在PHP中使用cURL的RAW POST

Question

如何使用cURL在PHP中执行RAW POST？

没有任何编码的原始帖子,我的数据存储在一个字符串中.数据应格式如下:

... usual HTTP header ...
Content-Length: 1039
Content-Type: text/plain

89c5fdataasdhf kajshfd akjshfksa hfdkjsa falkjshfsa
ajshd fkjsahfd lkjsahflksahfdlkashfhsadkjfsalhfd
ajshdfhsafiahfiuwhflsf this is just data from a string
more data kjahfdhsakjfhsalkjfdhalksfd

一种选择是手动编写正在发送的整个HTTP标头,但这似乎不太理想.

无论如何,我可以只将选项传递给curl_setopt(),说使用POST,使用text/plain,并从一个$variable？发送原始数据？

Answer 1

我刚刚找到了解决方案,有点回答我自己的问题,万一其他人偶然发现它.

$ch = curl_init();

curl_setopt($ch, CURLOPT_URL,            "http://url/url/url" );
curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1 );
curl_setopt($ch, CURLOPT_POST,           1 );
curl_setopt($ch, CURLOPT_POSTFIELDS,     "body goes here" ); 
curl_setopt($ch, CURLOPT_HTTPHEADER,     array('Content-Type: text/plain')); 

$result=curl_exec ($ch);

Answer 2

使用 Guzzle 库实现：

use GuzzleHttp\Client;
use GuzzleHttp\RequestOptions;

$httpClient = new Client();

$response = $httpClient->post(
    'https://postman-echo.com/post',
    [
        RequestOptions::BODY => 'POST raw request content',
        RequestOptions::HEADERS => [
            'Content-Type' => 'application/x-www-form-urlencoded',
        ],
    ]
);

echo(
    $response->getBody()->getContents()
);

PHP CURL 扩展：

$curlHandler = curl_init();

curl_setopt_array($curlHandler, [
    CURLOPT_URL => 'https://postman-echo.com/post',
    CURLOPT_RETURNTRANSFER => true,

    /**
     * Specify POST method
     */
    CURLOPT_POST => true,

    /**
     * Specify request content
     */
    CURLOPT_POSTFIELDS => 'POST raw request content',
]);

$response = curl_exec($curlHandler);

curl_close($curlHandler);

echo($response);

源代码