python中的requests模块

import requests

#post请求============================
# url='http://127.0.0.1:8000/api/update'#接口入参为str
# data = {'username':'123','password':'345'}
# r = requests.post(url,data)
# print(r.url)#get请求返回url #http://127.0.0.1:8000/api/update?username=123&password=345
# print((r.json()))#返回字典
#
# url_json='http://127.0.0.1:8000/api/check'#接口入参为json
# data = {'username':'123','password':'345'}
# r_json = requests.post(url_json,json=data)#json=data表示入参是json
# print((r_json.json()))#返回字典

#get请求================================
# url = 'http://xx.xx.xx.xx:8080/xx/user/login'
# headers = {'Cookie':'JSESSIONID=7C8052371BF7BFDA5865E895E6E49BEC; UAPSESSIONID=D1D0E1D83C344CCCB221712F79D4BBD4; _jfinal_captcha=aa9b345535f94af09886898ec1a2424a',
#            'Content-Type':'text/html;charset=UTF-8'}
# r = requests.get(url)
# if r.status_code==200:
#     print('访问正常')
# else:
#     r.raise_for_status()


# #下载===================
# url_p ='https://pics5.baidu.com/feed/d833c895d143ad4bb2226779dcd84da5a50f065a.jpeg?'
# r = requests.get(url_p)
# with open('a.jpg','wb') as f:
#     f.write(r.content)

# #上传===========================
# url = 'http://xxxxxx/api/file/file_upload'
# file = {'file':open('a.jpg','rb')}
# r = requests.post(url,files=file)
# if r.status_code==200:
#     print('成功')
# else:
#     r.raise_for_status()#非200 抛出错误异常





# r.encoding 获取网页编码格式
# r.text 获取响应类型为str格式 常见get请求获取网页数据
# r.content 以字节的方式去显示,中文显示为字符 常用文件/图片类型
# r.headers 获取响应头
# r.status_code 获取响应状态码
# r.cookies 获取cookies
# r.url 获取请求的url
# r.json() 获取响应为json的数据,前提是响应报文一定是json否则报错

 

posted @ 2022-03-30 14:56  小马哥007  阅读(66)  评论(0编辑  收藏  举报