CODEWITHSUNDEEP
bashperl
DemiImp
DemiImp

Reputation: 1018

How to escape semicolon?

I'm trying to run a command similar to this with bash and can't figure out how to escape the semicolon:

./script.pl data='!'"data;more_data"'!'

When my script reads the data, all it gets is "!data".

./script.pl data='!'"data\;more_data"'!'

The value I see is "!data\".

How should I be escaping the semicolon?

The Perl script uses CGI and param to get the data.

my $data = param("data");
print "$data\n";

Upvotes: 1

Views: 2334

Answers (1)

DemiImp
DemiImp

Reputation: 1018

The problem is in Perl. The CGI library's param() function will treat semicolons as delimiters, even when a script is called on the command line.

To get param() to work correctly, escape the semicolon as %3b.

Upvotes: 2

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