PHP - 无法使用包含变量的URL打开流

Question

I'm trying to use file_get_contents but it tells me failed to open stream.

My code:

$user="[e-mail]";
$user_id=str_replace(array('@', '#'), array('%40', '%23'), $user);
print $user_id;

$url=('http://admin:password@172.16.214.133/@api/users/=$user_id/properties');
$xmlString=file_get_contents($url);

This is what I get when I try to run it:

Warning: file_get_contents(http://...@172.16.214.133/@api/deki/users/=$user_id/properties): failed to open stream: HTTP request failed! HTTP/1.1 500 Internal Server Error

However, if I manually type in the $user_id first_last%40ourwiki.com then it works! What am I doing wrong? Shouldn't I be able to just use the variable name?

Remaining code:

$delete = "http://admin:password@172.16.214.133/@api/users/=$user_id/properties/%s";
$xml = new SimpleXMLElement($xmlString);

function curl_fetch($url,$username,$password,$method='DELETE')
{
    $ch = curl_init();
    curl_setopt($ch, CURLOPT_URL, $url);
    curl_setopt($ch, CURLOPT_CUSTOMREQUEST, $method);
    curl_setopt($ch, CURLOPT_RETURNTRANSFER, TRUE);
    curl_setopt($ch,CURLOPT_USERPWD,"$username:$password");
    return  curl_exec($ch);
}

foreach($xml->property as $property) {
  $name = $property['name'];
  $name2 =str_replace(array('@', '#'), array('%40', '%23'), $name);
  print $name2;
  curl_fetch(sprintf($delete, $name2),'admin','password');
}

Answer 1

This is because you have used single quotes. The content within single quotes is not parsed, so:
echo '$test';
won't display the value of the $test variable, but just the "$test" string. You can use double quotes instead, but anyway this is the best way to do it:

$url=('http://admin:password@172.16.214.133/@api/users/='.$user_id.'/properties');

Special characters such as , \t or also won't be parsed in single quotes.