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perl--logical help needed

13 New Member
I need a help with my code...
What I want to do is:
if there exists a directory_no, and found in the directories array , want to copy the contents of that array to a new array from where the directory_no is found to till the end of that array.
In the end copy the new_directory array to the directories array;

Ex:directories array: 1 2 3 4 5
directory_no is 3.
Output :new_directory 3 4 5
in the end directories array: 3 4 5


Expand|Select|Wrap|Line Numbers
  1. if($directory_no)
  2.   {
  3. foreach $directory (@directories)
  4.       {
  5.         $count++;
  6.         print "$count";
  7.         if($directory = $directory_no)
  8.         {
  9.               for($count;$count<=@directories;$count++)
  10.               {              
  11.               $new_directory[i]=$directories[$count-1];
  12.               $i++;              
  13.               }              
  14.           exit; 
  15.         }
  16.       }
  17. }
@directories=@n ew_directory;
Oct 13 '07 #1
9 1147
eWish
971 Recognized Expert Contributor
if($directory = $directory_no)
If you are comparing numerically then you would use ==. If you are comparing a a string then use eq.

Expand|Select|Wrap|Line Numbers
  1. if($directory eq $directory_no)
perlop
Oct 13 '07 #2
KevinADC
4,059 Recognized Expert Specialist
A very common error to use the assignment operator "=" instead of a comparison operator "==":

if ($directory = $directory_no)
Oct 13 '07 #3
KevinADC
4,059 Recognized Expert Specialist
Expand|Select|Wrap|Line Numbers
  1. @array = (1 .. 5);
  2. $n = 3;
  3. @new = grep {$_ >= $n} @array;
  4. print "@new";
Oct 13 '07 #4
bhumikas
13 New Member
Expand|Select|Wrap|Line Numbers
  1. @array = (1 .. 5);
  2. $n = 3;
  3. @new = grep {$_ >= $n} @array;
  4. print "@new";
can u xplain the third line....
Oct 13 '07 #5
Xmassey
6 New Member
@new = grep {$_ >= $n} @array;

Only values that are more than or equal to $n (which is 3) are taken from @array and placed into the new array @new
Oct 13 '07 #6
KevinADC
4,059 Recognized Expert Specialist
perldoc: grep Function
Oct 13 '07 #7
bhumikas
13 New Member
Hi,
a small clarification,e ven if i put && <=$p its gonna evaluate rite as its an expression.

Expand|Select|Wrap|Line Numbers
  1. @new = grep {$_ >= $n && <= $p} @array;
then it gonna put the values in between certain range into the new array...
Oct 16 '07 #8
KevinADC
4,059 Recognized Expert Specialist
Hi,
a small clarification,e ven if i put && <=$p its gonna evaluate rite as its an expression.

Expand|Select|Wrap|Line Numbers
  1. @new = grep {$_ >= $n && <= $p} @array;
then it gonna put the values in between certain range into the new array...
Yes, you can put any valid code into the expression thats evaluatued. Your code looks better writeen as:

Expand|Select|Wrap|Line Numbers
  1. @new = grep {$_ >= $n && $_ <= $p} @array;
of course $p needs to be defined.
Oct 16 '07 #9
bhumikas
13 New Member
Yes, you can put any valid code into the expression thats evaluatued. Your code looks better writeen as:

Expand|Select|Wrap|Line Numbers
  1. @new = grep {$_ >= $n && $_ <= $p} @array;
of course $p needs to be defined.

thnks for the suggestion.....
Oct 16 '07 #10

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