl

[bluefrog]

thx dlr,

yes you right, I got the variables all mixed up!!!

I finally got it working using the following script:

export JOB_STREAM="SCC_DAILY_UPDATE"

export SERVICE_NAME=""

SERVICE_NAME=$(awk '$1==x {print $2}' x="${JOB_STREAM}" service_stream_mapping.dat)

echo $SERVICE_NAME


thx again!