Webonyx/Graphql-php 入门:如何在 cURL 中获取 API 的响应,而没有来自 API 实现的回显?

Getting started in Webonyx/Graphql-php: How get the response of the API in cURL without echoes from the API implementation?

我是 GraphQL 的新手,想构建一个简单的 API 来开始。阅读文档并尝试示例后,API 终于可以正常工作了……但是 !!!

API 的 php 实现以“回声”结束(这对 Graphiql 客户端来说工作正常!),但是当我尝试在 cURL 中获取响应时,出现此回声在我的页面源中...

拜托各位,如何避免这种回显并在 cURL 中得到结果?我求助于巨大的集体智慧来获得一些帮助。

以下是我使用的资源:

这是实现webonyx/graphql-php:

use GraphQL\GraphQL;
use GraphQL\Type\Schema;

require('types.php');
require('querys.php');
require('mutations.php');

$schema = new Schema([
    'description' => 'Available querys and mutations',
    'query' => $rootQuery,
    'mutation' => $rootMutation
]);

try
{
    $rawInput = file_get_contents('php://input');
    $input = json_decode($rawInput, true);
    $query = $input['query'];
    $result = GraphQL::executeQuery($schema, $query);

    $output = $result->toArray();
}
catch (\Exception $e){
    $output = [
      'error' => [
          'message' => $e->getMessage()
      ]
    ];
}

header('Content-Type: application/json');

echo json_encode($output); //<-- This echo appear when i try to get the response in cURL

这里是 GraphiQL 中的查询:

mutation{
  addUser(name:"user name",email:"user email",password:"some pass"){
    id
  }
}

这是 GraphQL 中的结果(一切正常!):

{
  "data": {
    "addUser": {
      "id": 97
    }
  }
}

这里使用 cURL 中的函数来获取响应:

    function addUser($name,$email,$password){
        $query = <<<'JSON'
                mutation{
                    addUser(name:"*name", email:"*email", password:"*password"){
                    id
                    name
                }}
        JSON;

        $trans = array(
            "*name" => $name,
            "*email" => $email,
            "*password" => $password
        );

        $query = strtr($query, $trans);

        $variables = "";

        $json = json_encode(['query' => $query, 'variables' => $variables]);

        $chObj = curl_init();

        curl_setopt($chObj, CURLOPT_URL, $this->endpoint);
        curl_setopt($chObj, CURLOPT_RETURNTRANSFER, false);
        curl_setopt($chObj, CURLOPT_CUSTOMREQUEST, 'POST');
        curl_setopt($chObj, CURLOPT_VERBOSE, true);
        curl_setopt($chObj, CURLOPT_POSTFIELDS, $json);

        $result = curl_exec($chObj);

        return $result;
    }

这是我使用函数时页面中的源代码:

{"data":{"addUser":{"id":98,"name":"user name"}}}[1]
<!doctype html>
<html lang="en">
<head>
    <meta charset="utf-8">
    <meta name="viewport" content="width=device-width, initial-scale=1">
    <meta name="description" content="">
    <meta name="author" content="Mark Otto, Jacob Thornton, and Bootstrap contributors">
    <meta name="generator" content="Hugo 0.87.0">
    <title>Dashboard Template · Bootstrap v5.1</title>

    <link rel="canonical" href="https://getbootstrap.com/docs/5.1/examples/dashboard/">

字符串:{"data":{"addUser":{"id":98,"name":"用户名"}}}来自API 实现 echo json_encode($output);

我尝试使用 get_file_content 代替 cURL 但无法取得好的效果,请提供任何帮助!

Set CURLOPT_RETURNTRANSFER to true to return the transfer as a string of the return value of curl_exec() instead of outputting it directly.

curl_setopt($chObj, CURLOPT_RETURNTRANSFER, true);