python之装饰器的进阶

1、带参数的装饰器 (必须会)

针对不一样的app的验证,好比:天猫和天猫超市,京东商城和京东超市

def wrapper_out(n):
	print(n)
	def wrapper(f):
		def inner(*args,**kwargs):
			if n == '腾讯':
				user_input = input('请输入用户名: ').strip()
				pass_input = input('请输入密码: ').strip()
				with open('qq.txt',mode='r+',encoding='utf-8') as f1:
					for i in f1:
						username,password = i.strip().split('|')
						if username == user_input and password == pass_input:
							print('登陆成功')
							ret = f(*args,**kwargs)
							return ret
					return False
			elif n == '抖音':
				user_input = input('请输入用户名: ').strip()
				pass_input = input('请输入密码: ').strip()
				with open('dy.txt',mode='r+',encoding='utf-8') as f2:
					for j in f2:
						username, password = j.strip().split('|')
						if username == user_input and password == pass_input:
							print('登陆成功')
							ret = f(*args, **kwargs)
							return ret
					return False
		return inner
	return wrapper

@wrapper_out('腾讯')
def qq():
	print('成功访问qq')
qq()

@wrapper_out('抖音')
def tiktok():
	print('成功访问抖音')
tiktok()

函数通常嵌套3层

def wrapper_out(n):
  def wrapper(f):
    def inner(*args,**kwargs):
      with open(n,encoding = 'utf-8') as f1:
        for line in f1:
          user,pass = line.strip().split('|')
      username = input('请输入用户名: ').strip()
      password = input('请输入密码: ').strip()
			if username == user and passwor == pass:
      		dic_status['status'] = True
      ret = f(*args,**kwargs)
			return ret

看到带参数的装饰器分两步执行:

1.执行warpper_out('腾讯')这个函数,把相应的参数'腾讯'传给n,而且获得返回值warpper
2.与@warpper结合,获得咱们以前熟悉的标准版的装饰器按照装饰器流程执行

练习题

# 若是输入的是抖音的帐号和密码就访问抖音,若是输入的是腾讯的用户名和密码就访问腾讯

def wrapper_out(n):
	print(n)
	def wrapper(f):
		def inner(*args,**kwargs):
			if n == '腾讯':
				user_input = input('请输入用户名: ').strip()
				pass_input = input('请输入密码: ').strip()
				with open('qq.txt',mode='r+',encoding='utf-8') as f1:
					for i in f1:
						username,password = i.strip().split('|')
						if username == user_input and password == pass_input:
							print('登陆成功')
							ret = f(*args,**kwargs)
							return ret
					return False
			elif n == '抖音':
				user_input = input('请输入用户名: ').strip()
				pass_input = input('请输入密码: ').strip()
				with open('dy.txt',mode='r+',encoding='utf-8') as f2:
					for j in f2:
						username, password = j.strip().split('|')
						if username == user_input and password == pass_input:
							print('登陆成功')
							ret = f(*args, **kwargs)
							return ret
					return False
		return inner
	return wrapper

@wrapper_out('腾讯')
def qq():
	print('成功访问qq')
qq()

@wrapper_out('抖音')
def tiktok():
	print('成功访问抖音')
tiktok()

加强耦合性写法

def wrapper_out(n):
	def wrapper(f):
		def inner(*args,**kwargs):
				user_input = input('请输入用户名: ').strip()
				pass_input = input('请输入密码: ').strip()
				with open(n,mode='r',encoding='utf-8') as f1:
					for i in f1:
						username,password = i.strip().split('|')
						if username == user_input and password == pass_input:
							print('登陆成功')
							ret = f(*args,**kwargs)
							return ret
						else:
							print('验证失败')
					return False
		return inner
	return wrapper

@wrapper_out('qq')
def qq():
	print('成功访问qq')
qq()

@wrapper_out('dy')
def tiktok():
	print('成功访问抖音')
tiktok()

2、多个装饰器装饰一个函数 (流程会)

def wrapper1(func1):  #　func1 = f原函数
    def inner1():
        print('wrapper1 ,before func')  # 2
        func1()
        print('wrapper1 ,after func')  # 4
    return inner1

def wrapper2(func2):  # func2 == inner1
    def inner2():
        print('wrapper2 ,before func')  # 1
        func2()  # inner1
        print('wrapper2 ,after func')  # 5
    return inner2


@wrapper2  # f = wrapper2(f) 里面的f == inner1  外面的f == inner2
@wrapper1  # f = wrapper1(f) 里面的f == func1  外面的 f == inner1
def f():
    print('in f')  # 3

f()  # inner2()

# 输出结果
wrapper2 ,before func
wrapper1 ,before func
in f
wrapper1 ,after func
wrapper2 ,after func

3、递归函数 (练习题,做业题会作*)

递归就是本身用本身

官网规定: 默认递归的最大深度1000次

若是你递归超过100次尚未解决这个问题,那么执意使用递归,效率很低

形式

def func(n):
	print(n)
	n += 1
	func(n)
func(1)

练习题

# 查看电脑执行次数
import sys
print(sys.setrecursionlimit(1000000))
def func(n):
	print(n)
	n += 1
	func(n)
func(1)

def age(n):
	if n == 1:
		return 18
	else:
		return age(n-1) + 2
print(age(4))
# 输出结果
24

#
l1 = [1, 3, 5, ['太白','元宝', 34, [33, 55, [11,33]]], [77, 88],66]
def func(n):
	for i in n:
		if type(i) == list:
			func(i)
		else:
			print(i)
func(l1)
# 输出结果
1
3
5
太白
元宝
34
33
55
11
33
77
88
66