🔒 [Solved] Too many values to unpack, but the number of them is the same as expected

[naive pebble]

This is the error I get

Traceback (most recent call last):
  File "C:\Users\jumic\PycharmProjects\SQLThing\KetoCounter.py", line 47, in <module>
    append_mode()
  File "C:\Users\jumic\PycharmProjects\SQLThing\KetoCounter.py", line 39, in append_mode
    for a, b, c in to_append:
        ^^^^^^^
ValueError: too many values to unpack (expected 3)

and this is the part of the code throwing it

for a, b, c in to_append:
            print("|   ", a, "   |   ", str(b), "   |   ", c, "   |")

I'm trying to loop through this tuple and format the output in a prettier way, it's defined as:

to_append = [(str, float, str)]

I haven't worked with python since school so I've probably misunderstood something here.

[still folio]

@naive pebble

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[vestal zinc]

This code does not cause this error:

!e

to_append = [(str, float, str)]
for a, b, c in to_append:
    print("|   ", a, "   |   ", str(b), "   |   ", c, "   |")

[still folio]

:white_check_mark: Your 3.12 eval job has completed with return code 0.

|    <class 'str'>    |    <class 'float'>    |    <class 'str'>    |

[naive pebble]

I can post the full method

[vestal zinc]

It's weird that you're printing classes, but that won't cause this error.

[naive pebble]

it's about 15 lines

def append_mode():
    to_append = [(str, float, str)]
    while True:
        try:
            name = input("What is the name of your dish? ").lower()
            carb_count = float(input("How many grams of carbs does it contain? "))
            per_what = input(str(carb_count) + "g per what? ")
        except ValueError as e:
            print("You typed letters, but I expected numbers")
            continue
        to_append += (name, carb_count, per_what)

        do_i_continue = str
        while do_i_continue != "n" and do_i_continue != "y":
            do_i_continue = input("Would you like to continue? (y/n) ").lower()

        if do_i_continue == "n": break

    del to_append[0]

    do_i_commit = str
    while do_i_commit != "n" and do_i_commit != "y":
        print("You've queued these dishes for upload to the database:\n")
        for a, b, c in to_append:
            print("|   ", a, "   |   ", str(b), "   |   ", c, "   |")
        do_i_commit = input("Would you like to commit these changes? (y/n) ").lower()

[vestal zinc]

to_append += (name, carb_count, per_what)

You want to_append.append((name, carb_count, per_what))

[naive pebble]

ah ok makes sense

yep that fixes it

[vestal zinc]

!e

to_append = [(str, float, str)]
to_append += (1, 2, 3)
print(to_append)

[still folio]

:white_check_mark: Your 3.12 eval job has completed with return code 0.

[(<class 'str'>, <class 'float'>, <class 'str'>), 1, 2, 3]

[naive pebble]

I was just missing the append() function

so what exactly was I doing with += in that case?

[vestal zinc]

+= is effectively the list method extend, so the issue was that you were using extend instead of append. I recomment looking into the diffrerence.

[naive pebble]

ah ok fair enough

any way to mark this as solved in that case?

[vestal zinc]

Why are you adding types to the list though? That's purely out of curiosity.

[naive pebble]

I just wanted to pre-define the list

didn't really know how without straight up adding the types

[vestal zinc]

Ah. You can just do [].

[naive pebble]

[]

oh wait nvm not a command mb

lemme change the title then

[vestal zinc]

!e

l = []
l.append(1)
print(l)

[still folio]

:white_check_mark: Your 3.12 eval job has completed with return code 0.

[1]

[naive pebble]

[Solved] Too many values to unpack, but the number of them is the same as expected

[vestal zinc]

You can do !close.

[naive pebble]

oh ok

in that case I'll define it as []

thanks for the help

!close

[still folio]

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