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Home: Perl Programming Help: Regular Expressions:
Help with an if regx statement

 


seahorse
Novice

Jan 30, 2007, 8:32 AM

Post #1 of 12 (7418 views)
Help with an if regx statement Can't Post
Please help me figure out how to write the 2nd part of the 2nd If statement below that I have written in plain english. The $in files represent examples of what is good input and what is bad input. Thanks in advance.

$in-bad1 = "[0],[0,1],[1,0,3,5]";

$in-bad2 = "[[0],[0,2]]";

$in-bad3 = "[[0]stuff,morestuff[0,4]]";

$in-good1 = "[0][0,1,2][0,1]";

$in-good2 = "[0]stuff[0,1][1,0,2,5]";

if ($in-bad1 =~ /\,/)) {


If there are , (comma) character(s) then they all must be located between a set of [ ] (bracket) characters.


}


KevinR
Veteran


Jan 30, 2007, 11:26 AM

Post #2 of 12 (7414 views)
Re: [seahorse] Help with an if regx statement [In reply to] Can't Post
Code
if ($line =~ /\],\[/) { 
   print "it's bad"; 
}
-------------------------------------------------


seahorse
Novice

Jan 30, 2007, 1:27 PM

Post #3 of 12 (7411 views)
Re: [KevinR] Help with an if regx statement [In reply to] Can't Post
Thanks for your respone KevinR, however $inbad3 fails (says it's good when it should be bad). Below is a better version of the program thus far so please ignore the 1st one. Set $test to any of the $in strings to see what is good or bad. What I have labeled 'stuff' and 'morestuff' can be any number of characters except [ or ] or , (comma) characters.

Also, another rule is that the $in strings cannot be modified to 'fix' the problem.

$inbad1 = "[0],[0,1],[1,0,3,5]";

$inbad2 = "[[0],[0,2]]";

$inbad3 = "[[0]stuff,morestuff[0,4]]";

$ingood1 = "[0][0,1,2][0,1]";

$ingood2 = "[0]stuff[0,1][1,0,2,5]";

$test = $inbad3;

if ($test =~ /\,/) {

# If there are , (comma) character(s) then they all must be located between a set of [ ] (bracket) characters.


if ($test =~ /\],\[/) { print "it's bad"; } else { print "it's good"; }


}


(This post was edited by seahorse on Jan 30, 2007, 1:32 PM)


ProBulletin
Novice

Jan 30, 2007, 1:53 PM

Post #4 of 12 (7406 views)
Re: [seahorse] Help with an if regx statement [In reply to] Can't Post
There are better ways than this but it is one way at least. Unless I've overlooked something simple it was trickier than it first appeared.

Code
   
my $open = 0;   
LOOP: while ($test =~ /(.)/g) {   
    my $char = $1;  
    if ($char eq '[' or $char eq ']') {    
        $open = $char eq '[' ? 1 : 0;    
    }    
    else {  
        if ($char eq ',' and $open == 0) {  
            print "INVALID";  
            last LOOP;  
        }  
    }  
}

Regards,
Paul Wilson
ProBulletin Board: http://www.probulletin.com/


(This post was edited by ProBulletin on Jan 30, 2007, 1:53 PM)


seahorse
Novice

Jan 30, 2007, 2:30 PM

Post #5 of 12 (7402 views)
Re: [ProBulletin] Help with an if regx statement [In reply to] Can't Post
Kudos ProBulletin! It seems that you nearly got it. However, it fails the following case.

$inbad6 = "[0]stuff[0,1][1,0,2,5][stuff,morestuff[";

Is there a quick fix for that case? Sorry about the confusion on stuff and morestuff. I know I wasn't real clear on what I wanted with them.


(This post was edited by seahorse on Jan 30, 2007, 3:45 PM)


KevinR
Veteran


Jan 30, 2007, 2:51 PM

Post #6 of 12 (7400 views)
Re: [seahorse] Help with an if regx statement [In reply to] Can't Post
"stuff" and "morestuff" is hardly a good description for coming up with pattern matches. Wink

Code
$inbad1 = "[0],[0,1],[1,0,3,5]"; 
 
$inbad2 = "[[0],[0,2]]"; 
 
$inbad3 = "[[0]stuff,morestuff[0,4]]"; 
 
$ingood1 = "[0][0,1,2][0,1]"; 
 
$ingood2 = "[0]stuff[0,1][1,0,2,5]"; 
 
$test = $inbad3; 
 
for ($inbad1,$inbad2,$inbad3,$ingood1,$ingood2) { 
  if (/\](\w*,\w*)+\[/) { print "$_: it's bad\n"; } else { print "$_: it's good\n"; } 
}
-------------------------------------------------


seahorse
Novice

Jan 30, 2007, 3:30 PM

Post #7 of 12 (7396 views)
Re: [KevinR] Help with an if regx statement [In reply to] Can't Post
Thanks for the 1-liner. Sorry about the confusion on stuff and morestuff. I know I wasn't real clear on what I wanted with them. Yours is close however there are still some cases where it fails, namely $inbad4 & $inbad5. Then there is the comma at the end and the [stuff,morestuff[ case. How would you fit these into your 1-liner? BTW, I also found ProBulletin's solution fails $inbad6

