竞争无处不在,青春永不言败!专业撸代码,副业修bug

Talk is cheap , show me the code!



python web 开发

第一个 简单的 WSGI demo

from wsgiref.simple_server import make_server

def application(environ, start_response):
    start_response('200 OK', [('Content-Type', 'text/html')])
    return [b'<h1>Hello, web!</h1>']
httpd = make_server('', 8000, application)
print('Serving HTTP on port 8000...')
# 开始监听HTTP请求:
httpd.serve_forever()



from wsgiref.simple_server import make_server

def application(environ, start_response):
    start_response('200 OK', [('Content-Type', 'text/html')])
    return ['<h1>Hello, {}!</h1>'.format(environ['PATH_INFO'][1:] or 'web').encode('utf-8')]
httpd = make_server('', 8000, application)
print('Serving HTTP on port 8000...')
# 开始监听HTTP请求:
httpd.serve_forever()


访问本地url

http://192.168.0.xxx:8000/Frank

然而, WSGI 来开发显然是耗时不讨好的事情,所以实际开发,都是应用框架 flask ,django 这种

from flask import Flask
from flask import request

app = Flask(__name__)

@app.route('/', methods=['GET', 'POST'])
def home():
    return '<h1>Home</h1>'

@app.route('/signin', methods=['GET'])
def signin_form():
    return '''<form action="/signin" method="post">
              <p><input name="username"></p>
              <p><input name="password" type="password"></p>
              <p><button type="submit">Sign In</button></p>
              </form>'''

@app.route('/signin', methods=['POST'])
def signin():
    # 需要从request对象读取表单内容:
    if request.form['username']=='admin' and request.form['password']=='password':
        return '<h3>Hello, admin!</h3>'
    return '<h3>Bad username or password.</h3>'

if __name__ == '__main__':
    app.run()
posted @ 2018-10-30 09:55  云雾散人  阅读(185)  评论(0编辑  收藏  举报

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