Question

我正在尝试从URL获取JSON对象。我尝试过使用PHP（file_get_contents和curl）和Python（都使用urllib2和requests）。所有错误，但如果我使用浏览器访问URL，则会显示JSON对象。这是我在Apache2上的应用程序。

以下是我获得的每个命令的错误：

file_get_contents ---无法打开流
cURL ---错误28：连接在10001毫秒后超时

Python请求 -

File "test.py", line 4, in <module>
    response= requests.get(url, timeout=2)
  File "/usr/lib/python2.7/dist-packages/requests/api.py", line 55, in get
    return request('get', url, * *kwargs)
  File "/usr/lib/python2.7/dist-packages/requests/api.py", line 44, in request
    return session.request(method=method, url=url, *kwargs)
  File "/usr/lib/python2.7/dist-packages/requests/sessions.py", line 455, in request
    resp = self.send(prep, * *send_kwargs)
  File "/usr/lib/python2.7/dist-packages/requests/sessions.py", line 558, in send
    r = adapter.send(request, * *kwargs)
  File "/usr/lib/python2.7/dist-packages/requests/adapters.py", line 387, in send
    raise Timeout(e)
requests.exceptions.Timeout: (<urllib3.connectionpool.VerifiedHTTPSConnection instance at 0x7fbb3ec6d440>, 'Connection to **url** timed out. (connect timeout=2)')

我也尝试使用php在线模拟器运行file_get_contents只是因为它与我的服务器有关但是我收到了这个错误

E_WARNING : type 2 -- file_get_contents(): SSL operation failed with code 1. OpenSSL Error messages: error:14090086:SSL routines:SSL3_GET_SERVER_CERTIFICATE:certificate verify failed -- at line 4
E_WARNING : type 2 -- file_get_contents(): Failed to enable crypto -- at line 4

卷曲代码是：

curl_setopt($ch, CURLOPT_TIMEOUT, 10);

curl_setopt($ch, CURLOPT_URL, $url2);
curl_setopt($ch, CURLOPT_RETURNTRANSFER, true);
curl_setopt($ch, CURLOPT_FOLLOWLOCATION, true);
$result = curl_exec($ch);
if ( $result==false ) {
  echo curl_errno($ch).' '.curl_error($ch);
}
curl_close($ch);

为什么我可以使用浏览器而不是代码访问URL？

0 个答案: