[SOLVED] Proper way to parse (prepare) URL for curl

Issue

What is proper way to parse URLs for curl?

Let's say i have given URL

$url='https://example.com/some url/something?x=1&y=foo/bar#abc';
ch=curl_init();
curl_setopt($ch, CURLOPT_URL, $url);
//set all needed more curl parameters
$result=curl_exce($ch);

if i put that king of URL in browser, it'll work find. But for me it always returns 404 if it has any non-standard URL characters

sure, i could manually replace spaces with %20 and so on, but that won't do.

curl_escape() will not do too, because it will escape also all slashes and other characters, that should not be escaped in URL, but should be in GET parameters list.

I guess i should probably split this string to parts and threat those separately with differently, but I'm stuck here.

Solution

Try this, I parsed the URL into its components, individually encoded the path and query parameters, and then reconstructed the URL for proper handling in a cURL request in PHP.

  $url = 'https://example.com/some url/something?x=1&y=foo/bar#abc';
    
    $parsedUrl = parse_url($url);
    
    $encodedPath = implode("/", array_map("rawurlencode", explode("/", $parsedUrl['path'])));
    
    if (isset($parsedUrl['query'])) {
        parse_str($parsedUrl['query'], $queryParams);
        $encodedQuery = http_build_query($queryParams);
    } else {
        $encodedQuery = '';
    }
    
    $encodedUrl = $parsedUrl['scheme'] . '://' . $parsedUrl['host'] . $encodedPath;
    if (!empty($encodedQuery)) {
        $encodedUrl .= '?' . $encodedQuery;
    }
    if (isset($parsedUrl['fragment'])) {
        $encodedUrl .= '#' . $parsedUrl['fragment'];
    }
    
    $ch = curl_init();
    curl_setopt($ch, CURLOPT_URL, $encodedUrl);
    $result = curl_exec($ch);
    if ($result === false) {
        echo 'cURL Error: ' . curl_error($ch);
    }
    curl_close($ch);

Answered By - TSCAmerica.com

Answer Checked By - Mary Flores (WPSolving Volunteer)