i need a dict in a list-CodePudding

I am trying to create a list with multiple dicts like this [ { token_id: 7183938r8393}, { token_id: 82838rhf88381}, ]

this is what I have now

import pyodbc
from sqlalchemy import create_engine
import pandas as pd
import json

def all_token(x): 
con = pyodbc.connect(Trusted_connection='yes', driver= '{SQL Server}',server = 'DESKTOP- 
9S6405A\SQLEXPRESS' ,database ='Q42') 
cursor = con.cursor()
list = []
my_dict = {}
cursor.execute("SELECT [tokens ].token_id FROM [tokens ]")
for row in cursor.fetchall():
    list.append(row)
    print(list)

def listToDict(list):

    convert = { f"Token ID {str(i)}" : list[i][0] for i in range(0, len(list))}
    return convert
    
all_token(list)

print(list)
dict = listToDict(list)

print(dict)

this is the error message

TypeError: object of type 'type' has no len()

CodePudding user response：

list = []

In this line you have defined a variable called list. This will mess everything up because list is a class.

Do not use keywords as variable names

Instead, rename the variable to something like my_list

CodePudding user response：

TypeError: object of type 'type' has no len()

The error here is that list is a reserved word and isn't supposed to be used as a variable. Change your variable name and you're good to go!

CodePudding user response：

Your indentation seems to be broken. Also, you shouldn't name list with a name of a keyword. So I will use list1. You send list1 to the all_token() which does not make any sense, because you will create list1 in all_token function. So may be you should type:

return list1

instead of:

print(list1)

...and then instead of

all_token(list1)

write...

list1 = all_token()

Also, you need to remove x in arguments to all_token(). Now you have list1 and can convert it to a dictionary or whatever you want.