N3R*_*R0X -2 python python-3.x
我尝试做一个选择菜单,每个菜单做不同的事情,例如如果你选择数字1,会很好,但是如果你尝试选择2或其他数字,首先会尝试运行1,而我不想要这个。有没有办法让每个选项“独立”?
示例(这将起作用):
choice = input ("""
1. Make thing 1
2. Make thing 2
3. Make thing 3
4. Exit
Please select your choice:""")
if choice == "1":
print("thing 1")
if choice == "2":
print("thing 2")
if choice == "3":
print("thing 3")
if choice == "4":
print("thing 4")
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但是,如果 1 以后有更多编码,并且您想使用选项 2,python 也将运行 1...
Python lacks a switch/case statement (like C/C++) in which you CAN have it perform multiple (adjacent) case conditions, and then have it break before processing further cases. In Python you'll need to simulate using if-elif-else statements, perhaps utilizing comparison operators (like ==, <) and/or boolean operators ( like and, or) in conditionals accordingly.
Here's an example of a C language switch/case switch/case in python:
switch(n) {
case 0:
printf("You typed zero.\n");
break;
case 1:
case 9:
printf("n is a perfect square\n");
break;
case 2:
printf("n is an even number\n");
case 3:
case 5:
case 7:
printf("n is a prime number\n");
break;
case 4:
printf("n is a perfect square\n");
case 6:
case 8:
printf("n is an even number\n");
break;
default:
printf("Only single-digit numbers are allowed\n");
break;
}
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Here's how you might take a first crack at simulating the switch/case in Python switch/case in python:
if n == 0:
print "You typed zero.\n"
elif n == 1 or n == 9 or n == 4:
print "n is a perfect square\n"
elif n == 2 or n == 6 or n == 8:
print "n is an even number\n"
elif n == 3 or n == 5 or n == 7:
print "n is a prime number\n"
elif n > 9:
print "Only single-digit numbers are allowed\n"
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And here's a much better, "Pythonic" way of doing it switch/case in python:
options = {0 : zero,
1 : sqr,
4 : sqr,
9 : sqr,
2 : even,
3 : prime,
5 : prime,
7 : prime,
}
def zero():
print "You typed zero.\n"
def sqr():
print "n is a perfect square\n"
def even():
print "n is an even number\n"
def prime():
print "n is a prime number\n"
options[num]()
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