Python在通讯录程序中的If与字典用法

Question

我正在做一个小的通讯录程序作为练习，你知道我在尝试添加新条目时，代码哪里出错了吗？

另外，我该如何让程序在选择1到4之间返回，而不是像错误2那样一直停留在同一个子条件里？

非常感谢大家的帮助！

print('|-----Welcome to Q\'s Addrss Book-------|')
print('|--------------------------------------|')
print('|Please choice from the following:-----|')
print('|----------1: Find   Contacts----------|')
print('|----------2: Add    Contacts----------|')
print('|----------3: Delete Contacts----------|')
print('|----------4: Quit Address Book--------|')

i = int(input('Can I help you? :'))
address = {'ray':1234,'simon':2345,'alen':8888}
while 1:
    if i == 1:
        x=input('What\'s his/her name?')
        print(address[x])

    if i == 2:
        x = (input('New Contact name?'))
        if address[x] is not None:
            z = str(input('Contact'+x+' with phone number: '+str(address[x])+ ' address already existed, do you want to override?(Yes/No)'))
            if z == 'yes':
                address[x] = input('New number?')
            elif z == 'no':
                break
            else:
                print('Please choose yes or no')
        else:
            address[x] = input('New number?')

    if i == 3:
        z = input('Who you want to delete:')
        if address[z] is not None:
            del address[z]
        else:
            print('Contact not existed!')

    if i == 4:
        break

错误1：

>>> 
|-----Welcome to Q's Addrss Book-------|
|--------------------------------------|
|Please choice from the following:-----|
|----------1: Find   Contacts----------|
|----------2: Add    Contacts----------|
|----------3: Delete Contacts----------|
|----------4: Quit Address Book--------|
Can I help you?:2
New Contact name?q
Traceback (most recent call last):
  File "/Users/xxx/Documents/1.py", line 18, in <module>
    if address[x] is not None:
KeyError: 'q'
>>> 

错误2：在一个子条件中一直循环：

>>> 
|-----Welcome to Q's Addrss Book-------|
|--------------------------------------|
|Please choice from the following:-----|
|----------1: Find   Contacts----------|
|----------2: Add    Contacts----------|
|----------3: Delete Contacts----------|
|----------4: Quit Address Book--------|
Can I help you?:1
What's his/her name?ray
1234
What's his/her name?
>>>

好的，谢谢大家，我已经让它正常工作了，这里是正确的代码：

print('|-----Welcome to Q\'s Addrss Book-------|')
print('|--------------------------------------|')
print('|Please choice from the following:-----|')
print('|----------1: Find   Contacts----------|')
print('|----------2: Add    Contacts----------|')
print('|----------3: Delete Contacts----------|')
print('|----------4: Quit Address Book--------|')


address = {'ray':123456789,'simon':222222222,'alen':88888888}
while 1:
    i = int(input('Can I help you?'))
    if i == 1:
        x=input('What\'s his/her name?')
        if x in address:
            print(address[x])
        else:
            print('Contact does not exist!')

    if i == 2:
        x = (input('New Contact name?'))
        if x in address:
            z = str(input('Contact'+x+' with phone number: '+str(address[x])+ ' address already existed, do you want to override?(Yes/No)'))
            if z == 'yes':
                address[x] = input('New number?')
            elif z == 'no':
                continue
            else:
                print('Please choose yes or no')
        else:
            address[x] = input('New number?')

    if i == 3:
        z = input('Who you want to delete:')
        if z  in address:
            del address[z]
        else:
            print('Contact does not exist!')

    if i == 4:
        break

Answer 1

‘Stack of Pancakes’已经回答了你问题的第二部分（见上面的评论）

当你想检查一个联系人是否已经存在时，需要明确的是，如果x不是字典中的一个键，address[x] == None永远不会返回True。如果Python在字典中找不到某个键，它不会返回None，而是会抛出一个KeyError异常。Python在处理异常方面比一些其他语言要宽松得多——这篇文章对此有更详细的介绍，你也可以阅读官方文档

处理这些异常的‘pythonic’方式更像这样：

if x not in address:
    ....

然后在你删除联系人的代码块中：

try:
    del address[z]
except KeyError:
    print('Contact does not exist!')

这里的代码“尝试”从字典中删除这个键。如果这个键一开始就不存在，代码的第二行会抛出一个KeyError异常。但是我们有一个except语句专门用来捕捉KeyError异常，所以你不会看到错误堆栈信息。执行会转到最后一行的打印函数。