$inbad6 = "[0]stuff[0,1][1,0,2,5][stuff,morestuff[";



$inbad1 = "[0],[0,1],[1,0,3,5]";

$inbad2 = "[[0],[0,2]]";

$inbad3 = "[[0]stuff,morestuff[0,4]]";

$inbad4 = ",[0][0,1,2][0,1]";

$inbad5 = "[0]stuff[0,1][1,0,2,5]]stuff,morestuff]";

$ingood1 = "[0][0,1,2][0,1]";

$ingood2 = "[0]stuff[0,1][1,0,2,5]";

for ($inbad1,$inbad2,$inbad3,$inbad4,$inbad5,$ingood1,$ingood2) {


if (/\](\w*,\w*)+\[/) { print "$_: it's bad\n"; } else { print "$_: it's good\n"; }


}


(This post was edited by seahorse on Jan 30, 2007, 3:44 PM)


KevinR
Veteran


Jan 30, 2007, 4:49 PM

Post #8 of 12 (7391 views)
Re: [seahorse] Help with an if regx statement [In reply to] Can't Post
I give up.Crazy
-------------------------------------------------


ProBulletin
Novice

Jan 31, 2007, 2:39 AM

Post #9 of 12 (7383 views)
Re: [seahorse] Help with an if regx statement [In reply to] Can't Post
Try:

Code
 my $open = 0;      
my $comma = 0;  
LOOP: while ($test =~ /(.)/g) {          
    my $char = $1;    
    if ($char eq '[' or $char eq ']') {               
        $open = $char eq '[' ? 1 : 0;           
        $comma = 0 if ($char = ']');  
    }           
    else {             
        if ($char eq ',' and $open == 0) {                     
            print "INVALID";                 
            last LOOP;             
        }     
        elsif ($char eq ',' and $open == 1) {  
            $comma = 1;  
        }      
    }     
}   
print "INVALID" if ($open and $comma);

Regards,
Paul Wilson
ProBulletin Board: http://www.probulletin.com/


(This post was edited by ProBulletin on Feb 1, 2007, 5:10 PM)


KevinR
Veteran


Jan 31, 2007, 4:10 PM

Post #10 of 12 (7374 views)
Re: [seahorse] Help with an if regx statement [In reply to] Can't Post
another approach:

Code
$s1 = "[0],[0,1],[1,0,3,5]"; 
$s2 = "[[0],[0,2]]"; 
$s3 = "[[0]stuff,morestuff[0,4]]"; 
$s4 = ",[0][0,1,2][0,1]"; 
$s5 = "[0]stuff[0,1][1,0,2,5]]stuff,morestuff]"; 
$s6 = "[0]stuff[0,1][1,0,2,5][stuff,morestuff["; 
$s7 = "[0][0,1,2][0,1]"; 
$s8 = "[0]stuff[0,1][1,0,2,5]"; 
my $n = 1; 
for ($s1,$s2,$s3,$s4,$s5,$s6,$s7,$s8) { 
   my $r = $_; 
   s/\[[^\]]*\]//g; 
   my $p = (/,/)? BAD: GOOD;   
   print "string \$s$n) $p: $r\n"; 
   $n++; 
}
output:
string $s1) BAD: [0],[0,1],[1,0,3,5]
string $s2) BAD: [[0],[0,2]]
string $s3) BAD: [[0]stuff,morestuff[0,4]]
string $s4) BAD: ,[0][0,1,2][0,1]
string $s5) BAD: [0]stuff[0,1][1,0,2,5]]stuff,morestuff]
string $s6) BAD: [0]stuff[0,1][1,0,2,5][stuff,morestuff[
string $s7) GOOD: [0][0,1,2][0,1]
string $s8) GOOD: [0]stuff[0,1][1,0,2,5]
Code
 
but if the occurances and frequencies of [] can be anywhere in the string in any combination this might prove to be less than 100% accurate. Pauls suggestion above might work better in that case. 
-------------------------------------------------


ProBulletin
Novice

Jan 31, 2007, 4:15 PM

Post #11 of 12 (7372 views)
Re: [KevinR] Help with an if regx statement [In reply to] Can't Post
Just a minor point, but I would use index() instead of (/,/)


Regards,
Paul Wilson
ProBulletin Board: http://www.probulletin.com/


KevinR
Veteran


Jan 31, 2007, 11:04 PM

Post #12 of 12 (7369 views)
Re: [ProBulletin] Help with an if regx statement [In reply to] Can't Post
using index() is a good suggestion. Cool

edit:

BTW, nice to have another person take interest in this forum and post replies.

Regards,
Kevin
-------------------------------------------------


(This post was edited by KevinR on Jan 31, 2007, 11:09 PM)

 
 


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