Php版本:5.6

Symfony版本:2.3

暴食:5.0

我最近有一次用Guzzle发送json的经历。我使用的是Symfony 2.3，所以我的暴饮暴食版本可以稍微老一点。

我还会展示如何使用调试模式，你可以在发送请求之前看到它，

当我提出如下所示的请求时，得到了成功的响应;

use GuzzleHttp\Client;

$headers = [
        'Authorization' => 'Bearer ' . $token,        
        'Accept'        => 'application/json',
        "Content-Type"  => "application/json"
    ];        

    $body = json_encode($requestBody);

    $client = new Client();    

    $client->setDefaultOption('headers', $headers);
    $client->setDefaultOption('verify', false);
    $client->setDefaultOption('debug', true);

    $response = $client->post($endPoint, array('body'=> $body));

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

你可以使用硬编码的json属性作为键，或者你可以方便地使用GuzzleHttp\RequestOptions:: json常量。

下面是使用硬编码的json字符串的例子。

use GuzzleHttp\Client;

$client = new Client();

$response = $client->post('url', [
    'json' => ['foo' => 'bar']
]);

见文档。

这适用于我的Guzzle 6.2:

$gClient =  new \GuzzleHttp\Client(['base_uri' => 'www.foo.bar']);
$res = $gClient->post('ws/endpoint',
                            array(
                                'headers'=>array('Content-Type'=>'application/json'),
                                'json'=>array('someData'=>'xxxxx','moreData'=>'zzzzzzz')
                                )
                    );

根据文档guzzle做json_encode

我使用下面的代码，工作非常可靠。

JSON数据在参数$request中传递，特定的请求类型在变量$searchType中传递。

该代码包含一个陷阱，用于检测和报告不成功或无效的调用，然后返回false。

如果调用成功，json_decode ($result->getBody()， $return=true)返回一个结果数组。

    public function callAPI($request, $searchType) {
    $guzzleClient = new GuzzleHttp\Client(["base_uri" => "https://example.com"]);

    try {
        $result = $guzzleClient->post( $searchType, ["json" => $request]);
    } catch (Exception $e) {
        $error = $e->getMessage();
        $error .= '<pre>'.print_r($request, $return=true).'</pre>';
        $error .= 'No returnable data';
        Event::logError(__LINE__, __FILE__, $error);
        return false;
    }
    return json_decode($result->getBody(), $return=true);
}

@user3379466是正确的，但在这里我重写了全文:

-package that you need:

 "require": {
    "php"  : ">=5.3.9",
    "guzzlehttp/guzzle": "^3.8"
},

-php code (Digest is a type so pick different type if you need to, i have to include api server for authentication in this paragraph, some does not need to authenticate. If you use json you will need to replace any text 'xml' with 'json' and the data below should be a json string too):

$client = new Client('https://api.yourbaseapiserver.com/incidents.xml', array('version' => 'v1.3', 'request.options' => array('headers' => array('Accept' => 'application/vnd.yourbaseapiserver.v1.1+xml', 'Content-Type' => 'text/xml'), 'auth' => array('username@gmail.com', 'password', 'Digest'),)));

$url = "https://api.yourbaseapiserver.com/incidents.xml"; $data = '<事件> <名称>事件Title2a < /名称> <优先>中> < /优先 <请求者> < >电子邮件dsss@mail.ca < /电子邮件> < /请求者> <描述> description2a > < /描述 > < /事件”;

    $request = $client->post($url, array('content-type' => 'application/xml',));

    $request->setBody($data); #set body! this is body of request object and not a body field in the header section so don't be confused.

    $response = $request->send(); #you must do send() method!
    echo $response->getBody(); #you should see the response body from the server on success
    die;

解决*暴饮暴食6 * - -你需要的包:

 "require": {
    "php"  : ">=5.5.0",
    "guzzlehttp/guzzle": "~6.0"
},

$client = new Client([
                             // Base URI is used with relative requests
                             'base_uri' => 'https://api.compay.com/',
                             // You can set any number of default request options.
                             'timeout'  => 3.0,
                             'auth'     => array('you@gmail.ca', 'dsfddfdfpassword', 'Digest'),
                             'headers' => array('Accept'        => 'application/vnd.comay.v1.1+xml',
                                                'Content-Type'  => 'text/xml'),
                         ]);

$url = "https://api.compay.com/cases.xml";
    $data string variable is defined same as above.


    // Provide the body as a string.
    $r = $client->request('POST', $url, [
        'body' => $data
    ]);

    echo $r->getBody();
    die;

对于《Guzzle 5》，《Guzzle 6》和《Guzzle 7》，你是这样做的:

use GuzzleHttp\Client;

$client = new Client();

$response = $client->post('url', [
    GuzzleHttp\RequestOptions::JSON => ['foo' => 'bar'] // or 'json' => [...]
]);

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