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Top Forums Shell Programming and Scripting Function ignoring global variables Post 302401579 by cfajohnson on Saturday 6th of March 2010 11:59:07 PM
Old 03-07-2010
Quote:
Originally Posted by mikesimone
Hi there.
I'm writing a function to which I want to pass a global variable. For some reason, it's ignoring the variable.

Code:
#!/bin/bash
#####################################
#Variable Declaration
#####################################

CURPATH=`dirname $0`


There's no need to use dirname in bash (or any POSIX shell); use parameter expansion:
Code:
CURPATH=${0##*/}

(But there's rarely a good reason to require the location of the script.)
Quote:
Code:
DEEP=$CURPATH/depth.txt

export $DEEP


Code:
export DEEP

Why are you exporting DEEP? Are you calling another script that will need it?
Quote:
Code:
function getdepth()


That is a bash hybrid, a combination of ksh syntax and standard syntax. Use standard syntax:
Code:
getdepth()

Quote:
Code:
{

  echo $DEEP

  cat $1  | egrep ^${2} | awk -vgroup=$1 -vdepth=$DEEP -F, '


UUOC
Quote:
Code:
  {
      #Bunch of stuff that works when I hardcode it, but fails with the variable.


What "stuff"? If that's where it fails, that's what you should be posting.
Quote:
Code:
  }'
}


######
# Main block
######

getdepth some_file some_element

Any idea why the output of "echo $DEEP" is just a blank? And how do I make it pay attention to the global variables?

Thanks in advance,
Mike

---------- Post updated at 11:18 PM ---------- Previous update was at 11:14 PM ----------

Nevermind, I figured it out.

it's
Code:
export DEEP=$CURPATH/depth.txt.

Now I feel like a tool.
 

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