PHP GuzzleHttp。如何使用参数进行发布请求？
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/27825667/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
PHP GuzzleHttp. How to make a post request with params?
提问by Arsen
How to make a post request with GuzzleHttp( version 5.0 ).
如何使用 GuzzleHttp（5.0 版）发出 post 请求。
I am trying to do the following:
我正在尝试执行以下操作：
$client = new \GuzzleHttp\Client();
$client->post(
'http://www.example.com/user/create',
array(
'email' => '[email protected]',
'name' => 'Test user',
'password' => 'testpassword'
)
);
But I am getting the error:
但我收到错误：
PHP Fatal error: Uncaught exception 'InvalidArgumentException' with the message 'No method can handle the email config key'
PHP 致命错误：未捕获异常“InvalidArgumentException”，消息为“No method can handle the email config key”
采纳答案by Marco
Try this
尝试这个
$client = new \GuzzleHttp\Client();
$client->post(
'http://www.example.com/user/create',
array(
'form_params' => array(
'email' => '[email protected]',
'name' => 'Test user',
'password' => 'testpassword'
)
)
);
回答by Samuel Dauzon
Since Marco's answer is deprecated, you must use the following syntax (according jasonlfunk's comment) :
由于Marco 的答案已弃用，您必须使用以下语法（根据 jasonlfunk 的评论）：
$client = new \GuzzleHttp\Client();
$response = $client->request('POST', 'http://www.example.com/user/create', [
'form_params' => [
'email' => '[email protected]',
'name' => 'Test user',
'password' => 'testpassword',
]
]);
Request with POST files
使用 POST 文件请求
$response = $client->request('POST', 'http://www.example.com/files/post', [
'multipart' => [
[
'name' => 'file_name',
'contents' => fopen('/path/to/file', 'r')
],
[
'name' => 'csv_header',
'contents' => 'First Name, Last Name, Username',
'filename' => 'csv_header.csv'
]
]
]);
REST verbs usage with params
带参数的 REST 动词用法
// PUT
$client->put('http://www.example.com/user/4', [
'body' => [
'email' => '[email protected]',
'name' => 'Test user',
'password' => 'testpassword',
],
'timeout' => 5
]);
// DELETE
$client->delete('http://www.example.com/user');
Async POST data
异步 POST 数据
Usefull for long server operations.
对长时间的服务器操作很有用。
$client = new \GuzzleHttp\Client();
$promise = $client->requestAsync('POST', 'http://www.example.com/user/create', [
'form_params' => [
'email' => '[email protected]',
'name' => 'Test user',
'password' => 'testpassword',
]
]);
$promise->then(
function (ResponseInterface $res) {
echo $res->getStatusCode() . "\n";
},
function (RequestException $e) {
echo $e->getMessage() . "\n";
echo $e->getRequest()->getMethod();
}
);
More information for debugging
有关调试的更多信息
If you want more details information, you can use debugoption like this :
如果您需要更多详细信息，可以使用如下debug选项：
$client = new \GuzzleHttp\Client();
$response = $client->request('POST', 'http://www.example.com/user/create', [
'form_params' => [
'email' => '[email protected]',
'name' => 'Test user',
'password' => 'testpassword',
],
// If you want more informations during request
'debug' => true
]);
Documentationis more explicits about new possibilities.
文档更明确地说明了新的可能性。
回答by Scott Yang
Note in Guzzle V6.0+, another source of getting the following error may be incorrect use of JSON as an array:
请注意，在 Guzzle V6.0+ 中，出现以下错误的另一个原因可能是不正确地使用 JSON 作为数组：
Passing in the "body" request option as an array to send a POST request has been deprecated. Please use the "form_params" request option to send a application/x-www-form-urlencoded request, or a the "multipart" request option to send a multipart/form-data request.
不推荐将“body”请求选项作为数组传递以发送 POST 请求。请使用“form_params”请求选项发送 application/x-www-form-urlencoded 请求，或使用“multipart”请求选项发送 multipart/form-data 请求。
Incorrect:
不正确：
$response = $client->post('http://example.com/api', [
'body' => [
'name' => 'Example name',
]
])
Correct:
正确：
$response = $client->post('http://example.com/api', [
'json' => [
'name' => 'Example name',
]
])
Correct:
正确：
$response = $client->post('http://example.com/api', [
'headers' => ['Content-Type' => 'application/json'],
'body' => json_encode([
'name' => 'Example name',
])
])
回答by Prakash D
$client = new \GuzzleHttp\Client();
$request = $client->post('http://demo.website.com/api', [
'body' => json_encode($dataArray)
]);
$response = $request->getBody();
Add
添加
openssl.cafilein php.inifile
openssl.cafile在php.ini文件中
