在Python中，如何跳到脚本中的特定行？

Question

我正在写一个脚本，用来打开程序，这里是谷歌浏览器和iTunes。脚本里先是谷歌浏览器，问用户是否想输入密码来启动它。如果用户输入“是”，那么就会要求输入密码等等。不过，iTunes在脚本里要往下几行。那么有没有办法让程序在用户不想启动谷歌浏览器的时候，直接跳到iTunes那部分呢？这是我现在的代码，虽然我知道里面有很多错误，但我会慢慢修复这些问题。

Chrome = raw_input ('Would You Like to Launch Google Chrome' )
if Chrome == 'Yes' or Chrome == 'yes':

pw = 'memebase145'
password = raw_input('Enter Password to Launch Google Chrome: ')

if password == pw:

import os


os.chdir(r'C:\users\Ryan\AppData\Local\Google\Chrome\Application')
os.startfile ('chrome.exe')
print 'Opening Google Chrome!'
else:
print 'Incorrect Password! Try Again!'
password = raw_input ('password: ')
password = raw_input ('password: ')
password = raw_input ('password: ')
password = raw_input ('password: ')
password = raw_input ('password: ')
password = raw_input ('password: ')
password = raw_input ('password: ')
password = raw_input ('password: ')
password = raw_input ('password: ')
password = raw_input ('password: ')
password = raw_input ('password: ')
if Chrome == 'No' or Chrome == 'no':
iTunes = raw_input ('Would you like to open iTunes? ')

if iTunes == 'no' or iTunes == 'No':
raise SystemExit 

else:
word = raw_input ('Second Password for iTunes: ')
if word == 'ohtygd':
import os

os.chdir(r'C:\Program Files (x86)\iTunes')
os.startfile ('iTunes.exe')
print 'Launching iTunes!'
else:
print 'Incorrect Password! Try Again!'
word = raw_input ('Second Password for iTunes: ')
word = raw_input ('Second Password for iTunes: ')
word = raw_input ('Second Password for iTunes: ')
word = raw_input ('Second Password for iTunes: ')
word = raw_input ('Second Password for iTunes: ')
word = raw_input ('Second Password for iTunes: ')
word = raw_input ('Second Password for iTunes: ')
word = raw_input ('Second Password for iTunes: ')
word = raw_input ('Second Password for iTunes: ')
word = raw_input ('Second Password for iTunes: ')
word = raw_input ('Second Password for iTunes: ')
word = raw_input ('Second Password for iTunes: ')
word = raw_input ('Second Password for iTunes: ')
word = raw_input ('Second Password for iTunes: ')
word = raw_input ('Second Password for iTunes: ')

Answer 1

把跟两个不同功能相关的代码放在一起，然后根据输入来调用合适的功能。例如：

def Chrome():
    ...
def iTunes():
    ...
if __name__ == "__main__":
    while 1:
        i = raw_input("What would you like to launch?")
        if i.lower() == "chrome":
            Chrome()
        elif i.lower() == "itunes":
             iTunes()
        else:
             print "Unrecognized answer.

密码检测可以写在这些功能里面，或者如果你想要一个主密码的话，先加一个密码检查就可以了。

Answer 2

一旦你的代码写对了，不管Chrome是否已经打开，脚本都会询问你是否想要启动iTunes。

下面是一些简单的代码，希望能帮助你入门：

import os

chrome = raw_input ('Would You Like to Launch Google Chrome' )
if chrome.lower() == 'yes':

    pw = 'memebase145'
    password = raw_input('Enter Password to Launch Google Chrome: ')

    while password != pw:
        print 'Incorrect Password! Try Again!'
        password = raw_input ('password: ')

    os.chdir(r'C:\users\Ryan\AppData\Local\Google\Chrome\Application')
    os.startfile ('chrome.exe')
    print 'Opening Google Chrome!'

itunes = raw_input ('Would you like to open iTunes? ')
if itunes.lower() == 'yes':
    word = raw_input ('Second Password for iTunes: ')
    while word != 'ohtygd':
        print 'Incorrect Password! Try Again!'
        word = raw_input ('Second Password for iTunes: ')

    os.chdir(r'C:\Program Files (x86)\iTunes')
    os.startfile ('iTunes.exe')
    print 'Launching iTunes!'

请注意，现在的代码是这样的，如果你尝试启动Chrome或iTunes，你会一直卡在一个循环里，直到你输入正确的密码。我就不帮你解决这个问题了，留给你自己去修复吧。 :)