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Old 04-01-2005, 11:31 AM   #1
Xris718
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Registered: May 2003
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Perl string comparisons


Hi

I need to know how to do string comparisons in perl just like shell does. How do you convert the following into perl?
ie:

if [-n "$string"]; then
statement.....

if [-z "$string"]; then
statement...

if [$var -gt 0]; then
statement...
 
Old 04-01-2005, 12:20 PM   #2
david_ross
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I think this is what you want:
Code:
if($string){
 print "Variable exists\n";
}
if(!$string){
 print "Variable is empty\n";
}
if($string>0){
 print "Variable greater than 0\n";
}
 
Old 04-01-2005, 01:42 PM   #3
Xris718
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thank you i think thats it!
 
Old 04-01-2005, 02:06 PM   #4
Xris718
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oh what about this convert into perl?
ie:

if [! -z $string]; then
something....
 
Old 04-01-2005, 03:42 PM   #5
puffinman
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You might be interested in the "defined" function.

Code:
if($string) {
    # string exists and evaluates as true
} elsif(defined $string) {
    # string exists but evaluates as false
} else {
    # string doesn't exist
}
Also, you can use numerical comparisons directly on strings, if they consist of just a number Perl will treat them as a number. On the other hand, you can use the "eq" operator to test equality as a string. (All the numerical comparison operators have string versions)

Code:
$number = 2;
$number_string = "02";

if ($number == $number_string) {
    print "Variables are numerically equal.\n";
}
if ($number eq $number_string) {
    print "Variables are equivalent strings.\n";
}
Run this and you'll see that the first statement is printed but not the second.
 
Old 04-03-2005, 11:43 AM   #6
MetaPhase
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Question How are "!-z" and "-n" different?

Quote:
Originally posted by Xris718
oh what about this convert into perl?
ie:

if [! -z $string]; then
something....
How is the test:
Code:
[! -z $string]
different from the test
Code:
 [-n $string]
?
 
  


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