Hmm. It should. If you do a print $var, it doesn't print correctly? Or are you trying to do something else that fails? What is the error message?
-----Original Message----- From: Matt Fuerst [mailto:[EMAIL PROTECTED]] Sent: Tuesday, April 23, 2002 1:06 PM To: '[EMAIL PROTECTED]' Subject: Dumb question I have a variable that needs to store a string such as $var = "te$t"; This doesn't work since perl thinks the $ means I want some variable substitution in there. I tried doing: $var = "te\$t"; but no luck there either. What's the secret? Thanks! Matt -=-=-=-=-=-=-=-=-=-=-=-=-=-= Matt Fuerst Developer, Telecorp Products [EMAIL PROTECTED] 248,960,1000 -- To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] -- To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED]
