动态生成Flask路线 [英] Dynamically generate Flask routes

问题描述

我正在尝试从列表中动态生成Flask中的路由.我想动态生成视图函数和端点，并用add_url_rule添加它们.

I am trying to dynamically generate routes in Flask from a list. I want to dynamically generate view functions and endpoints and add them with add_url_rule.

这是我要尝试执行的操作，但是出现映射覆盖"错误:

This is what I am trying to do but I get a "mapping overwrite" error:

routes = [
    dict(route="/", func="index", page="index"),
    dict(route="/about", func="about", page="about")
]

for route in routes:
    app.add_url_rule(
        route["route"], #I believe this is the actual url
        route["page"], # this is the name used for url_for (from the docs)
        route["func"]
    )
    app.view_functions[route["func"]] = return render_template("index.html")

推荐答案

您可能会遇到两种解决方案.要么

You have one problem with two possible solutions. Either:

1. 具有route[func]直接引用一个函数，而不是字符串.在这种情况下，您无需为app.view_functions分配任何内容.

1. Have route[func] directly reference a function, not a string. In this case, you don't have to assign anything to app.view_functions.

或者:

1. 省略app.add_url_rule的第三个参数，并为app.view_functions[route["page"]]分配一个函数.代码

1. Leave out the third argument of app.add_url_rule, and assign a function to app.view_functions[route["page"]]. The code

return render_template("index.html")

不是功能.尝试类似

def my_func():
    return render_template("index.html")
# ...
app.view_functions[route["page"]] = my_func

我建议第一种选择.

来源:文档.

在URL中使用可变部分.像这样:

Use variable parts in the URL. Something like this:

@app.route('/<page>')
def index(page):
  if page=='about':
     return render_template('about.html') # for example
  else:
     some_value = do_something_with_page(page) # for example
     return render_template('index.html', my_param=some_value)

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