Applying a regular expression to a string

mw has asked for the wisdom of the Perl Monks concerning the following question:

Greetings Monks, This must be an extremely easy question to answer, but much searching has not yielded the answer I seek. I am parsing messages from TSM, which have a format like this:

ANE4954I Total number of objects backed up: 3

I need the "ANE4951I" part of the string, which can be found by simply applying the following regular expression: s/\s.*// However, I don't wish to change the original text, so I reasoned that as $a=+1 is equivalent to $a=$a+1, so $a=~s/\s.*// should be equivalent to $a = $a ~ s/\s.*//, enabling me to express my needs thus:

$message_id=$message ~ s/\s.*//;
[download]

Sadly, this gives me a syntax error. What is the correct syntax for this? For now, I have coded it like this:

$message_id=$message;
$message_id=~s/\s.*//;
[download]

But I suspect there is a better way. Thanks,
Menno

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Re: Applying a regular expression to a string by ysth (Canon) on Apr 04, 2005 at 11:54 UTC
=~ is not related to the ~ operator; the assignment variant of ~ would be ~=, except that ~ is a unary operator and so has no assignment variant. What you want is to select everything before the first \s, and assign that to $message_id; use a match, not a substitution, to do that: `($message_id) = $message =~ /^(\S*)/;` [download] (Note the parentheses around $message_id that make it a list assignment; you need to do the ~~assignment~~match in list context to have it return anything besides success/failure.)	[reply] [d/l]
Re^2: Applying a regular expression to a string by mw (Sexton) on Apr 07, 2005 at 07:50 UTC
Simple and elegant. I will start using this construct as soon as I next come across this type of problem. Thank you all.	[reply]
Re: Applying a regular expression to a string by ambs (Pilgrim) on Apr 04, 2005 at 11:42 UTC
If $a=+1 is not the same as $a=$a+1. This is just true in case $a is undef or zero. So, you might want to say $a+=1 is the same as $a=$a+1. In any case, $a=$a+1 changes $a, so that is not what you want. Your second code snipet is correct. You can also use `($message_id = $message)=~s/\s.//;` [download] Alberto Simões*	[reply] [d/l]
Re: Applying a regular expression to a string by BazB (Priest) on Apr 04, 2005 at 11:46 UTC
Try something like: `my $message_id = $1 if $message =~ m/([\w\d]+)\s+.*/;` [download] If the information in this post is inaccurate, or just plain wrong, don't just downvote - please post explaining what's wrong. That way everyone learns.	[reply] [d/l]
Re^2: Applying a regular expression to a string by Smylers (Pilgrim) on Apr 04, 2005 at 13:22 UTC
`my $message_id = $1 if $message =~ m/([\w\d]+)\s+./;` Putting `my` (a compile-time directive) with an `if` statement modifier (which can only take place at run-time) is confusing, and therefore deprecated. Also, there's no point at all in having `.` at the end of the regexp: since one of the things it would happily match is nothing, it isn't making any restrictions at all about what can follow the `\s+`, so is just adding noise. Finally, the `\w` class already includes digits, so writing `[\w\d]` doesn't add anything. Here's how I might have written this: `$message =~ /^(\w+)/ or die "Message '$message' doesn't start with an identifier.\n"; my $message_id = $1;` [download] `die` might not be the most appropriate thing to do here; adjust as appropriate. But if you're adamant that `$message` could never not match the pattern, then a `die` test is better than nothing: if you're right, then it'll never happen, and if you're wrong then your assumption was wrong so there's a problem elsewhere and at least you found out about it. Smylers	[reply] [d/l] [select]
Re: Applying a regular expression to a string by Zaxo (Archbishop) on Apr 04, 2005 at 11:45 UTC
How about this? `my $message_id = do { local $_ = $message; (split)[0]; };` [download] After Compline, Zaxo	[reply] [d/l]

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