Curl Post +错误处理+设置头[感谢@mantas-d]:

function curlPost($url, $data=NULL, $headers = NULL) {
    $ch = curl_init($url);
    curl_setopt($ch, CURLOPT_RETURNTRANSFER, true);

    if(!empty($data)){
        curl_setopt($ch, CURLOPT_POSTFIELDS, $data);
    }

    if (!empty($headers)) {
        curl_setopt($ch, CURLOPT_HTTPHEADER, $headers);
    }

    $response = curl_exec($ch);

    if (curl_error($ch)) {
        trigger_error('Curl Error:' . curl_error($ch));
    }

    curl_close($ch);
    return $response;
}


curlPost('google.com', [
    'username' => 'admin',
    'password' => '12345',
]);

可以通过以下方式轻松到达:

<?php

$post = [
    'username' => 'user1',
    'password' => 'passuser1',
    'gender'   => 1,
];
$ch = curl_init();
curl_setopt($ch, CURLOPT_URL, 'http://www.domain.com');
curl_setopt($ch, CURLOPT_RETURNTRANSFER, true);
curl_setopt($ch, CURLOPT_POSTFIELDS, http_build_query($post));
$response = curl_exec($ch);
var_export($response);

<?php
//
// A very simple PHP example that sends a HTTP POST to a remote site
//

$ch = curl_init();

curl_setopt($ch, CURLOPT_URL,"http://www.example.com/tester.phtml");
curl_setopt($ch, CURLOPT_POST, 1);
curl_setopt($ch, CURLOPT_POSTFIELDS,
            "postvar1=value1&postvar2=value2&postvar3=value3");

// In real life you should use something like:
// curl_setopt($ch, CURLOPT_POSTFIELDS, 
//          http_build_query(array('postvar1' => 'value1')));

// Receive server response ...
curl_setopt($ch, CURLOPT_RETURNTRANSFER, true);

$server_output = curl_exec($ch);

curl_close($ch);

// Further processing ...
if ($server_output == "OK") { ... } else { ... }
?>

最简单的是以application/json的形式发送数据。这将接受一个数组作为输入，并正确地将其编码为json字符串:

$data = array(
    'field1' => 'field1value',
    'field2' => 'field2value',
)

$ch = curl_init($url);
curl_setopt($ch, CURLOPT_POST, true);
curl_setopt($ch, CURLOPT_POSTFIELDS, json_encode($data));

curl_setopt($ch, CURLOPT_HTTPHEADER, array(
    'Content-Type:application/json',
));
curl_setopt($ch, CURLOPT_RETURNTRANSFER, true);
$resultStr = curl_exec($ch);
return json_decode($resultStr, true);

如果表单使用重定向、身份验证、cookie、SSL (https)或其他任何东西，而不是期望POST变量的完全开放脚本，那么您将很快开始咬牙切齿。看看Snoopy，它完全按照您的想法来做，而且不需要设置大量的开销。

我很惊讶没有人建议file_get_contents:

$url = "http://www.example.com";
$parameters = array('username' => 'user1', 'password' => 'passuser1', 'gender' => '1');
$options = array('http' => array(
    'header'  => 'Content-Type: application/x-www-form-urlencoded\r\n',
    'method'  => 'POST',
    'content' => http_build_query($parameters)
));

$context  = stream_context_create($options);
$result = file_get_contents($url, false, $context);

这很简单，很有效;我把它用在两端都能控制代码的环境中。

更好的方法是使用json_decode(并设置代码以返回JSON)

$result = json_decode(file_get_contents($url, false, $context), TRUE);

这种方法在幕后调用curl，但不需要经历那么多困难。

从Stack Overflow上其他地方的原始答案提炼出来的答案: PHP将变量发送到file_get_contents()