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IPstacks has asked for the wisdom of the Perl Monks concerning the following question:
I have a script that reads a file one line at a time and
searches for "--foo" and print's everything after that. I
have that part working fine, but, when it prints the output
it has everything on the line after --foo
At the end of the line in the file is a TAB followed by a \
I Think that I need to use a regular expression to tell it
to ignore the TAB \ but I'm not sure exactly how to word
it as I'm new to this. Also, the output is in quoted (") text
and I would like to strip off the quotes around the text.
Is regex what I want to use? |
Re: Regex question
by monkfish (Pilgrim) on Nov 14, 2001 at 22:47 UTC
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If you don't want to print a line with a tab and a slash put this in your loop...
print $_ unless ($_ =~ m@^\t\\$@);
If you want to stop printing when you get to a line like that you could use:
last if ($_ =~ m@^\t\\$@);
Edit Changed / to \\
-monkfish (The Fishy Monk)
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Re: Regex question
by jeroenes (Priest) on Nov 14, 2001 at 23:28 UTC
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If the tab and the '\' are always at the end of a line, you can use the lvalue of substr:
substr($line,-3,2)=''; #assuming your line ends with a newline
$line =~ tr/"//d; #deletes all quotes in your text
If you want to take care of nested quotes, take a look at
Parse::RecDescent or the likes.
Cheers,
Jeroen
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Re: Regex question
by c (Hermit) on Nov 14, 2001 at 23:37 UTC
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one other path you may want to head down is getting rid of your tab and following slash using the s/// operator:
my $string = "hello\t\\";
$string =~ s/\t\\//g;
this should return the value of $string as simply hello
humbly -c | [reply]
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Re: Regex question
by atlantageek (Monk) on Nov 15, 2001 at 09:47 UTC
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Regex will work
assumming that the content is in $line
$line =~ /"(.*)\t\\"/;
I always wanted to be somebody... I guess I should have been more specific. | [reply]
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and also use $1 now for your value.
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