Regex Expressions

[John_L]

Just started using regex expressions but i'm stumped when it comes to the following. I'm parsing data and I want to strip away the following tag found in xml files but leave the text it surrounds intact.

<![CDATA[ text ]]>

It's the text inside I would like to keep. I tried searching the net for the proper way but all i've come up with is a way to strip the tag along with what is contained inside it. If I break up my single expression into two and looking it as two regex substitution calls, I get compile time errors...Thanks in advance.

what I have now, that strips away the tag and what's inside it...

$string =~ s/<![CDATA[]]//i; #substitute it with nothing (strip it away)

[WingedPanther]

The way I would approach this is as follows:
There are three sections of your code: "<![CDATA[ ", text, and " ]]>"
If you wrap the middle in parenthesis, as (.*), you can return it using $1 (or similar)

[John_L]

won't that only work if i'm trying to find a match? I can't do that with the substitution expression can I? What if I have multiple CDATA tags to worry about? I'll try playing around with it but i'm not entirely sure I will be writing this out correctly.

[KevinADC]

$string =~ s/<!\[CDATA\[(.*?)\]\]>/$1/gis;

[John_L]

I tried what you suggested, but it still seems to just remove the tag along with what's inside it. Are you sure that's what it's supposed to be?

[KevinADC]

The regexp works.

$string = '<![CDATA[this is a test]]>';
$string =~ s/<!\[CDATA\[(.*?)\]\]>/$1/gis;
print $string;

The problem is more than likely because your pattern is spread over multiple lines. Post a sample of the real data.

[John_L]

your right, I got it working correctly. Thanks! Can that same code segment for the substitution be used on lets say a string with quotation marks? Cause I would like to just use it again but for a string that has something like

<some element url = "www.gosomewhere.com" type = "some type" />

and use the expression you gave me to grab the url between the first set of parenthesis. Would I have to encode the quotation marks though?

$string =~ s/"(.*?)"/$1/is; #wouldn't work like this, would it?

[KevinADC]

try it and see.

[John_L]

nah, doesn't seem to work. I thought putting a couple slashes in '\' would let me include the quotation marks, but that doesn't seem to work either. Should I be using some kind of special characters instead of quotation marks?

[KevinADC]

unless you want to modify the string there is no need to use a s/// regexp, just use m// to find the pattern and assign it to $1 or to a scalar:

$string = '<some element url = "www.gosomewhere.com" type = "some type" />';
$string =~ /"(.*?)"/;
print  $1;

$string = '<some element url = "www.gosomewhere.com" type = "some type" />';
($domain) = $string =~ /"(.*?)"/;
print  $domain;

The parenthesis are important in the above code.