Actually, "defined" is all I needed. I don't know why I couldn't find that myself, but I was really starting to tear my hair out. Thanks!
- Jesse -----Original Message----- From: Andrew Curry [mailto:[EMAIL PROTECTED] Sent: Tuesday, August 14, 2007 3:23 PM To: Jesse Farrell Cc: beginners@perl.org Subject: RE: Recognizing integer as string for if statement Unless im being very stupid in what your requirements are If(defined $variable && $variable ne "0") { } -----Original Message----- From: Jesse Farrell [mailto:[EMAIL PROTECTED] Sent: 14 August 2007 21:18 To: beginners@perl.org Subject: Recognizing integer as string for if statement Is there a way to have a variable recognized as a string even if it contains a valid integer? I have a script that is taking strings from a file that can be numbers, including zero. I have an if statement to simply see if the variable exists (otherwise I need to throw an error), and if the variable contains 0, the if statement is failing. Right now, my code is essentially: if ($variable) { # act on $variable } else { # throw error } Is there an easy way to get the if statement to recognize a 0, or do I need to add a special case? I don't care to keep the 0 as an integer, if I can change it to a string somehow, that would work great. --- Jesse Farrell This e-mail is from the PA Group. For more information, see www.thepagroup.com. This e-mail may contain confidential information. Only the addressee is permitted to read, copy, distribute or otherwise use this email or any attachments. If you have received it in error, please contact the sender immediately. Any opinion expressed in this e-mail is personal to the sender and may not reflect the opinion of the PA Group. Any e-mail reply to this address may be subject to interception or monitoring for operational reasons or for lawful business practices. -- To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] http://learn.perl.org/
