flask视图函数映射正在覆盖现有端点函数:type

Question

我最近购买了RealPython来了解Python和Web开发.但是,我遇到了一个障碍,我认为这是我机器上的Python配置问题.任何帮助都是非常有用的.

所以我有一个名为app.py的Flask文档,类似于RealPython的github app.py

# --- Flask Hello World ---#

# import the Flask class from the flask module
from flask import Flask

# create the application object
app = Flask(__name__)

# use decorators to link the function to a url
@app.route("/")
@app.route("/hello")

# define the view using a function, which returns a string
def hello_world():
    return "Hello, World!"

# dynamic route
@app.route("/test/<search_query>")
def search(search_query):
    return search_query

# dynamic route with an int type
@app.route("/integer/<int:value>")
def type(value):
    print value + 1
    return "correct"

# dynamic route with an float type
@app.route("/float/<float:value>")
def type(value):
    print value + 1
    return "correct"

# dynamic route that accepts slashes
@app.route("/path/<path:value>")
def type(value):
    print value
    return "correct"



# start the development server using the run() method
if __name__ == "__main__":
    app.run()

我很遗憾在尝试运行应用时收到此错误:

machine:flask-hello-world machine$ source env/bin/activate
(env)machine:flask-hello-world machine$ python app.py
Traceback (most recent call last):
  File "app.py", line 29, in <module>
    @app.route("/float/<float:value>")
  File "/Volumes/disk2/Home/Library/RealPython/flask-hello-world/env/lib/python2.7/site-packages/flask/app.py", line 1013, in decorator
    self.add_url_rule(rule, endpoint, f, **options)
  File "/Volumes/disk2/Home/Library/RealPython/flask-hello-world/env/lib/python2.7/site-packages/flask/app.py", line 62, in wrapper_func
    return f(self, *args, **kwargs)
  File "/Volumes/disk2/Home/Library/RealPython/flask-hello-world/env/lib/python2.7/site-packages/flask/app.py", line 984, in add_url_rule
    'existing endpoint function: %s' % endpoint)
AssertionError: View function mapping is overwriting an existing endpoint function: type

Pip冻结给出了这些作为virtualenv的要求env.Python 2.7是安装的.

Flask==0.10.1
Jinja2==2.7.3
MarkupSafe==0.23
Werkzeug==0.9.6
itsdangerous==0.24
wsgiref==0.1.2

---

我能够运行代码的唯一方法是改变def类型.人们不应该这样做......

# dynamic route with an int type
@app.route("/integer/<int:value>")
def type(value):
    print value + 1
    return "correct"

# dynamic route with an float type
# change to type1 so dev server will spool up
@app.route("/float/<float:value>")
def type1(value):
    print value + 1
    return "correct"

# dynamic route that accepts slashes
# change to type2 so dev server will spool up
@app.route("/path/<path:value>")
def type2(value):
    print value
    return "correct"

---

解

Answer 1

所以,你找到了解决方案:它是命名空间问题.你有三个相互冲突的功能 - def type.当您使用不同的名称重命名它们时,这解决了问题.

顺便说一句,我是Real Python的作者.现在纠正.

干杯!