我试图提供HTTP OPTIONS方法的响应与Access-Control-Allow-Origin标头复制请求中的Origin标头的内容.
由于我无法弄清楚的原因,这显然不起作用.
tl; dr: 来自OPTIONS的回应说:
Access-Control-Allow-Origin: http://10.0.0.105:9294
随后的GET有:
Origin:http://10.0.0.105:9294
Chrome说:
Origin http://10.0.0.105:9294 is not allowed by Access-Control-Allow-Origin
WTF不是吗？
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通过查看Chrome的开发者工具窗口,请求标头是:
OPTIONS /user/kris HTTP/1.1
Host: 10.0.0.104:8080
Connection: keep-alive
Access-Control-Request-Method: GET
Origin: http://10.0.0.105:9294
User-Agent: Mozilla/5.0 (X11; Linux i686) AppleWebKit/537.1 (KHTML, like Gecko) Chrome/21.0.1180.75 Safari/537.1
Access-Control-Request-Headers: origin, x-requested-with, content-type, accept
Accept: */*
Referer: http://10.0.0.105:9294/
Accept-Encoding: gzip,deflate,sdch
Accept-Language: en-GB,en-US;q=0.8,en;q=0.6
Accept-Charset: ISO-8859-1,utf-8;q=0.7,*;q=0.3
响应标头是:
HTTP/1.0 200 OK
Date: Mon, 13 Aug 2012 11:23:45 GMT
Server: WSGIServer/0.1 Python/2.7.3
Content-Length: 0
Access-Control-Allow-Methods: GET, PUT, POST, DELETE, HEAD, …我正在使用urllib.request.urlopen()从我正在尝试测试的Web服务获取GET.
这将返回一个HTTPResponse对象,然后我读取()以获取响应主体.
但我总是看到一个关于来自socket.py的未封闭套接字的ResourceWarning
这是相关的功能:
from urllib.request import Request, urlopen
def get_from_webservice(url):
""" GET from the webservice """
req = Request(url, method="GET", headers=HEADERS)
with urlopen(req) as rsp:
body = rsp.read().decode('utf-8')
return json.loads(body)
这是程序输出中出现的警告:
$ ./test/test_webservices.py
/Library/Frameworks/Python.framework/Versions/3.3/lib/python3.3/socket.py:359: ResourceWarning: unclosed <socket.socket object, fd=5, family=30, type=1, proto=6>
self._sock = None
.s
----------------------------------------------------------------------
Ran 2 tests in 0.010s
OK (skipped=1)
如果我可以对HTTPResponse(或Request？)做任何事情让它干净地关闭它的套接字,我真的很想知道,因为这个代码是我的单元测试; 我不喜欢在任何地方忽略警告,但尤其不在那里.