php – 将cURL转换为Guzzle POST

2023-01-08 23:07:02

我正在调整一个例子来上传一个图像到imgur.该示例使用curl,我使用guzzle ^ 6.1.卷曲的例子是：

<html>
  <h3>Form</h3>
  <form method="post" enctype="multipart/form-data">
    <input type="hidden" name="MAX_FILE_SIZE" value="50000" />
    Image (< 50kb): <input type="file" name="upload" /><br/>
    ClientID: <input type="text" name="clientid" /><br/>
    <input type="submit" value="Upload to Imgur" />
  </form>
</html>
<?php

if (empty($_POST['clientid']) || @$_FILES['upload']['error'] !== 0 || @$_FILES['upload']['size'] > 50000) {
    exit;
}

$client_id = $_POST['clientid'];

$filetype = explode('/',mime_content_type($_FILES['upload']['tmp_name']));
if ($filetype[0] !== 'image') {
    die('Invalid image type');
}

$image = file_get_contents($_FILES['upload']['tmp_name']);

$ch = curl_init();
curl_setopt($ch, CURLOPT_URL, 'https://api.imgur.com/3/image.json');
curl_setopt($ch, CURLOPT_POST, TRUE);
curl_setopt($ch, CURLOPT_RETURNTRANSFER, TRUE);
curl_setopt($ch, CURLOPT_HTTPHEADER, array( 'Authorization: Client-ID ' . $client_id ));
curl_setopt($ch, CURLOPT_POSTFIELDS, array( 'image' => base64_encode($image) ));

$reply = curl_exec($ch);

curl_close($ch);

$reply = json_decode($reply);

echo "<h3>Image</h3>";
printf('<img height="180" src="%s" >', @$reply->data->link);

echo "<h3>API Debug</h3><pre>";
var_dump($reply);

我尝试使用下一个代码转换为Guzzle：

use GuzzleHttp\Client;
use GuzzleHttp\Psr7\Request as gRequest;   
//....Clases and functions ...

        $url = "https://api.imgur.com/3/image.json";
        $client_id = "miclientid";            
        $client = new Client([
            // Base URI is used with relative requests
            'base_uri' => $url,
            // You can set any number of default request options.
            'timeout'  => 15.0,
        ]);    
        $gRequest  = new gRequest('POST', 'https://api.imgur.com/3/image.json', [
                        'headers' => [
                            'Authorization: Client-ID' => $client_id
                        ],
                        'image' => "data:image/png;base64,iVBORw0K..."

        ]);

        $gResponse = $client->send($gRequest, ['timeout' => 2]);

但我收到了400个不好的请求;我的代码有什么问题？

解决方法:

乍一看,我看到两个问题：

>授权标题.在Guzzle版本中,您使用Authorization：Client-ID作为标题名称,$client_id作为标题值.这将生成一个(错误的)HTTP标头,如下所示：

Authorization: Client-ID: myclientid

解决方案：像这样传递你的标题：

"headers" => [
    "authorization" => "Client-ID " . $clientId
]

>请求正文.您原始的基于cURL的版本包含带有图像参数的URL查询编码正文.该参数包含图像文件的base64编码原始内容.在Guzzle版本中,您实际上根本不发送正文,因为您正在使用不存在的图像选项(请查看Guzzle documentation以获取所有支持选项的列表).此外,您的原始示例不包含数据：image / png; base64,prefix(通常只是浏览器的提示).

尝试传递参数如下：

"form_params" => [
    "image" => base64_encode(/* image content here */)
]

码农公寓

相关文章