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Top Forums Shell Programming and Scripting Add to constant fields at the end of every line Post 302407767 by alister on Friday 26th of March 2010 10:54:37 AM
Old 03-26-2010
Quote:
Originally Posted by poova
Thanks all... It's working. I was supposed to pass variable for 1000 and XYZ-1234 there. So I changed the command to ,
sed "s/$/,$id1,$id2/" file

When using single quote it was not working.
If the replacement text in your variables contains characters or sequences which are special to sed, this approach will not work. Examples of such characters/sequences include '&', '/' (if default regular expression delimiters are being used), '\' followed by anything (it's either a defined escape sequence or an undefined sequence which may result in the loss of a backslash in your replacement text).

Assuming that your two values are stored in the shell variables $a and $b:
Code:
a='your 1st value'
b='your 2nd value'

The following are safe alternatives, one is pure shell (printf is usually a built-in these days) and the other is AWK:
Code:
while IFS='' read -r line; do printf '%s,%s,%s\n' "$line" "$a" "$b"; done < data

awk '{printf("%s,%s,%s", $0, a, b RS)}' a="$a" b="$b" file

Regards,
Alister

Last edited by alister; 03-26-2010 at 12:02 PM..
 

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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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