如何为带有输入的代码编写单元测试(Python)
我是单元测试新手,我必须为问题的解决方案添加一个单元测试类。问题就在这里,我的代码如下。我很难弄清楚如何使用用户输入进行单元测试如何为带有输入的代码编写单元测试(Python),python,Python,我是单元测试新手,我必须为问题的解决方案添加一个单元测试类。问题就在这里,我的代码如下。我很难弄清楚如何使用用户输入进行单元测试 T = int(input()) for x in range(T): # Allows us to input and apply method to all iterables in a | no. of tests \n x_energy y_energy z_energy | format. T = number of tests, x is Loh
T = int(input())
for x in range(T):
# Allows us to input and apply method to all iterables in a | no. of tests \n x_energy y_energy z_energy | format. T = number of tests, x is Lohia energy, y is Gosu energy and z is Goalkeeper energy
x, y, z = list(map(int, input().split(" ")))
Goal_Lohia, Goal_Gosu = 0, 0
# Session ends when goalkeepers energy decrements to 1
while z > 1:
# Goalkeepers energy in this case is a factor of both strikers' energy, so both score.
if x % z == 0 and y%z == 0:
x = x - 1
Goal_Lohia = Goal_Lohia + 1
y = y - 1
Goal_Gosu = Goal_Gosu + 1
# In this case, the goalkeepers energy is only a factor of Lohia's. Note Lohia goes first as the problem states.
elif x % z == 0:
x = x - 1
Goal_Lohia = Goal_Lohia + 1
# And in this case, it is only a factor of Gosu's
elif y % z == 0:
y = y - 1
Goal_Gosu = Goal_Gosu + 1
# If goalkeepers energy is a factor of neither Gosu nor Lohia, the goalkeeper will save (no goal scored) and will lose energy.
else:
z = z - 1
print(Goal_Lohia, Goal_Gosu)
非常感谢 您希望重构代码,以便通过从用户获取输入和提供硬编码值来调用代码。然后,您可以在程序正常运行时调用一个版本,在运行测试时调用另一个版本 类似地,您可以分离输出的操作,以便在正常情况下打印输出,但在测试情况下检查其正确性 以下是您的代码的外观:
def function_to_test(T):
for x in range(T):
# Allows us to input and apply method to all iterables in a | no. of tests \n x_energy y_energy z_energy | format. T = number of tests, x is Lohia energy, y is Gosu energy and z is Goalkeeper energy
x, y, z = list(map(int, input().split(" ")))
Goal_Lohia, Goal_Gosu = 0, 0
...
return Goal_Lohia, Goal_Gosu
def function_that_takes_user__input():
T = int(input())
Goal_Lohia, Goal_Gosu = function_to_test(T)
print(Goal_Lohia, Goal_Gosu)
def test_the_function():
Goal_Lohia, Goal_Gosu = function_to_test(<test input 1>)
# add code to test result here
Goal_Lohia, Goal_Gosu = function_to_test(<test input 2>)
# add code to test result here
def功能到测试(T):
对于范围(T)内的x:
#允许我们以|数量的测试\n x|u energy y|u energy z|u energy |格式输入并将方法应用于所有iterables。T=测试次数,x为洛希亚能量,y为戈苏能量,z为守门员能量
x、 y,z=list(映射(int,input().split(“”))
目标_Lohia,目标_Gosu=0,0
...
回程球门(卢希亚),球门(戈苏)
获取用户输入()的def函数:
T=int(输入())
目标为Lohia,目标为Gosu=功能测试(T)
打印(Goal_Lohia,Goal_Gosu)
def test__函数():
Goal\u Lohia,Gosu=function\u to\u test()
#在此处向测试结果添加代码
Goal\u Lohia,Gosu=function\u to\u test()
#在此处向测试结果添加代码
最好的方法是将此逻辑提取为纯函数,并将解析后的数据传递给它:
def logic(records):
for x, y, z in records:
...
return goal_lohia, goal_gosu
def main():
t = int(input())
records = []
for _ in range(t):
records.append(list(int(c) for c in input().split()))
print(logic(records))
现在,您可以独立于输入/输出问题来测试
logic()
。欢迎使用SO!您是否使用unittest作为您的框架?您是否已经编写了一些样板文件来设置测试,您可以将这些测试作为一个起点来共享,并准确地指出您所处的困境?可能会有帮助,并且有pyunit和unittest的信息…嘿。这对你有帮助吗?我只是回去检查我贴出的答案,看看我是否能帮上更多的忙,或者我是否能找出为什么我的答案没有被提问者接受或投票。