[SOLVED] Using regex to get data between two quotation marks

[BenGoldberg]

I'm really horrible with regex but I need to use it for a site that I'm building, so that's why I'm here. Basically, I need to extract data between two quotation marks and store the value in a variable. An example of a string that I need to work on is...

 

deg_F = "---.-"

 

But I need to do this for quite a few lines. The cool thing is for every one, the data is inside the quotation marks. Another example...

 

percent = "--"

 

So the quotation marks don't always start at the same point since the strings are different, but the data I need is always inside the quotation marks. Any help would be greatly appreciated!

[Adam]

This will return an array off all the values within quotes:

 

preg_match_all('/"[^"]"/', $str, $matches);

print_r($matches);

[BenGoldberg]

I entered in a valid string with quotation marks, but I'm just getting an empty array for $matches. I guess it's not matching anything.  ???

[sasa]

try

preg_match_all('/"[^"]*"/', $str, $matches);

print_r($matches);

[Adam]

Ah yes my mistake, missed the * or +!

[BenGoldberg]

Thanks you guys, it works just how I expect it to! The only thing is, is it possible to not copy the quotation marks? I just want the data inside them.

[.josh]

preg_match_all('/"([^"]*)"/', $str, $matches);

print_r($matches[1]);

[BenGoldberg]

Ah yes, perfect. Thanks everyone!