Very Computer

I've searched the archives of this group, but I couldn't find an
answer, so either it's dumb question or I'm using the wrong search
keywords.

I'm trying to create a bash script that takes as command line
parameters a bunch of variables and their values.  Something like this:

Here's the 't' script:

#!/bin/sh
echo ${VAR}

And here's what I run

$ ./t VAR=4

I was hoping it would display "4", but it displays nothing.  What do I
need to do to get the script to set VAR to 4?  I guess I need to loop
through all the command-line parameters, and then "execute" each one,
but I don't know how to do that.  Obviously, this could be a security
hole, but I'm not concerned about that.  Any help is appreciated.

It's very unclear what you're trying to do. You say you want to "get the
script to set VAR to 4" but then your example shows you attempting to
set VAR to 4 outside the script.

Are you trying to set a variable that can be used within "t" or trying
to get "t" to set a variable that can be used in the calling shell? If
it's the former, do this:

    VAR=4 ./t

or this:

    export VAR
    VAR=4
    ./t

If it's the latter, do this in "t"

    #!/bin/sh
    VAR="$1"

then call "t" as something like (depending on what you're really up to):

    . ./t 4
    echo "$VAR"

Note the "<dot><space>" at the start of the line.

If you're trying to do something else, do tell....

        Ed.

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[snip]

Put the VAR=4 /before/ the invocation of the script.

Like this...

  #!/bin/sh
  echo ${VAR}

  -rwxr--r--    1 pitchl   users          23 Jul 13 12:52 t

  1

  4

- --

Lew Pitcher, IT Specialist, Enterprise Data Systems
Enterprise Technology Solutions, TD Bank Financial Group

(Opinions expressed here are my own, not my employer's)
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The program should know about its variable names. The call...

   ./t 4

The content of file t...

   VAR=$1
   echo ${VAR}

If you indeed want to handle an arbitrary and unknow variable name
use eval; for example...

   eval $1

(which is indeed a bad idea!)

Janis

Quote:> I've searched the archives of this group, but I couldn't find an
> answer, so either it's dumb question or I'm using the wrong search
> keywords.

> I'm trying to create a bash script that takes as command line
> parameters a bunch of variables and their values.  Something like this:

> Here's the 't' script:

> #!/bin/sh
> echo ${VAR}

> And here's what I run

> $ ./t VAR=4

> I was hoping it would display "4", but it displays nothing.  What do I
> need to do to get the script to set VAR to 4?  I guess I need to loop
> through all the command-line parameters, and then "execute" each one,
> but I don't know how to do that.  Obviously, this could be a security
> hole, but I'm not concerned about that.  Any help is appreciated.

    You are passing VAR=4 to your script as an argument; there is no
    assignment to the variable VAR.

    If you put the line "echo $1" in your script, it will print "VAR=4".

    You can assign the variable on the same line as the cal to your
    script, by putting it before the call:

VAR=4 ./t

    Or define it separately and export it:

VAR=4
export VAR
./t

--
    Chris F.A. Johnson                     <http://cfaj.freeshell.org>
    ==================================================================
    Shell Scripting Recipes: A Problem-Solution Approach, 2005, Apress
    <http://www.torfree.net/~chris/books/cfaj/ssr.html>