Python 3获取HTTP页面

Question

如何让python获取HTTP页面的内容？到目前为止,我所有的是请求,我已经导入了http.client.

Answer 1

使用urllib.request可能是最简单的方法:

import urllib.request
f = urllib.request.urlopen("http://stackoverflow.com")
print(f.read())

Answer 2

用法内置模块"http.client"

import http.client

connection = http.client.HTTPSConnection("api.bitbucket.org", timeout=2)
connection.request('GET', '/2.0/repositories')
response = connection.getresponse()
print('{} {} - a response on a GET request by using "http.client"'.format(response.status, response.reason))
content = response.read().decode('utf-8')
print(content[:100], '...')

结果:

200 OK - 使用"http.client"{"pagelen":10,"values":[{"scm":"hg","website":"","has_wiki":true,对GET请求的响应) "name":"tweakmsg","链接......

用法第三方库"请求"

response = requests.get("https://api.bitbucket.org/2.0/repositories")
print('{} {} - a response on a GET request by using "requests"'.format(response.status_code, response.reason))
content = response.content.decode('utf-8')
print(content[:100], '...')

结果:

200 OK - 使用"http.client"{"pagelen":10,"values":[{"scm":"hg","website":"","has_wiki":true,对GET请求的响应) "name":"tweakmsg","链接......

用法内置模块"urllib.request"

response = urllib.request.urlopen("https://api.bitbucket.org/2.0/repositories")
print('{} {} - a response on a GET request by using "urllib.request"'.format(response.status, response.reason))
content = response.read().decode('utf-8')
print(content[:100], '...')

结果:

200 OK - 使用"http.client"{"pagelen":10,"values":[{"scm":"hg","website":"","has_wiki":true,对GET请求的响应) "name":"tweakmsg","链接......

笔记:

1. Python 3.4
2. 响应的结果很可能只是内容不同