将 PHP HTTP 请求转换为 Guzzle

Question

我目前有一个旧的 PHP 页面，它对外部 API 执行 post 请求，但我想将其转换为 Guzzle 以整理它向上，但我不确定我是否在正确的路线上。

PHP POST

 function http_post($server, $port, $url, $vars)
    {
        // get urlencoded vesion of $vars array
        $urlencoded = "";
        foreach ($vars as $Index => $Value)
            $urlencoded .= urlencode($Index) . "=" . urlencode($Value) . "&";
        $urlencoded = substr($urlencoded, 0, -1); 

        $headers = "POST $url HTTP/1.0\r\n";
        $headers .= "Content-Type: application/x-www-form-urlencoded\r\n";
        $headers .= "Host: secure.test.com\r\n";
        $headers .= "Content-Length: " . strlen($urlencoded)

        $fp = fsockopen($server, $port, $errno, $errstr, 20);  // returns file pointer
        if (!$fp) return "ERROR: fsockopen failed.\r\nError no: $errno - $errstr";  // if cannot open socket then display error message

        fputs($fp, $headers);

        fputs($fp, $urlencoded);

        $ret = "";
        while (!feof($fp)) $ret .= fgets($fp, 1024);
        fclose($fp);
        return $ret;
    }

检索PHP响应

以下是您检索响应的方式，您可以在其中使用 $_POST 字段

 $response = http_post("https://secure.test", 443, "/callback/test.php", $_POST);

Guzzle 尝试POST

$client = new Client();

$request = $client->request('POST','https://secure.test.com/callback/test.php', [
    // not sure how to pass the $vars from the PHP file
    'headers' => [
        'Content-Type' => 'application/x-www-form-urlencoded',
        'Host' => 'secure.test.com'
    ]
]);

$request->getBody()->getContents();

Guzzle 尝试 GET

$client = new Client();

    $request = $client->get('https://secure.test.com/callback/test.php');
    
    $request->getBody()->getContents();

然后我如何从响应中获取特定字段？

根据我上面的尝试，我在正确的路线上吗？

Answer 1

如果你想发送$vars作为POST请求的主体，你需要设置body 属性.

$client = new Client();

// get urlencoded vesion of $vars array
$urlencoded = "";
foreach ($vars as $Index => $Value)
    $urlencoded .= urlencode($Index) . "=" . urlencode($Value) . "&";
$urlencoded = substr($urlencoded, 0, -1); 

$response = $client->request('POST','https://secure.test.com/callback/test.php', [
    'headers' => [
        'Content-Type' => 'application/x-www-form-urlencoded',
        'Host' => 'secure.test.com'
    ],
    'body' => $urlencoded,
]);

如果您使用 form_params 而不是 body，Guzzle 可以为您进行 URLencode $vars。

$response = $client->request('POST','https://secure.test.com/callback/test.php', [
    'headers' => [
        'Content-Type' => 'application/x-www-form-urlencoded',
        'Host' => 'secure.test.com'
    ],
    'form_params' => $vars,
]);

要读取服务器的响应，您需要在 $response 上调用 getBody 并将其转换为 string。您将获得与原始 http_post 函数中的 return 值完全相同的值。

$resultFromServer = (string) $response->getBody();

将 PHP HTTP 请求转换为 Guzzle

Convert PHP HTTP request to Guzzle

php

http

guzzle