in reply to Grabbing tokens

I ran a test and verified that this will work:
#!/usr/local/bin/perl
$prg = "testers.sh";
$ret_var = <`$prg`>;
print "\n\n$ret_var\n\n";

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I got what testers.sh echoed back in $ret_val. As well, I verified that the command that testers.sh executed (`/bin/cp file1 file2`) worked too. So, the execution was not hindered, and the output of testers.sh was brought back into my Perl program for parsing. This should do it.
HTH,
Chris

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Re: Re: Grabbing tokens
by Skeeve (Parson) on May 14, 2003 at 08:58 UTC
    Are you sure? I tried and it failed. To my understanding 
    $ret_var = <`$prg`>;

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    would first run $prg because of the backticks, and then try to handle this output as a filehandle.
    $ret_var = `$prg`;

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    will do what you stated.
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      Yes, I'm sure...otherwise, I would not have stated that I tested it.
      Here are the exact programs so that you can verify. I will admit, I did mistype one thing...I forgot to put the './' in the $prg scalar:
      ###Begin Perl###
#!/usr/local/bin/perl
$prg = "./testers.sh";
$ret_val = <`$prg`>;
print "\n\n$ret_val:From Perl Not BASH Script\n\n";

###End Perl####
###Begin Bash###
#!/usr/local/bin/bash
`/bin/cp /wantstats.sh /newstats.sh`
echo Success
###End Bash###

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      What is printed:


      Success:From Perl Not BASH Script


      Try these scripts Skeeve. I wouldn't just claimed to have tested something...
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        Not here! 
        $ ./p1


`./testers.sh`:From Perl Not BASH Script

$

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        I think you don't know what thos `` mean! In your testers.sh I see: 
        `/bin/cp /wantstats.sh /newstats.sh`

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        Luckily /bin/cp doesn't have any STDOUTput! Just go ahead and enter: 
        `echo ls`

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        You will get the output of ls not the text "ls"!

        this is because `xxx` first execute xxx and then execute the output of xxx!

        No don't execute but try to imagine what this will do: 
        `echo rm -rf \*`

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