如何在 Guzzle 上获得 XML 响应

Question

我一直在尝试将正文 return 的文本解析为 XML（应该 return 在 XML 中）但到目前为止它只是 return就像文字一样。这是代码以及应该是什么样子的。

    require 'vendor/autoload.php';
    use GuzzleHttp\Client;
    use GuzzleHttp\Psr7\Response;
    use GuzzleHttp\Exception\RequestException;
    use GuzzleHttp\Promise\Promise;
    use GuzzleHttp\Psr7\Request;
    $client = new GuzzleHttp\Client();
    $options = [
        'headers' => ['Accept'=>'application/xml'],
        'auth' => ['Deversor-Channel', '?1_sIWVZ((ujV)kDBM!6O5!AFAQM*K*yr(?E.(g='],
        'body' => "<request>
                        <hotel_id>345786
                        </hotel_id>
                        <last_change>2021-07-19</last_change>
                    </request>",
    ];
    $res = $client->request('POST', 'https://secure-supply-xml.booking.com/hotels/xml/reservations', $options);
    echo $res->getStatusCode();
    // "200"
    echo $res->getBody();
    // {"type":"User"...'
    $xml = simplexml_load_string($res->getBody(),'SimpleXMLElement',LIBXML_NOCDATA);
    echo $xml;

This is how it should look like:

This is how it looks

Answer 1

您现在正在执行的 echo 将触发您使用 simplexml_load_string() 创建的 SimpleXMLElement 的 __toString() 方法，这实际上是 [=] 的字符串内容21=].

你要找的是$xml->asXML()，它会输出一个well-formed XML字符串。您可能还需要为某些浏览器添加 Content-Type header 以获取 XML 内容。尝试以下操作：

    require 'vendor/autoload.php';
    use GuzzleHttp\Client;
    use GuzzleHttp\Psr7\Response;
    use GuzzleHttp\Exception\RequestException;
    use GuzzleHttp\Promise\Promise;
    use GuzzleHttp\Psr7\Request;
    $client = new GuzzleHttp\Client();
    $options = [
        'headers' => ['Accept'=>'application/xml'],
        'auth' => ['Deversor-Channel', '?1_sIWVZ((ujV)kDBM!6O5!AFAQM*K*yr(?E.(g='],
        'body' => "<request>
                        <hotel_id>345786
                        </hotel_id>
                        <last_change>2021-07-19</last_change>
                    </request>",
    ];
    $res = $client->request('POST', 'https://secure-supply-xml.booking.com/hotels/xml/reservations', $options);
    echo $res->getStatusCode();
    // "200"
    echo $res->getBody();
    // {"type":"User"...'
    $xml = simplexml_load_string($res->getBody(),'SimpleXMLElement',LIBXML_NOCDATA);

    header('Content-Type: application/xml');
    echo $xml->asXML();

如何在 Guzzle 上获得 XML 响应

How to get XML response on Guzzle

php

xml

https

guzzle