ansaurus

Question

Answer 1

A:

Something like this will work:

awk -v x="$READ_LINE" -v y="SOME_WORD" 'BEGIN { split(x, a); if (a[2] != y) print "not match";}'

But where does $READ_LINE come from? What are you trying to accomplish? There might just also is a good plain sh or ksh solution.

I highly doubt your claim that echo (which might be a shell builtin) takes more time than awk. Here is a plain sh version:

set -- $READ_LINE
[ x$2 != xSOME_WORD] && echo "not match"

But the ksh solution of Dennis Williamson looks the best for your situation.

schot 2010-08-30 08:22:57

the target is to save time (echo take time awk not)

lidia 2010-08-30 09:07:10

Answer 2

A:

read ign val && [ X$val != XSOME_WORD ] && echo "not match"

reinierpost 2010-08-30 08:23:05

Answer 3

A:

This splits the line by words and tests the second word.

var=($READ_LINE)
[[  ${var[1]} != SOME_WORD  ]] && print "not match"

What is it you're trying to accomplish? Several of your questions are nearly identical.

Dennis Williamson 2010-08-30 09:39:25

var=($READ_LINE) isnt valid syntax in ksh!

lidia 2010-08-30 09:59:51

yes it is. you must be using very old ksh

ghostdog74 2010-08-30 13:48:05

@lidia: What version of ksh? Are you using `#!/bin/ksh`, `#!/usr/bin/ksh` or `#!/bin/sh` as the first line of your script (the shebang)?

Dennis Williamson 2010-08-30 14:23:13

Answer 4

A:

if you are using ksh, then just use the shell without having to call external commands

set -- $READ_LINE
case "$2" in
  "SOME_WORD" ) echo "found";;
esac

Why are you still tacking this problem since I see you have other threads similar to this.?

ghostdog74 2010-08-30 13:49:19

awk/sed for test command