得到放大器 使用http_build_query的符号

Question

我使用curl从website1向website2提交一些数据.

当我通过接收端提交数据时,我得到它

Array
(
    [ip] => 112.196.17.54
    [amp;email] => test@test.com
    [amp;user] => test123,
    [amp;type] => point
    [amp;password] => password
)

根据我的http_build_query()产生错误的结果.

"ip"字段是正确的休息是不正确的.

请让我知道为什么会这样.

curl函数如下:http_build_query($ config)

function registerOnPoints($username ,$password,$email,$ip ,  $time )
{

    $ch = curl_init("http://website2c.com/curl-handler");

    curl_setopt(

    $ch, CURLOPT_RETURNTRANSFER, 1);



    $config = array( 'ip' => $ip, 
                     'user' => $username, 
                     'email' => $email,
                     'password'=> $password,
                     'time' =>  $time,
                     'type' => 'point') ;

    # add curl post data                 
    curl_setopt($ch, CURLOPT_POSTFIELDS, http_build_query($config));
    curl_setopt($ch, CURLOPT_POST, true);

    # execute 
    $response = curl_exec($ch);

    # retreive status code
    $http_status = curl_getinfo($ch , CURLINFO_HTTP_CODE);

   if($http_status == '200')
   {
      $response = json_decode($response);

    } else {
       echo $http_status;
    }


   // Close handle
   curl_close($ch);

}

如果它是php版本问题那么,说清楚地说我没有权限更改php的版本因为只有curl函数产生错误休息项目完成并按预期工作.请帮我.

Answer 1

我想你可以试试:

http_build_query($config, '', '&');

或替代方案:

$paramsArr = array();

foreach($config  as $param => $value) {
   $paramsArr[] = "$param=$value";
}

$joined = implode('&', $paramsArr);
//and use
curl_setopt($ch, CURLOPT_POSTFIELDS, $joined);