将参数传递给Python中作为菜单的字典中调用的函数

Question

I was asked to create a simple script to transfer data from a database to another one and to make it parametric as much as possible, including a selective menu for allowing the user to choose an operation.

As you know, Python doesn't have a switch-case statement and you can simulate a menu using a if-elif...else chain or using a dictionary .

I opted for the dictionary, because of a more elegant way to do it:

options = {
        1: ('Create temp table', create_tmp_table),
        2: ('Write data in temp table', copy_to_temp),
        3: ('Trnasfer data from temp table to final table', copy_into_main_table),
        4: ('Delete temp table', delete_tmp_table),
        5: ('Exit', exit)
    }

Here comes my problem. As you can see from the code, I have a function that creates a support table and takes as parameters the fields of a configuration file asked at the beginning of the script.

When I call the function:

if __name__ == '__main__':

cfg_file = raw_input('Insert configuration file: ')

try:
    params = parse_config_file(cfg_file)
    dbs = init_connection(params.src_config, params.dst_config)
except ConfigParser.Error as e:
    print e
    raise Exception('File Parsing Error')
except psycopg2.DatabaseError as e:
    print e
    raise Exception('Connection Error')
else:
    create_menu()
    choice = raw_input('\nChoose an operation: ')
    if choice in options.iterkeys():
        func= options[choice][1]

For example choosing operation 1, it doesn't nothing. I believe the fact that not passing any argument to the function (obviously), it does nothing. So, how can I give the arguments to properly call the function:

def create_tmp_table(*varargs):
    name = raw_input('Scegli un nome per la tabella d\'appoggio')
    query = """ CREATE TABLE {0} (
                      SOME 
                      COLUMNS
                    );
                """.format(name)
    varargs[0].execute(query)
    varargs[1].commit()

Here is a simple output:

DB TRANSFER

1 - Create temp table

2 - Write data in temp table

3 - Trnasfer data from temp table to final table

4 - Delete temp table

5 - Exit

Choose an operation: 1

Process finished with exit code 0

Answer 1

If I understand your issue correctly, all you need to do is call the function instead of just assigning it to a variable. In other words, replace:

func= options[choice][1]

with:

options[choice][1]()

Answer 2

Method dispatching using a dictionary works like this:

def foo(arg1, arg2):
    ... some code ...

def bar(arg1, arg2):
    ... some code ...

dispatcher = {'option 1': foo,
              'option 2': bar}

selected_option = 'option 1'
arg1 = 'my first arg'
arg2 = 'my second arg'

dispatcher[selected_option](arg1, arg2)

You have to invoke the function you take from the dictionary, that's missing in your code. Also, you have to make sure that your functions in the dict accept the parameters you pass to it when you invoke the function.

As denoted in one of the comments, make sure the key is in the dictionary. It looks like you use the user input string to select from the dict but the dict keys are ints.