Php 如何在Guzzle 5中发送PUT请求的参数？

我有此代码用于发送POST请求的参数，它可以工作：

$client = new GuzzleHttp\Client();
$request = $client->createRequest('POST', 'http://example.com/test.php');
$body = $request->getBody();

$request->getBody()->replaceFields([
    'name' => 'Bob'
]);

但是，当我将POST更改为PUT时，会出现以下错误：

Call to a member function replaceFields() on a non-object

这是因为getBody返回null

在主体中发送PUT参数是否正确？还是应该在URL中执行此操作？

根据

主体选项用于控制封闭实体的主体 请求（例如，PUT、POST、补丁）

put

的文件化方法是：

$client = new GuzzleHttp\Client();

$client->put('http://httpbin.org', [
    'headers'         => ['X-Foo' => 'Bar'],
    'body'            => [
        'field' => 'abc',
        'other_field' => '123'
    ],
    'allow_redirects' => false,
    'timeout'         => 5
]);

编辑 根据您的评论：

---

您缺少

createRequest

函数的第三个参数-组成

post

或

put

数据的键/值对数组：

$request = $client->createRequest('PUT', '/put', ['body' => ['foo' => 'bar']]);

---

当服务等待json原始数据时

$request = $client->createRequest('PUT', '/yourpath', ['json' => ['key' => 'value']]);

或

如果您使用的是版本6，则可以通过以下方式发出请求：

$client = new \GuzzleHttp\Client();

$response = $client->put('http://example.com/book/1', [
    'query' => [
        'price' => '50',
    ]
]);

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

---

在Guzzle 6中，如果您希望将JSON数据传递给PUT请求，则可以按如下方式实现：

           $aObj = ['name' => 'sdfsd', 'language' => 'En'];

            $headers = [
                "User-Agent"    => AGENT,
                "Expect"        => "100-continue",
                "api-origin"    => "LTc",
                "Connection"    => "Keep-Alive",
                "accept"        => "application/json",
                "Host"          => "xyz.com",
                "Accept-Encoding"=> " gzip, deflate",
                "Cache-Control"=> "no-cache",
                "verify"        => false,
                "Content-Type" => "application/json"
            ];

          $client = new GuzzleHttp\Client([
            'auth'  => ['testUsername', 'testPassword'],
            'timeout'   => '10000',
            'base_uri'  => YOUR_API_URL,
            'headers' => $headers
        ]);

        $oResponse = $client->request('PUT', '/user/UpdateUser?format=json', ['body' => json_encode( $aObj, JSON_UNESCAPED_SLASHES)]);

        $oUser = json_decode( $oResponse->getBody());

---

问题是我希望能够使用createRequest（）方法，因为我正在确定以编程方式使用哪种方法。您缺少

createRequest

函数的第三个参数-一个组成

post

或

put

数据的键/值对数组：

$request=$client->createRequest（'put'，'put'，['json'=>['foo'=>'bar']]；

答案中的链接不再有效，但这是Guzzle的最新版本：或者更准确地说。

body

已被弃用，应该是Guzzle 6中的

form params

           $aObj = ['name' => 'sdfsd', 'language' => 'En'];

            $headers = [
                "User-Agent"    => AGENT,
                "Expect"        => "100-continue",
                "api-origin"    => "LTc",
                "Connection"    => "Keep-Alive",
                "accept"        => "application/json",
                "Host"          => "xyz.com",
                "Accept-Encoding"=> " gzip, deflate",
                "Cache-Control"=> "no-cache",
                "verify"        => false,
                "Content-Type" => "application/json"
            ];

          $client = new GuzzleHttp\Client([
            'auth'  => ['testUsername', 'testPassword'],
            'timeout'   => '10000',
            'base_uri'  => YOUR_API_URL,
            'headers' => $headers
        ]);

        $oResponse = $client->request('PUT', '/user/UpdateUser?format=json', ['body' => json_encode( $aObj, JSON_UNESCAPED_SLASHES)]);

        $oUser = json_decode( $oResponse->getBody());