Question

我有一个API端点，可根据某些传递数据动态生成图像。我想调用API并将响应下载到文件中。理想情况下，我希望避免在PHP中调用cURL而是使用标准库。什么是在PHP中实现这一目标的最佳方法？

请求在cURL中显示如下：

curl https://localhost:4000/bananas/12345.png \
  -O \
  -X POST \
  -d '[ 1, 2, 3, 4 ]'

Answer 1

您必须使用标题：

header("Content-Type: image/jpeg");

$url = 'https://example.com/images/12345.jpg';
$ch = curl_init();
curl_setopt($ch, CURLOPT_URL, $url);
curl_setopt($ch, CURLOPT_HEADER, false);
curl_setopt($ch, CURLOPT_RETURNTRANSFER, true);
curl_setopt($ch, CURLOPT_SSL_VERIFYPEER, false);
curl_setopt($ch, CURLOPT_USERAGENT, 'Mozilla/5.0 (Windows NT 6.1) 
AppleWebKit/537.11 (KHTML, like Gecko) Chrome/23.0.1271.1 Safari/537.11');
$res = curl_exec($ch);
$rescode = curl_getinfo($ch, CURLINFO_HTTP_CODE); 
curl_close($ch) ;
echo $res; // This response can be stored in any type of file using below link

http://php.net/manual/en/function.file-put-contents.php

通过PHP中的POST请求下载文件

1 个答案: