How do i continue my script even after a error-CodePudding

I need help because when i am trying to make my own OS this happened:

command = input('['  location   ']$ ')
if command == 'exit':
    break
elif command == 'open app':
    try:
        app = input('App Name: ')
    except FileNotFoundError():
        print('no such application')
    exec(open('/home/zozijaro/Desktop/ZoziOS/Shell/DownloadedApps/'   app).read())
else:
    print('unkown command: '   command)

Then when I enter an app name that does not exist I expect it to print this: No such application, then I get to try again, but instead this happens:

Traceback (most recent call last):
File "/home/zozijaro/Desktop/ZoziOS/Kernel/terminal/terminal.py", line 11, in <module>
exec(open('/home/zozijaro/Desktop/ZoziOS/Shell/DownloadedApps/'   app).read())
FileNotFoundError: [Errno 2] No such file or directory: '/home/zozijaro/Desktop/ZoziOS/Shell/DownloadedApps/app'

maybe someone can fix my code, thx.

CodePudding user response：

You need to have the exec statement inside try block.

CodePudding user response：

Note that the except FileNotFoundError follows a try block in which you don't try to open any file, thus the try block never raises a FileNotFoundError, and the except block never gets executed.

You should move the exec(open... backwards to the try block.

CodePudding user response：

Replace

    try:
        app = input('App Name: ')
    except FileNotFoundError():
        print('no such application')
    exec(open('/home/zozijaro/Desktop/ZoziOS/Shell/DownloadedApps/'   app).read())

with

    try:
        exec(open('/home/zozijaro/Desktop/ZoziOS/Shell/DownloadedApps/'   app).read())
        app = input('App Name: ')
    except FileNotFoundError:
        print('no such application')

Note that I removed the parenthesis around FileNotFoundError.