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is it possible to use a CGI script to create other CGI scripts?

Posted on 2000-03-21
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Last Modified: 2011-04-14
I want to create a CGI script from within a CGI script.  How do I get around the following:

open(NEW_CGI_SCRIPT, ">$address");

print NEW_CGI_SCRIPT "cgi code";

close(NEW_CGI_SCRIPT);

What can I do to prevent the current CGI script from interpreting the cgi code I want to write to the new script?

Any pointers would be appreciated!
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Question by:flahertd
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Expert Comment

by:maneshr
ID: 2642053
"What can I do to prevent the current CGI script from interpreting the cgi code I want to write to the new script? "

you dont have to do anything. As long as your CGI code is within " and you have escaped any special chars you are safe.

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Expert Comment

by:PC_User321
ID: 2642367
$cgi_code = 'Your CGI code goes here;
$ signs get ignored because of the single quotes
Double quotes ignored "too".
End of CGI script';


print NEW_CGI_SCRIPT $cgi_code;



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Expert Comment

by:maneshr
ID: 2645121
flahertd,

were you able to find a solution to your problem? if so, pl. let this forum know of the solution.

thanks
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Author Comment

by:flahertd
ID: 2645525
First, thanks for the suggestions!

What I did was a mixture of both -

I put some of the code within single
quotes, so that no special characters would be interpreted:

$cgi_code = '#!/usr/local/bin/perl
$ENV{"PATH"} .=
":/usr/local/bin:/usr/local/lib/perl";
require "/path/to/cgi-bin/cgi.pl";';

print NEW_CGI_SCRIPT $cgi_code;

Then more code that I needed interpreted, I put in double quotes, backslashing the stuff I didn't want interpreted - it looks pretty ugly, but it does the job:

print NEW_CGI_SCRIPT "\$filename = \"$filepath_I_need_interpreted\";";
           
So then in NEW_CGI_SCRIPT I would end up with:
#!/usr/local/bin/perl
$ENV{"PATH"} .= ":/usr/local/bin:/usr/local/lib/perl";
require "/path/to/cgi-bin/cgi.pl";
$filename = "/path/to/file";

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Accepted Solution

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maneshr earned 300 total points
ID: 2645672
glad to know that your problem was solved. :)
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