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iphantom
V2EX  ›  Python

flask route()能否加载动态网页?

  •  
  •   iphantom · Jun 14, 2016 · 3321 views
    This topic created in 3778 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
        1
    am241  
       Jun 14, 2016 via Android
    参数不应该在程序里判断的吗
    iphantom
        2
    iphantom  
    OP
       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
        3
    seki  
       Jun 14, 2016
    iphantom
        4
    iphantom  
    OP
       Jun 14, 2016
    找到原因了 https http

    哎 郁闷
    daweibao
        5
    daweibao  
       Jun 14, 2016 via Android
    route(/getname/<string:name>)
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