编写出色的Twisted Web资源 [英] Writing excellent Twisted web Resources

问题描述

我写了第一篇Twisted 10.1.0网页资源，我正在寻求反馈，因为我觉得这并不完全符合最佳做法，可能包含新手错误。

I wrote my very first Twisted 10.1.0 web Resource and I am seeking for feedback, because I feel this isn't exactly following the best practice and may contain newbies bugs.

资源响应 /?url=http://www.foo.baz/abc123 和依赖于返回 dict 的服务。如果出现任何问题（例如，无效或不存在 url ，则返回 400 ）。

The resource responds to /?url=http://www.foo.baz/abc123 and relies on a service that returns a dict. If anything goes wrong (e.g., invalid or non-existing url, then a 400 is returned).

有何评论？有什么需要修改，改进

Any comment? Anything to fix, to improve

class ProcessedUrl(resource.Resource):
    isLeaf = True

    def __init__(self, service):
        resource.Resource.__init__(self)
        self.service = service

    def _cancel(self, err, deferred):
        deferred.cancel()

    def _write(self, value, request):
        request.setResponseCode(http.OK)
        request.write(json.dumps(value))
        request.finish()

    def _cleanUrl(self, url):
        return cleanUrl(url)

    def _checkUrl(self, url):
        if url is not None:
            if isValidUrl(url):
                return True
        return False

    def render_GET(self, request):
        request.setResponseCode(http.BAD_REQUEST)
        url = request.args.get('url', [None])[0]

        if self._checkUrl(url):
            url = self._cleanUrl(url)
            d = self.service.processUrl(url)
            request.notifyFinish().addErrback(self._cancel, d)
            d.addCallback(_write)
            d.addErrback(log.err)
        else:
            return 'Invalid or no URL.'
        return server.NOT_DONE_YET

    def getChild(self, name, request):
        return self

推荐答案

我想你不会如果设置 isLeaf = True

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