php Laravel - Guzzle 请求/cURL 错误 6：无法解析主机
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Laravel - Guzzle Request / cURL error 6: Could not resolve host
提问by bobbybackblech
I try to make an API Request to the Github API just for testing. I installed the latest Guzzle Version ( "guzzle/guzzle": "^3.9" ) on my Laravel 5.1 APP.
In my routes.phpi have the following Code:
我尝试向 Github API 发出 API 请求，仅用于测试。我在 Laravel 5.1 APP 上安装了最新的 Guzzle 版本 ( "guzzle/guzzle": "^3.9" )。在我的routes.php我有以下代码：
Route::get('guzzle/{username}', function($username) {
$client = new Client([
'base_uri' => 'https://api.github.com/users/',
]);
$response = $client->get("/users/$username");
dd($response);
});
If i now visit the URL domain.dev/github/kayyyy i get the Error cURL error 6: Could not resolve host: users.
如果我现在访问 URL domain.dev/github/kayyyy 我得到错误cURL error 6: Could not resolve host: users。
Why am i getting this Error?
为什么我会收到此错误？
If i visit https://api.github.com/users/kayyyyi can see the jsonoutput.
如果我访问https://api.github.com/users/kayyyy，我可以看到json输出。
I am also using Homestead / Vagrant is this maybe a Problem that the host cant be resolved?
我也在使用 Homestead / Vagrant 这可能是主机无法解决的问题吗？
EDITIf i try this without the base_uri it works.
编辑如果我在没有 base_uri 的情况下尝试此操作，它会起作用。
Route::get('guzzle/{username}', function($username) {
$client = new GuzzleHttp\Client();
$response = $client->get("https://api.github.com/users/$username");
dd($response);
});
回答by M. Sabev
Actually a variable interpolation is not possible within single quotes. This means that you currently are calling users/$usernameand the $usernamevariable gets not replaced with its value.
实际上，单引号内不可能进行变量插值。这意味着您当前正在调用users/$username并且该$username变量不会被其值替换。
In order to get it, you should use it in one of the following ways:
为了获得它，您应该通过以下方式之一使用它：
$response = $client->get("users/$username")->send();
$response = $client->get('users/' . $username)->send();
I personally prefer the second one as it is assumed to be faster.
我个人更喜欢第二个，因为它被认为更快。
回答by Ben Johnson
Why are you calling $client->get->()->send()? In particular, why are you chaining the send() method at the end? The documentation does not append the send()method when demonstrating what seems to be the same action:
你为什么打电话$client->get->()->send()？特别是，为什么要在最后链接 send() 方法？send()在演示似乎是相同的操作时，文档没有附加该方法：
http://guzzle.readthedocs.org/en/latest/quickstart.html#creating-a-client
http://guzzle.readthedocs.org/en/latest/quickstart.html#creating-a-client
Also, did you consider the implications of this statement on the above-cited manual page?
另外，您是否考虑过上述手册页中此声明的含义？
When a relative URI is provided to a client, the client will combine the base URI with the relative URI using the rules described in RFC 3986, section 2.
当向客户端提供相对 URI 时，客户端将使用 RFC 3986 第 2 节中描述的规则将基本 URI 与相对 URI 组合。
回答by bobbybackblech
Okay i solved it, stupid mistake by me.
I used new Client.
好的，我解决了，我犯了一个愚蠢的错误。我用过new Client。
And it should be of course new GuzzleHttp\Client
当然应该是 new GuzzleHttp\Client
As it is just for testing in my routes.phpi did not the Namespace
因为它只是为了测试我的routes.php我没有Namespace
Thanks for your help everybody.
谢谢大家的帮助。
回答by Sarmad Baloch
Thanks to Paratron https://github.com/googleapis/google-api-php-client/issues/1184#issuecomment-355295789In my case, on cakephp 3.8 & docker 19.03.5, I was facing curl error due to some network issue. I restarted my cake-server docker container, & it worked like a charm.
感谢 Paratron https://github.com/googleapis/google-api-php-client/issues/1184#issuecomment-355295789就我而言，在 cakephp 3.8 和 docker 19.03.5 上，由于某些网络，我遇到了 curl 错误问题。我重新启动了我的 cake-server docker 容器，它工作起来很有吸引力。
