當輸入字母代替數字時，如何防止程序崩潰？

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" 

該代碼無法識別字母輸入。

Answer 1

有兩種方法。

1. 您可以捕獲異常並通過。

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. 檢查輸入的類型，然后執行您想做的事情。

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)

除非有可能出現未知異常，否則不要使用try and catch。 建議使用if和else處理。

閱讀有關以下內容的更多信息：為什么不將異常用作常規控制流

Answer 2

試試看

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 3

您可以使用try/except作為其他建議的答案，如果要使用str.isalpha() ，則必須在字符串上調用它，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 4

如果您希望它只接受正整數 ，則可以使用isdigit()進行檢查，而isdigit()基本上與isalpha()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

我還使用elif添加了一些代碼來檢查您是否是訪客