php scraping excat search with quotation marks "batman kill a human" - php

Question

I have a little PHP script for scraping google. I wnat to google the exact search "batman kill a human" I pass the param with simple quoation marks "batman kill a human"
and I can see a correct URL generated.
I call the script with the param '"batman kill a human"', and I get >Error 400 (Bad Request)
example
root#ubuntu:/var/www/html# php ejemplo.php '"batman kill a human"'
------------- [https://www.google.es/search?q="batman kill a human"]
------------- [
Error 400 (Bad Request)!!1 *{margin:0;padding:0}html,code{font:15px/22px arial,sans-serif}html{background:#fff;color:#222;padding
....
...
-------------------------------------------
if i try to use this url in firefox, https://www.google.es/search?q="batman kill a human" y get a correct answer from google.
Why i don't get a correct answer when I'm trying it from the php scritp
this is the source code.
<?php
include('simple_html_dom.php');
function file_get_contents_curl($url) {
/*
This is a file_get_contents replacement function using cURL
One slight difference is that it uses your browser's idenity
as it's own when contacting google.
*/
$ch = curl_init();
// curl_setopt($ch, CURLOPT_USERAGENT, $_SERVER['HTTP_USER_AGENT']);
curl_setopt($ch, CURLOPT_HEADER, 0);
curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1);
curl_setopt($ch, CURLOPT_FOLLOWLOCATION , 1);
curl_setopt($ch, CURLOPT_URL, $url);
$data = curl_exec($ch);
curl_close($ch);
return $data;
}
$busca=$argv[1];
$cadena="https://www.google.es/search?q=$busca";
print "\n------------- [$cadena]";
$data=file_get_contents_curl($cadena);
$html = str_get_html($data);
print "\n------------- [$html]";
$html->clear(); exit();
?>

Answer 1

Check out the PHP manual on urlencode()
$busca=$argv[1];
becomes
$busca=urlencode($argv[1]);
Which renders as:
https://www.google.es/search?q=batman+kills+a+human