Help with regex

[ShaolinF]

Hi Guys,

 

I have the following URL:

 

<a href="test.test"></a>

 

I want to remove everything except the href="test.test" bit so I did the following:

 

[^(href=\"(.*)\")]

 

but this returns href="e.e".

 

Any ideas?

[premiso]

Why not just pull out the url and then build your own <a href statement. This would surely be easier and then you do not have to worry about having a complex regex to attempt to do it.

[GoneNowBye]

[] means a class, or posible options [abc] means a or b or c

 

so....your telling regex to match not match alot of stuff....

anyway what you want is:

 

(href="(.*?)")

 

that should work fine (as your entire expression, you dont need / / at the start and the end)

 

the data between quotes will be at result_array[1] btw

 

and the * is greedy, it matches any character AS MUCH AS IT CAN you want lazy to match it as little as posible (up until the next quote) otherwise it would match for example

 

href="stuff" more "stuff" "doo" "foo" "bar"

 

greedy regex matches:

stuff" more "stuff" "doo" "foo" "bar

 

Lazy matches:

stuff

 

the *? makes it 0 or more matches - but as little as posible.

 

[ShaolinF]

[] means a class, or posible options [abc] means a or b or c

 

so....your telling regex to match not match alot of stuff....

anyway what you want is:

 

(href="(.*?)")

 

that should work fine (as your entire expression, you dont need / / at the start and the end)

 

the data between quotes will be at result_array[1] btw

 

and the * is greedy, it matches any character AS MUCH AS IT CAN you want lazy to match it as little as posible (up until the next quote) otherwise it would match for example

 

href="stuff" more "stuff" "doo" "foo" "bar"

 

greedy regex matches:

stuff" more "stuff" "doo" "foo" "bar

 

Lazy matches:

stuff

 

the *? makes it 0 or more matches - but as little as posible.

Thanks - That does work but what Im trying to do is remove everything other than the href="test.htm" bit. How would I do that ? The only way I can think of doing this is using something like [^(expression here)] but that will match on a per letter basis.

[GoneNowBye]

check your PM i think you want PREG_REPLACE

 

check your pm and i'll probably be able to help

[sasa]

<?php
$test = '<a href="test.test"></a>';
$test = preg_replace('/^.*(href="[^"]*").*$/','$1',$test);
Echo $test;
?>