awk

kalyanofb · 02-07-2007, 05:53 AM

hai friends,

i am reading a file in awk and check whether $1 is linux command or not.

I don't know how to check the $1 is linux command or not.

Please help me

jschiwal · 02-07-2007, 07:09 AM

In an awk program, $# refers to a field in the line read. $0 refers to the entire line.

Check if you can install a "gawk-doc" package if available. I contains an excellent book: "Gawk: Effective AWK Programming"

rizhun · 02-07-2007, 07:27 AM

umm.. if you're $PATH is setup propperly, you could do it quite easily in shell-scirpt....

Code:

#!/usr/bin/ksh
INPUT=$1

cat ${INPUT} | while read app
do
  which ${app} >/dev/null 2>&1
  [[ $? -eq 0 ]] && {
    echo "${app} is a linux command!"
    continue
  }
  echo "${app} is not a linux command!"
done

So, if you saved the above code in a file called "check.sh" and you had a file with commands (one per line) called commands.txt you would run:

/path/to/check.sh /path/to/commands.txt

Nick_Battle · 02-07-2007, 07:37 AM

If you must do this in awk, then the system() call might help. It would be easiest to invoke something like the scripts already shown, but if you have to make the call in-line, then something like this should work:

system("IFS=:; for dir in $PATH; do if test -x $dir/"$1"; then exit 0; fi; done; exit 1")

or just

system("which "$1)

That returns 0 if $1 is on $PATH and executable, or 1 otherwise. Only tested on HP-UX, but it should work anywhere :-)

colucix · 02-07-2007, 07:51 AM

Here is an awk solution:

Code:

{ 
( "which --skip-alias " $1 " 2> /dev/null" ) | getline command ;
  if ( command == "" )
     print $1,"is not a linux command"
  else
     print $1,"is",command
  command = ""
}