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Top Forums Shell Programming and Scripting Script takes too long to complete Post 302976287 by mohtashims on Monday 27th of June 2016 12:31:54 PM
Old 06-27-2016
Quote:
Originally Posted by MadeInGermany
Yes, your inner loop (in post#1) should fill both variables in one stroke, use the correct InputFileSeparator
Code:
    input="alter.txt"
    while IFS="=" read -r searchterm replaceterm
    do
      echo "ST:$searchterm"
      echo "RT:$replaceterm"
    done < "$input"

If you look at the modified script in my last post .. it takes the same time with or without this suggestion.

I was able to bring down the execution time from 25 mins to -> just 7 mins.

Please suggest if there is anything else that can be done to optimize this ?

@RudiC:

I m sorry for not able to completely understand your suggestion.

Can you please elaborate the below only if the same is not covered in my last post with the updated script.

Quote:
Why don't you leave the looping to one single instance of e.g. awk ?
Create a list of all file candidates ( find can have several paths as starting points) and run awk , first reading all the search/replacement pairs, and then working those on all files presented.
 

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SHELL-QUOTE(1)						User Contributed Perl Documentation					    SHELL-QUOTE(1)

NAME
shell-quote - quote arguments for safe use, unmodified in a shell command SYNOPSIS
shell-quote [switch]... arg... DESCRIPTION
shell-quote lets you pass arbitrary strings through the shell so that they won't be changed by the shell. This lets you process commands or files with embedded white space or shell globbing characters safely. Here are a few examples. EXAMPLES
ssh preserving args When running a remote command with ssh, ssh doesn't preserve the separate arguments it receives. It just joins them with spaces and passes them to "$SHELL -c". This doesn't work as intended: ssh host touch 'hi there' # fails It creates 2 files, hi and there. Instead, do this: cmd=`shell-quote touch 'hi there'` ssh host "$cmd" This gives you just 1 file, hi there. process find output It's not ordinarily possible to process an arbitrary list of files output by find with a shell script. Anything you put in $IFS to split up the output could legitimately be in a file's name. Here's how you can do it using shell-quote: eval set -- `find -type f -print0 | xargs -0 shell-quote --` debug shell scripts shell-quote is better than echo for debugging shell scripts. debug() { [ -z "$debug" ] || shell-quote "debug:" "$@" } With echo you can't tell the difference between "debug 'foo bar'" and "debug foo bar", but with shell-quote you can. save a command for later shell-quote can be used to build up a shell command to run later. Say you want the user to be able to give you switches for a command you're going to run. If you don't want the switches to be re-evaluated by the shell (which is usually a good idea, else there are things the user can't pass through), you can do something like this: user_switches= while [ $# != 0 ] do case x$1 in x--pass-through) [ $# -gt 1 ] || die "need an argument for $1" user_switches="$user_switches "`shell-quote -- "$2"` shift;; # process other switches esac shift done # later eval "shell-quote some-command $user_switches my args" OPTIONS
--debug Turn debugging on. --help Show the usage message and die. --version Show the version number and exit. AVAILABILITY
The code is licensed under the GNU GPL. Check http://www.argon.org/~roderick/ or CPAN for updated versions. AUTHOR
Roderick Schertler <roderick@argon.org> perl v5.16.3 2010-06-11 SHELL-QUOTE(1)
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