多文件上传Curl转PHP Guzzle报500错误，求验证转换正确性

原Curl命令：

curl -X POST "https://uploadexample.net/api/upload" -H "accept: application/json" -H "Authorization: bearer: 928292992qwg" -H "Content-Type: multipart/form-data" -F "front_photo=@photo-1-214x300.jpg;type=image/jpeg" -F "back_photo=@photo-2-214x300.jpg;type=image/jpeg"

你写的错误代码：

$headers = [
    'Content-type' => 'application/json',
    'Content-type' => 'multipart/form-data',
    'Accept' => 'application/json',
    "Authorization" => "Bearer 928292992qwg"
];   
$client = new Client([
     // Base URI is used with relative requests
    'base_uri' => 'https://exampleuploadrx.net',
]);
  
$response = $client->request('POST', '/api/upload',  [  
 'json' => [
    'front_photo' => new CURLFile('photo-1-214x300.jpg;type=image/jpeg'),
    'back_photo' => new CURLFile('photo-2-214x300.jpg;type=image/jpeg'),
    ],
                'headers' => $headers,
             
    ]
    );

代码里的错误点： #

• 重复且错误的Content-Type头：数组里重复定义了Content-type，后一个会覆盖前一个，但更关键的是，Guzzle处理multipart请求时会自动生成带boundary的正确Content-Type头，手动设置反而会干扰逻辑；另外你错误添加了application/json的Content-Type，和文件上传的multipart格式完全冲突。
• 用错请求参数类型：原Curl是multipart/form-data格式上传文件，你却用了Guzzle的json选项——这个选项是用来发送JSON格式请求体的，和文件上传完全不兼容，应该用multipart选项。
• CURLFile构造错误：CURLFile的第一个参数是纯文件路径，不能带;type=image/jpeg，文件类型应该作为第二个参数传入，格式为new CURLFile($filePath, $mimeType)。
• Authorization头格式不符：原Curl里是bearer: 928292992qwg，你的代码写成了Bearer 928292992qwg，少了冒号，可能导致认证失败。
• base_uri域名错误：原Curl的目标域名是uploadexample.net，你写成了exampleuploadrx.net，请求发到错误服务器会直接引发500错误。

正确的Guzzle实现代码： #

use GuzzleHttp\Client;
use GuzzleHttp\Psr7\Utils;

// 初始化客户端，域名和原Curl保持一致
$client = new Client([
    'base_uri' => 'https://uploadexample.net',
]);

$response = $client->request('POST', '/api/upload', [
    'headers' => [
        'Accept' => 'application/json',
        'Authorization' => 'bearer: 928292992qwg' // 和原Curl格式完全匹配
    ],
    // 使用multipart参数处理文件上传
    'multipart' => [
        [
            'name'     => 'front_photo',
            'contents' => Utils::tryFopen('photo-1-214x300.jpg', 'r'),
            'filename' => 'photo-1-214x300.jpg',
            'headers'  => [
                'Content-Type' => 'image/jpeg'
            ]
        ],
        [
            'name'     => 'back_photo',
            'contents' => Utils::tryFopen('photo-2-214x300.jpg', 'r'),
            'filename' => 'photo-2-214x300.jpg',
            'headers'  => [
                'Content-Type' => 'image/jpeg'
            ]
        ]
    ]
]);

// 获取响应内容
$body = $response->getBody()->getContents();

如果习惯用CURLFile，也可以这么写：

use GuzzleHttp\Client;

$client = new Client([
    'base_uri' => 'https://uploadexample.net',
]);

$response = $client->request('POST', '/api/upload', [
    'headers' => [
        'Accept' => 'application/json',
        'Authorization' => 'bearer: 928292992qwg'
    ],
    'multipart' => [
        [
            'name'     => 'front_photo',
            'contents' => new \CURLFile('photo-1-214x300.jpg', 'image/jpeg')
        ],
        [
            'name'     => 'back_photo',
            'contents' => new \CURLFile('photo-2-214x300.jpg', 'image/jpeg')
        ]
    ]
]);

内容的提问来源于stack exchange，提问作者Bob Hijt