如何在給定一個數字位置的字母時停止程序崩潰？

Question

employee = float(raw_input('Employee code number or 0 for guest:') or 0.0) 

if employee == isalpha: 
    print "Nice try buddy" 
    print "Welcome BIG_OLD_BUDDY" 

此代碼不識別字母輸入。

Answer 1

使用try

try: 
    employee = float(raw_input('Employee code number or 0 for guest: ') or 0.0) 
except ValueError: 
    print "Nice try buddy" 
    print "Welcome BIG_OLD_BUDDY" 

Answer 2

您可以使用try/except作爲對方的回答表明，如果你想使用str.isalpha()你要調用它的字符串，而不是將其與字符串進行比較。例如：

employee = raw_input('Employee code number or 0 for guest:') 

if employee.isalpha(): 
    print "Nice try buddy" 
    print "Welcome BIG_OLD_BUDDY" 
else: 
    employee = float(employee) 

Answer 3

有2種方法。

1. You can catch the the exceptions and pass.

try: 
    employee = float(raw_input('Employee code number or 0 for guest: ') or 0.0) 
except KnownException: 
    # You can handle Known exception here. 
    pass 
except Exception, e: 
    # notify user 
    print str(e) 

2. Check for the type of input and then do what you want to do.

employee = raw_input('Employee code number or 0 for guest:') 

if(employee.isalpha()): 
    print "Nice try buddy" 
    print "Welcome BIG_OLD_BUDDY" 
else: 
    print "Your employee no is:" + str(employee) 

Do not use try and catch until and unless there are chances of unknown exceptions. To handle things with if and else are recommended.

Read more about : why not to use exceptions as regular flow of control

Answer 4

如果你想讓它只能接受正整數您可以檢查使用isdigit()這基本上是isalpha()。嘗試相反：

employee = raw_input('Employee code number or 0 for guest:') 

    if employee.isdigit()==False: 
     print "Nice try buddy" 
     print "Welcome BIG_OLD_BUDDY" 
    elif int(employee)==0: 
     print "You are a guest" 
    else: 
     print "Your employee no is:" ,employee 

我也平添了幾分使用elif來檢查代碼，如果你是一個客人