Question

我正试图将一个卷曲转换为guzzle请求，这是卷曲请求。

curl https://{subdomain}.zendesk.com/api/v2/tickets.json \
  -d '{"ticket": {"subject": "My printer is on fire!", "comment": { "body": "The smoke is very colorful." }}}' \
  -H "Content-Type: application/json" -v -u {email_address}:{password} -X POST

这是JSON部分：

{
    "ticket": {
        "requester": {
            "name": "The Customer",
            "email": "thecustomer@domain.com"
        },
        "subject": "My printer is on fire!",
        "comment": {
            "body": "The smoke is very colorful."
        }
    }
}

这是我破解的PHP代码。

$client = new GuzzleHttp\Client();

$res = $client->post('https://midnetworkshelp.zendesk.com/api/v2/tickets/tickets.json', [

            'query' => [

                     'ticket' => ['subject' => 'My print is on Fire'], [ 

                     'comment' => [

                                'body' => 'The smoke is very colorful'] ],  'auth' =>  ['email', 'Password']]);

echo $res->getBody();

我一直未经授权访问用户，但是当我触发curl命令时，它工作正常。

对于我在这里可能缺少的东西有任何想法吗？

谢谢

Answer 1

参考：

http://curl.haxx.se/docs/manpage.html
http://guzzle.readthedocs.org/en/latest/clients.html
https://github.com/guzzle/log-subscriber
http://guzzle.readthedocs.org/en/latest/clients.html#json

您最大的问题是您没有正确转换卷曲请求。

-d =正在发布的数据。换句话说，这是您的要求的主体。
-u =用于验证您的请求的用户名：pw。
-H =您要在请求中使用的额外标头。
-v =详细输出。
-X =指定请求方法。

我建议您按照以下方式对客户进行实例化：

$client = new GuzzleHttp\Client([
        'base_url'      => ['https://{subdomain}.zendesk.com/api/{version}/', [
                'subdomain'     => '<some subdomain name>',
                'version'       => 'v2',
        ],
        'defaults'      => [
                'auth'  => [ $username, $password],
                'headers' => ['Content-Type' => 'application/json'],    //only if all requests will be with json
        ],
        'debug'         => true,                                        // only for debugging purposes
]);

这将：

确保对api发出的多个后续请求将具有身份验证信息。使您不必将其添加到每个请求中。
确保使用此客户端进行的multipl后续（实际上所有）请求将包含指定的标头。使您不必将其添加到每个请求中。
提供一定程度的未来校对（将子域和api版本移动到可编辑字段中）。

如果您选择记录您的请求和响应对象，您也可以这样做：

// You can use any PSR3 compliant logger in space of "null".
// Log the full request and response messages using echo() calls.
$client->getEmitter()->attach(new GuzzleHttp\Subscriber\Log\LogSubscriber(null, GuzzleHttp\Subscriber\Log\Formatter::DEBUG);

您的请求将变为：

$json = '{"ticket": {"subject": "My printer is on fire!", "comment": { "body": "The smoke is very colorful." }}}';
$url = 'tickets/tickets.json';

$request = $client->createRequest('POST', $url, [
        'body' => $json,
]);
$response = $client->send($request);

或

$json = '{"ticket": {"subject": "My printer is on fire!", "comment": { "body": "The smoke is very colorful." }}}';
$url = 'tickets/tickets.json';

$result = $client->post(, [
        'body'  => $json,
]);

编辑：在进一步阅读Ref 4之后，应该可以做到以下几点：

$url = 'tickets/tickets.json';
$client = new GuzzleHttp\Client([
    'base_url'      => ['https://{subdomain}.zendesk.com/api/{version}/', [
        'subdomain'     => '<some subdomain name>',
        'version'       => 'v2',
    ],
    'defaults'      => [
        'auth'  => [ $username, $password],
    ],
    'debug'         => true,                                        // only for debugging purposes
]);
$result = $client->post($url, [
    'json' => $json,            // Any PHP type that can be operated on by PHP’s json_encode() function.
]);

Answer 2

您不应该使用查询参数，因为您需要将原始json作为请求的主体发送（不在您正在执行的参数中。）检查here以获取有关如何使用的信息完成这个。此外，请务必尝试enable debugging找出请求未按您的要求发布的原因。（您可以比较curl和guzzles调试输出以验证它们是否匹配）。

将JSON转换为guzzle php librairy请求

2 个答案: