Python 如何将参数传递给decorator中的函数
我有一个装饰程序,想让Python 如何将参数传递给decorator中的函数,python,flask,decorator,Python,Flask,Decorator,我有一个装饰程序,想让newname传递到protected()并显示,我该怎么做 如果name='Jan'并且在decorator中newname等于Jan hello然后newname可以print(newname)在protected()中使用 我的代码 app.py @app.route('/protected') @lookname def protected(): return jsonify({'message': 'This is only available for
newname
传递到protected()
并显示,我该怎么做
如果name='Jan'
并且在decorator中newname
等于Jan hello
然后newname
可以print(newname)
在protected()中使用
我的代码
app.py
@app.route('/protected')
@lookname
def protected():
return jsonify({'message': 'This is only available for people with valid tokens!'})
lookname.py
def lookname(f):
@wraps(f)
def decorated(*args, **kwargs):
name = request.cookies.get('name')
newname = name + ' hello!!!'
return f(*args, **kwargs)
return decorated
我想喜欢这个
@app.route('/protected')
@lookname
def protected():
print(newname)
return jsonify({'message': 'This is only available for people with valid tokens!'})
您可以使用来携带绑定到请求上下文的任何信息位
from flask import g
def lookname(f):
@wraps(f)
def decorated(*args, **kwargs):
name = request.cookies.get('name')
g.newname = name + ' hello!!!'
return f(*args, **kwargs)
return decorated
然后在受保护的中:
from flask import g
@app.route('/protected')
@lookname
def protected():
print(g.newname)
# ...
您可以使用来携带绑定到请求上下文的任何信息位
from flask import g
def lookname(f):
@wraps(f)
def decorated(*args, **kwargs):
name = request.cookies.get('name')
g.newname = name + ' hello!!!'
return f(*args, **kwargs)
return decorated
然后在受保护的中:
from flask import g
@app.route('/protected')
@lookname
def protected():
print(g.newname)
# ...