python Tornado获取请求url
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python tornado get request url
提问by Lee_Prison
Here is my code:
这是我的代码：
class MainHandler(tornado.web.RequestHandler):
def get(self):
self.write(self.request.url)
def main():
settings = {"template_path": "html","static_path": "static"}
tornado.options.parse_command_line()
application = tornado.web.Application([
(r"/story/page1", MainHandler),
],**settings)
I want to get the string "/story/page1". how ?
我想得到字符串“/story/page1”。如何 ？
回答by stalk
You can get current url inside RequestHandlerusing self.request.uri:
您可以RequestHandler使用self.request.uri以下方法获取当前网址：
class MainHandler(tornado.web.RequestHandler):
def get(self):
self.write(self.request.uri)
回答by Kemal Ahmed
I think what you're looking for is self.request.path. Look at the functions available for HTTPServerRequest.
我想你要找的是self.request.path. 查看可用于HTTPServerRequest.
class MainHandler(tornado.web.RequestHandler):
def get(self):
self.write(self.request.path)
回答by Mike N
There is no method that gives you access to the full URL, but you can construct it using the protocol, host and uri, e.g.
没有任何方法可以让您访问完整的 URL，但您可以使用协议、主机和 uri 构建它，例如
url = '{}://{}{}'.format(self.request.protocol,self.request.host,self.request.uri)
回答by Andreas Gajdosik
Use self.request.full_url()if you want to access the whole URL of the request (protocol + subdomain + domain + path + query). It will return, for example: https://myserver.com/profiles?id=115.
使用self.request.full_url()如果您要访问的请求（协议+子域名+域名+路径+查询）的整个URL。它将返回，例如：https://myserver.com/profiles?id=115。
Use self.request.uriif you want to access URI (path + query). It will give, for example: /profiles?id=115.
使用self.request.uri如果您要访问的URI（路径+查询）。它将给，例如：/profiles?id=115。
Use self.request.pathif you want to access just the path without query string. It will give, for example: /profiles.
self.request.path如果您只想访问没有查询字符串的路径，请使用。它将给，例如：/profiles。
