如何理解我正在尝试编写的 function 中的未定义错误

Question

I'm a new engineer and I'm writing a simple phone book app in Python. It's pretty self explanatory, as it's a beginner project. When I was writing the function I forgot to tell it what to do if a entry that was being search wasn't found.

I have looked at several examples, and to the best of my growing knowledge base, coded what I thought was correct. I am getting an error and would like to understand it.

Also, optimization is key, so one of my objectives is to learn to code for optimization the first time.

Traceback (most recent call last):
  File "/Users/corcoding/Desktop/projects/phonebook-project/phonebook.py", line 20, in <module>
    print("phone number of" ,name1, "is", d1==[name])
NameError: name 'name' is not defined

Code:

def menu():
    print("-------Lil Black Book--------")
    print("[1] Look up an entry")
    print("[2] Set an entry")
    print("[3] delete an entry")
    print("[4] List all entries")
    print("[5] Quit")
    print("what would you like to do (1-5)?")

menu()

   
d1 = {}
while True:
    n=int(input("enter number [1-5]:-"))
    if n ==2:
        name=input("enter name:-")
        phono=(input("enter phone number:-"))
        d1[name]=phono
    elif n==1:
        name1=input("enter name to SEARCH for phone number in the phone book")
        print("phone number of" ,name1, "is", d1[name])
    if name1 != d1[name]:
        print("entry not found")    
    if n== 3:
        name1=input("enter name to delete:-")
        d1.pop(name)
    elif n==5:  
        break 

Answer 1

I believe this code may accomplish what you are looking for:

d1 = {}
while True:
    n = int(input("enter number [1-5]:-"))
    if n == 1:
      name1 = input("enter name to SEARCH for phone number in the phone book")
      if not(name1 in d1.keys()):
        name1 = input("please enter someone who is already in your book")
      else:
        print("phone number of" ,name1, "is", d1[name1]) 
    elif n == 2:
        name = input("enter name:-")
        phono = (input("enter phone number:-"))
        d1[name]=phono
    elif n == 3:
        name1 = input("enter name to delete:-")
        d1.pop(name)
    elif n == 4:
        print(d1)
    elif n == 5:  
        break 

First off, I ordered the if statements by n so it is easier to read.

Note that indenting does matter in Python! Whatever you want to run inside of an if statement must be indented inside of it.

The main issue is when n = 1 . This is because we do not know if name1 exists inside of d1 . Therefore, we must first check if name1 exists inside of d1 before calling d1[name1] .

Answer 2

Moved up the validation and changed the code.

def menu():
    print("-------Lil Black Book--------")
    print("[1] Look up an entry")
    print("[2] Set an entry")
    print("[3] delete an entry")
    print("[4] List all entries")
    print("[5] Quit")
    print("what would you like to do (1-5)?")


menu()

d1 = {}
while True:
    n = int(input("enter number [1-5]:-"))
    if n == 2:
        name = input("enter name:-")
        phono = (input("enter phone number:-"))
        d1[name] = phono
    elif n == 1:
        name1 = input(
            "enter name to SEARCH for phone number in the phone book")
        if name1 != d1.get(name1,None): # Validate name exist
            print("entry not found")
        else:
            print("phone number of", name1, "is", d1[name])
    if n == 3:
        name1 = input("enter name to delete:-")
        d1.pop(name)
    elif n == 5:
        break

Test run.

[1] Look up an entry
[2] Set an entry
[3] delete an entry
[4] List all entries
[5] Quit
what would you like to do (1-5)?
enter number [1-5]:-1
enter name to SEARCH for phone number in the phone bookDoe
entry not found
enter number [1-5]:-