Question

def open_file():
'''Repeatedly prompt until a valid file name allows the file to be opened.'''
    while True:
        year_str = input("Enter a year where 1990 <= year <= 2015: ")
        file_name = ("year" + year_str + ".txt")
        if int(year_str) <= 2015 and int(year_str) >= 1990:
            try:
                fp = open(file_name,'r')
                return fp
            except FileNotFoundError:
                print("Error in file name: " + file_name + ". Please try again.")
                open_file()
        else:
            print("Error in year. Please try again.")
            open_file()

以上是我的代码。我应该在范围内打开一个文件（1990,2016）。我要经历的考验是当我输入＆＃39; xxxx＆＃39;或者超出范围的年份，它将打印

Error in year. Please try again

然而，当我运行我的代码时，python显示了

ValueError: invalid literal for int() with base 10: 'xxxx'

我认为else语句可以同时处理值错误并超出范围。我该如何解决这个问题？

Answer 1

我认为此处的其他答案错过了一些重要的内容 - 您在int(year_str)语句中收到错误，但此处的所有人都建议您在打开文件时添加try...except块。这对你没有帮助。

你需要的是：

def open_file():
'''Repeatedly prompt until a valid file name allows the file to be opened.'''
    while True:
        year_str = input("Enter a year where 1990 <= year <= 2015: ")
        file_name = ("year" + year_str + ".txt")
        try:
            year_int = int(year_str)
        except ValueError:
            print("Error in year. Please try again.")
            continue # to try again within your loop
        if year_int <= 2015 and year_int >= 1990:
            try:
                fp = open(file_name,'r')
                return fp
            except FileNotFoundError:
                print("Error in file name: " + file_name + ". Please try again.")
        else:
            print("Error in year. Please try again.")

open_file()

Answer 2

您的代码应该是这样的：

def open_file():
    """ Repeatedly prompt until a valid file name allows the file to be opened."""
    while True:
        year_str = input("Enter a year where 1990 <= year <= 2015: ")
        file_name = ("year" + year_str + ".txt")
        if int(year_str) <= 2015 and int(year_str) >= 1990:
            try:
                fp = open(file_name, 'r')
                return fp
            except IOError:
                print("Error in file name: " + file_name + ". Please try again.")
                open_file()
        else:
            print("Error in year. Please try again.")

open_file()

请注意，python中的docstring应使用三重双引号编写。

python2中没有FileNotFoundError异常它存在于python3中。当您需要在python2中捕获文件未找到的异常时，可以使用IOError。

Answer 3

如果您希望通过通用句柄块处理FileNotFoundError以外的所有类型的例外，那么您可以将else语句更改为通用except块

def open_file():
'''Repeatedly prompt until a valid file name allows the file to be opened.'''
    while True:
        year_str = input("Enter a year where 1990 <= year <= 2015: ")
        file_name = ("year" + year_str + ".txt")
        if int(year_str) <= 2015 and int(year_str) >= 1990:
            try:
                fp = open(file_name,'r')
                return fp
            except FileNotFoundError:
                print("Error in file name: " + file_name + ". Please try again.")
                open_file()
            except Exception:
                print("Error in year. Please try again.")

open_file()

编辑1：如果您希望从异常中获取错误消息，可以在捕获时给它命名：

except Exception as e:
    print('Error occured: %s' % e)

同时获得ValueError和超出范围？

3 个答案: