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Substituting leading and trailing blanks

Posted on 1999-06-30
Medium Priority
Last Modified: 2008-02-26
I'm trying to substitute leading and trailing blanks in order to enable boolean searching.  I use the following code:

if ($input{'indexinput'}=~/or/) {
      foreach $i(@input) {
            s/^\s+//; # remove leading blanks
            s/\s+$//; # remove trailing blanks
The error log shows that the substitution is not carried out and in fact the array INPUTOR remains empty.  Why?
Question by:wilfrieds
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LVL 19

Accepted Solution

Kim Ryan earned 200 total points
ID: 1213666
You are not binding your regular expression to your loop variable $i, so it defualts to the $_ place holder. Either use:
$i =~ s/^\s+//; # remove leading blanks
$i =~ s/\s+$//; # remove trailing blanks

OR to default to $_

foreach (@input) {
s/^\s+//; # remove leading blanks
s/\s+$//; # remove trailing blanks


LVL 84

Expert Comment

ID: 1213667
#note that @input is modified, so you could have said
foreach( @input ){
     s/^\s+//; # remove leading blanks
     s/\s+$//; # remove trailing blanks
push @INPUTOR, @input;

#by the way, did you mean to say
LVL 19

Expert Comment

by:Kim Ryan
ID: 1213668
Just noticed a typo, should have said:




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LVL 84

Expert Comment

ID: 1213669
I noticed the missing @ too, but when I tried it without, it worked anyway.
Aren't prototypes amazing?

Author Comment

ID: 1213670
The sequence push(INPUTOR,$_) isn't a prototype.  The camel book says the @ is optional (p.169).
Thanks for answering my question.
LVL 84

Expert Comment

ID: 1213671
#But the new camel book doesn't say the @ is optional, because it's not quite true.
push(@time,$_); #vs push(time,$_);
push(@s,$_); #vs push(s,$_);

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