php正则表达式在文件中查找字符串

Question

I am trying to make a php script to output all values inside an hypehref=" * * " from a text file.

I'm having a hard time with the regular expression part.

This is what I have

$Vdata = file_get_contents('file.txt');
preg_match( '/hyperef="(.*?)"/', $Vdata, $match );
echo '<pre>'; print_r($match); echo '</pre>'

My result is this :

Array
(
    [0] => hyperef="http://lookbook.nu/look/5709720-Choies-Silvery-Bag-Rosewholesale-Punk-Style/hype"
    [1] => http://lookbook.nu/look/5709720-Choies-Silvery-Bag-Rosewholesale-Punk-Style/hype
)

The [0] is incorrect, it includes the part I am searching for... all I want is the result after the hypehref="

The second result [1] is correct

and my file should have given me about 10 results.. not just 2...

Any ideas why ? Thx

Answer 1

You can use preg_match_all to find all matches. There you will also have the full part and only the value of the hyperef - but you can only use the former.

if (preg_match_all('/hyperef="(.*?)"/i', $Vdata, $result)) {
    $matches = $result[1]; //only values inside quotation marks
    print_r($matches);
} else
    print "nothing found";

I added the if for obvious reasons and the i delimiter, so the pattern will ignore case sensitivity.

Answer 2

$pattern = "/\bo'reilly\b/i"; // only O'Reilly
$ora_books = preg_grep($pattern, file('filed.txt'));

var_dump($ora_books);

example 2

$fh = fopen('/path/to/your/file.txt', 'r') or die($php_errormsg);
while (!feof($fh)) {
$line = fgets($fh);
if (preg_match($pattern, $line)) { $ora_books[ ] = $line; }
}
fclose($fh);