flask route()能否加载动态网页？

推荐学习书目

› Learn Python the Hard Way

Python Sites

› PyPI - Python Package Index

› http://diveintopython.org/toc/index.html

› Pocoo

值得关注的项目

› PyPy

› Celery

› Jinja2

› Read the Docs

› gevent

› pyenv

› virtualenv

› Stackless Python

› Beautiful Soup

› 结巴中文分词

› Green Unicorn

› Sentry

› Shovel

› Pyflakes

› pytest

Python 编程

› pep8 Checker

Styles

› PEP 8

› Google Python Style Guide

› Code Style from The Hitchhiker's Guide

This topic created in 3643 days ago, the information mentioned may be changed or developed.

route(/getname) 可以
route(/getname?abc) 不可以

求教解决办法·······

5 replies • 2016-06-14 12:57:11 +08:00

am241

Jun 14, 2016 via Android

参数不应该在程序里判断的吗

iphantom

Jun 14, 2016

@am241 什么意思
https://127.0.0.1:5000/Assess/getPriceByKeyWord?access_token=xxxx&sCity=xxxx&sKeyWord=xxxx&sUserKey=xxxx

这样的链接返回一个固定的值怎么实现？

seki

Jun 14, 2016

问号后是参数
http://flask.pocoo.org/docs/0.11/quickstart/#accessing-request-data

iphantom

Jun 14, 2016

找到原因了 https http

哎郁闷

daweibao

Jun 14, 2016 via Android

route(/getname/<string:name>)