set command

consty · 02-22-2007, 06:22 AM

Hi all,
I am writing a shell script with positionnal parameters
and a set command (set `date`), the positionnal parameters take the value of fields of date command after the set command so that the arguments used when I execute my shell script are changed. How can I call them back inside my shell script (ksh).
Thanks for help
Consty

matthewg42 · 02-22-2007, 07:08 AM

It's not clear to me what you are asking. You have a shell script called "set"? You want to pass the output of the date command as arguments to a script?

consty · 02-22-2007, 08:06 AM

Thanks for your answer,
I execute my script with arguments, example : prog arg1 arg2 arg.
In my script prog there is the command "set `date`" which reset (change) the
values of $1, $2, $3, previously coreesponding to the value of arg1, arg2, arg3.
I would like to call back the values of arg1 arg2 and arg3 at the end of my program.
I hope this is clear.
Thanks

matthewg42 · 02-22-2007, 08:36 AM

Do you mean you want to do something like this:

Code:

prog one two three
echo "one is $one"
echo "two is $two"
echo "three is $three"

...and expect to see:

Code:

one is The
two is Feb
three is 22

Is that what you mean?

You can't have a script which you call directly modify the environment of the script which called it. Environment variables are inheritied by child processes, but there is no mechanism for sending changes back to the parent environment (without using mechanism like shared memory anyhow).

What you can do is take the standard output of your program and assign that to variables. For example:

Code:

date |while read a b c; do echo a=$a b=$b c=$c; done

consty · 02-22-2007, 09:03 AM

THanks a lot
Djessi

cfaj · 02-22-2007, 07:06 PM

Quote:

Originally Posted by matthewg42

What you can do is take the standard output of your program and assign that to variables. For example:

Code:

date |while read a b c; do echo a=$a b=$b c=$c; done

That will not work in most shells if you want to access the variables after the loop has finished. Elements of a pipeline are executed in subshells, and subsequent commands will not be able to access the assignments made in the pipeline.

This would work:

Code:

date | {
 read a b c
 echo a=$a b=$b c=$c
 ## All assignments made within the braces will be available
 ## only within the braces
}

A better way is to use eval. For example, I like to store the various date and time components in variables like this:

Code:

eval "$( date "+
   DATE=%Y-%m-%d
   YEAR=%Y
   MONTH=%m
   DAY=%d
   TIME=%H:%M:%S
   HOUR=%H
   MINUTE=%M
   SECOND=%S
   datestamp=%Y-%m-%d_%H.%M.%S
   DayOfWeek=%a
   DayOfMonth=%d
   DayOfYear=%j
   DayNum=%w
   MonthAbbrev=%b
   MonthFull=%B
   MonthName=%B
   DayName=%A
" )"

gnashley · 02-23-2007, 02:32 AM

Pull your initial values out first:
ARG1=$1 ARG2=$2 ARG3=$3

#then use a for loop instead of while/until:

for arg in "$@" ; do
commands
done

bigearsbilly · 02-23-2007, 02:55 AM

you don't need the parenthesis which make it a sub shell, so
only inside will they be visible i.e:

Code:

$ (set `date`;echo $4 $3 $2 $1)      
08:52:48 23 Feb Fri
$ echo got:$1 $2
got:

just use the plain
set `date`

Code:

$ set `date`
$  echo $1 $2
Fri Feb