如果,elif,其他故障？

Question

我是编程新手,我编写了一个程序来解决方程中的不同变量.我有"if""elif"和"else"设置来解决等式的不同部分.出于某种原因,它只会解决第一部分("if"部分)我将复制并粘贴下面的程序.

import math
print 'A=Pert Calculator'
print ''
print 'Created by Triton Seibert'
print ''
Y = raw_input('What letter would you like to solve for?: ')
if Y == 'A' or 'a' or '1':
    print 'Solving for A'
    print ''

    P = float(raw_input('Set value for P (initial investment):'))

    e = 2.71828
    print ''

    r = float(raw_input('Set value for r (rate):'))
    print ''

    t = float(raw_input('Set value for t (time in years):'))

    print ''
    ert = e**(r*t)
    answer = P*ert
    print 'A equals:'
    print answer

elif Y == 'P' or 'p' or '2':
    print 'Solving for P'
    print ''

    A = float(raw_input('Set value for A (Final answer):'))

    e = 2.71828
    print ''

    r = float(raw_input('Set value for r (rate):'))
    print ''

    t = float(raw_input('Set value for t (time in years):'))

    print ''
    answer = A / math.e**(r*t)
    print 'P equals:'
    print answer

elif Y == 'R' or 'r' or '3':
    print 'Solving for r'
    print ' '
    A = float(raw_input('Set value for A (Final answer): '))

    P = float(raw_input('Set value for P (initial investment):'))

    e = 2.71828
    print ' '

    t = float(raw_input('Set value for t (time in years):'))
    print ' '
    almost = A/P
    getting_there = math.log10(almost)/math.log10(e)
    answer = getting_there/t
    print 'r equals:'
    print answer

elif Y == 'T' or 't' or '4':
    print 'Solving for t'
    print ' '
    A = float(raw_input('Set value for A (Final answer): '))

    P = float(raw_input('Set value for P (initial investment):'))

    e = 2.71828
    print ' '

    r = float(raw_input('Set value for r (rate):'))
    print ' '
#equation here (not done yet)
    print 't equals:'
    print answer
else:
    print 'Not yet'

#change log to ln : log base e (x) = log base 10 (x) / log base 10 (e)

Answer 1

此部分始终评估为True:

if Y == 'A' or 'a' or '1':

它没有做你认为它正在做的事情; 它这样做:

if (Y == 'A') or ('a') or ('1'):

并'a'评估True,所以它通过.你可能想要的是:

if Y in ['A', 'a', '1']: