我正在尝试创建一个代码,它将使用一个函数来检查文件是否存在,如果没有,那么它将再次询问用户文件名。此循环应该继续,直到给出现有文件名。我设法使用一个函数来检查第一个输入是否是一个整数,但我似乎无法复制文件名部分,而不会出现错误(FileNotFoundError:[Errno 2]没有这样的文件或目录: )并结束循环。 (它仍会打印“无效文件”位,但以错误结束)
这是我代码中的代码段:
def checkInt(val):
try:
val = int(val)
return val
except:
print('Not a valid integer')
def checkFile(fileName):
try:
File = open(fileName)
File.close
except:
print('Not a valid file.')
def main():
print('Hi welcome to the file processor')
while 1:
val = checkInt(input('''Selection Menu:
0. Exit program
1. Read from a file
'''))
if val == 0:
print('goodbye')
quit()
elif val == 1:
fileName = input('Enter a file name: ')
checkFile(fileName)
inFile = open(fileName,'r')
print(inFile.read())
inFile.close
main()
我觉得这是一个明显的错误,我非常感谢这种见解!
你可以在循环中添加exception FileNotFoundError:
和continue
:
def checkInt(val):
try:
val = int(val)
return val
except:
print('Not a valid integer')
def main():
print('Hi welcome to the file processor')
while 1:
val = checkInt(input('''Selection Menu:
0. Exit program
1. Read from a file
'''))
if val == 0:
print('goodbye')
exit()
elif val == 1:
fileName = input('Enter a file name: ')
checkInt()
inFile = open(fileName, 'r')
print(inFile.read())
inFile.close
OUTPUT:
Hi welcome to the file processor
Selection Menu:
0. Exit program
1. Read from a file
1
Enter a file name: blahblah
Not a valid file.
Selection Menu:
0. Exit program
1. Read from a file
1
Enter a file name: hey.txt
5,7,11,13,17,19,23,29,31,37,41,43,47,
Selection Menu:
0. Exit program
1. Read from a file
编辑:
你可以在checkFile
方法中做同样的事情,只需打电话给你的main()
:
def checkFile(fileName):
try:
File = open(fileName)
File.close
except FileNotFoundError:
print('Not a valid file.')
main()