get an input value in url?

i’am trying to get the input value in a url but the methord i use is not working can any one help pls.

<form action="" method="get">
            <input name="parrinName" type="text"/>
        </form>
        <center>
        <a target="_blank" href="index.php?page=impression/lettre_carte_cadeau&IDClient=<?php echo $_GET['id'], $_GET['parrinName']; ?>&print&output=pdf">
        <img src="<?=IMAGES_URL?>fleche.jpg" border="0" align="middle"> Imprimer lettre carte cadeau</a>
        </center>


Source: stackoverflow-php

Extract URL’s from a string using PHP if any of two special symbols in the URL should be treated as delimiters (the first characters followed by URL)?

To extract URLs (not a perfect solution but I’m almost satisfied as performance counts) I use

preg_match_all('#bhttps?://[^,s()<>]+(?:([wd]+)|([^,[:punct:]s]|/))#', $string, $match);

code to extract URLs.
However, it’s not a perfect solution for me as URLs should be forced to cut up to ] or "|" if any of these two symbols met in the extracted URL.

I know these symbols are valid symbols in URLs, however for my case they should be invalid.
How should the preg_match_all above be slightly modified to know about these two delimiters?
Thank you.


Source: stackoverflow-php

"open a website without really opening" on Javascript

atm I am trying to make an Brute Force password cracker for testing. I decided to use Javascript and I have an idea how to make the password generating thing.
The test page I try to enter uses the GET method, so I can easily enter the generated password in the URL. And the break condition would be something like comparing the URL with the password inserted and the actual URL after submitting it.
But my problem is I have no clue how to check the URL without open the actual website, because this option would slower the hole brute force process so much down that it would be as fast as entering the password myself. So how could I check if the URL is redirecting you to another URL without actual opening it?


Source: stackoverflow-javascript