Laravel 中的 cURL 请求

发布于01月16日

我很难在拉维尔提出这个卷发要求

curl -d '{"key1":"value1", "key2":"value2"}' -H "Content-Type: application/json"   -X GET http://my.domain.com/test.php

I've been trying this:

$endpoint = "http://my.domain.com/test.php";

$client = new \GuzzleHttp\Client();

$response = $client->post($endpoint, [
                GuzzleHttp\RequestOptions::JSON => ['key1' => $id, 'key2' => 'Test'],
            ]);

$statusCode = $response->getStatusCode();

但是我收到了一个错误Class 'App\Http\Controllers\GuzzleHttp\RequestOptions' not found

Any suggestions?

EDIT

我需要从$response中的API获得响应，然后将其存储在DB中…我怎么才能做到这一点呢？:/

推荐答案

Give the query-option from Guzzle a try:

$endpoint = "http://my.domain.com/test.php";
$client = new \GuzzleHttp\Client();
$id = 5;
$value = "ABC";

$response = $client->request('GET', $endpoint, ['query' => [
    'key1' => $id, 
    'key2' => $value,
]]);

// url will be: http://my.domain.com/test.php?key1=5&key2=ABC;

$statusCode = $response->getStatusCode();
$content = $response->getBody();

// or when your server returns json
// $content = json_decode($response->getBody(), true);

I use this option to build my get-requests with guzzle. In combination with json_decode($json_values, true) you can transform json to a php-array.