我有以下pycurl代码:
curl = pycurl.Curl()
foo = StringIO()
curl.setopt(pycurl.WRITEFUNCTION, foo.write)
curl.setopt(pycurl.POST, 1)
curl.setopt(pycurl.URL, finalURL)
curl.setopt(pycurl.POSTFIELDS, encodedArgs)
curl.perform()
responseCode = curl.getinfo(pycurl.RESPONSE_CODE)
effectiveURL = curl.getinfo(pycurl.EFFECTIVE_URL)
curl.close()
当命令行curl命令返回时,我看到:
HTTP/1.1 200 OK
Server: Apache-Coyote/1.1
Content-Type: text/xml;charset=UTF-8
Content-Length: 216
Date: Thu, 06 Jan 2011 15:49:36 GMT
Some XML Error Here: Something you are trying to do is not permitted.
但我从pycurl身上看不到这一点。
使用pycurl时如何提取此警报/错误消息?在
来自服务器的响应是使用curl选项
pycurl.WRITEFUNCTION
写入的。在在您的例子中,由于您传递给它一个
StringIO
对象,所以响应数据应该在foo
变量中:foo.getvalue()
参考号:http://pycurl.sourceforge.net/doc/curlobject.html
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