In general, chomp() can be trusted to remove a trailing carriage return from your string. If chomp() doesn't work, then perhaps your string doesn't contain what you think it does.

A few obvious problems with your regular expression:

• you are quoting it AND using the forward-slash characters to delimit your regex
• you seem to be looking for a literal '$' character at the end of your string

If I was looking for a series of word characters, I might prefer something like this, which looks for one or more characters in the word character class at the beginning of the string, and greedily captures them all into the $1 variable:

  my $trimmed_user;
  if ($user =~ /^(\w+)/) {
    $trimmed_user = $1;
  }
[download]

Of course, this assumes that you won't have any characters in a user name that are not in the word character class (A-Za-z0-9_).

Thank you ptum. Your recommendation did the trick. Thank you for your help.

if ( $user =~ /(.*)\^/ ) {
  my $user_name = $1;
}
[download]

Phil

For whatever reason when I paste the example into the thread it doesn't display right..

But the results that I am getting is the user name "SmithB" then on the end of that it adds a square or carriage return, then adds a carrot "^" on the end of that.

I need to extract the username (without the carriage return and ^) and write that to a text file.

Here the whole script... I am creating a list of the parameters being passed. Then I need to take ARGV 7 and extract the user name minus the carriage return and /^

open (TEST,">c:/temp/argtest6.txt") or die "Failed to open argtest.txt
+ file";

$count = 0;
foreach $element (@ARGV) {
print TEST $count,$element;
$count += 1;
}

$user = lc("$ARGV[7]");
chomp ($user);
if ($user =~ "/^\........\d\$/\n") {
        print TEST ".. You have a match for $user /\n";
}
 else {
         print TEST "... NO match for $user /\n";
}

print TEST $user;
close(TEST)
[download]

Sorry guys, I am new at this, so please bear with me.... And thanks for the help

Code tags and general formatting added by GrandFather

This looks a little better

open (TEST,">c:/temp/argtest6.txt") or die "Failed to open argtest.txt
+ file"; 
$count = 0; 
foreach $element (@ARGV) { print TEST $count,$element; $count += 1;
}
$user = lc("$ARGV7");
chomp ($user);
if ($user =~ "/^\........\d\$/\n")
{ print TEST "
[download]

howcome my thread is compacting so much? Is there something I am missing to make it easier to read?

If you want to remove an end of line character from the end of a string then chomp is your best bet. If that's not working then you need to tell us more about the data you're getting. Perhaps there's something else there other than your operating system's end of line character.

And I'm slightly confused by the rest of your question. You say that you want to remove '^' from the end of the string, but the sample data you give doesn't have that character in it.

I think that in order for us to be any real help to you, you'll need to expand a bit on what you're trying to do. Give us a complete (but small) example program that we can run and explain what unexpected behaviour you are seeing.

Not such a smart thing generally. Chop will remove the last character regardless of what it was. If it was a multiple character sequence then you have stuffed things for chomp - it won't remove the mutliated line end sequence. If it was a single character then the chomp is not required. In neither case is chomp going to do anything useful following chop.

chomp first to remove the line end sequence (which may comprise many characters if $/ has been altered) and then chop if you really want to always remove the last character (doesn't happen often actually).

---

DWIM is Perl's answer to Gödel

I don't agree. From the OP's example it is clear his string is built as follows: 'some_name' + 'EOL-character(s)' + '^'.

chop removes the '^' and then chomp can do its usual job of handling the 'EOL-character(s)'.

CountZero

"If you have four groups working on a compiler, you'll get a 4-pass compiler." - Conway's Law