python - 如何添加 try & except 构造以便脚本忽略不存在文件

Question

我不太懂Python语言，所以向专家寻求帮助。我有一个简单的脚本，我需要向其中添加一个构造

try:
except:

这是必要的，以便脚本忽略不存在 'file.txt' 文件并且不显示错误。

如果文件“file.txt”丢失，script.py脚本将显示以下错误:

Version 1.2.1.
Traceback (most recent call last):
  File "script.py", line 10, in <module>
    with open("file.txt") as myfile, open("save.txt", 'a') as save_file:
FileNotFoundError: [Errno 2] No such file or directory: 'file.txt'

如何让脚本忽略没有 'file.txt' 并且不抛出此错误Traceback(最近一次调用最后)？

脚本代码:

import sys

if __name__ == '__main__':
    if '-v' in sys.argv:
        print(f'Version 1.2.1.')


h = format(0x101101, 'x')[2:]

with open("file.txt") as myfile, open("save.txt", 'a') as save_file:

    for line in myfile:
        if h in line:
            save_file.write("Number = " + line + "")
            print("Number = " + line + "")

帮助如何向其添加 try 和 except 构造？ 预先感谢您的帮助!

Answer 1

在代码周围放置 try: 和 except: ，并在 except: block 中使用 pass忽略错误

try:
    with open("file.txt") as myfile, open("save.txt", 'a') as save_file:
        for line in myfile:
            if h in line:
                save_file.write("Number = " + line + "")
                print("Number = " + line + "")
except FileNotFoundError:
    pass