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我收到一个错误,不知道这是否为头错误?饲料工程100%,当我使用邮差用相同的标题和URL用CURL发送标题并接收Json数据 - PHP
{"error":{"message":"The content version specified in the request is not supported.","code":101}}
这里是我想,我的PHP代码
$url = 'http://x.x.x.x/api/slot/0/io/';
$headers = array(
'Accept:vdn.v1',
'Content-Type:application/json'
);
$ch = curl_init();
// Now set some options (most are optional)
// Set URL to download
curl_setopt($ch, CURLOPT_URL, $url);
curl_setopt($ch, CURLOPT_HTTPHEADER, $headers);
// Set a referer
curl_setopt($ch, CURLOPT_REFERER, "http://www.example.org/yay.htm");
// User agent
curl_setopt($ch, CURLOPT_USERAGENT, "MozillaXYZ/1.0");
// Include header in result? (0 = yes, 1 = no)
curl_setopt($ch, CURLOPT_HEADER, 0);
$headers
// Should cURL return or print out the data? (true = return, false = print)
curl_setopt($ch, CURLOPT_RETURNTRANSFER, true);
// Timeout in seconds
curl_setopt($ch, CURLOPT_TIMEOUT, 10);
// Download the given URL, and return output
$output = curl_exec($ch);
// Close the cURL resource, and free system resources
curl_close($ch);
print_r($output);exit;
错误表明您的Accept头存在问题。检查字符串“Accept:vdn.v1”中是否有任何拼写错误。 – jorgonor
@jorgonor我再次检查邮差和我的PHP代码接受:vdn.v1是正确的 – user580950
如果它在邮差工程中,单击右上角的'代码'链接,在“生成代码段”弹出式选择“PHP曲线”。生成的代码可以很容易粘贴到脚本中,并生成相同的http请求。 –