🔒 Why doesnt the python code returns the track_number?

[neon sky]

@valid violet

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[valid violet]

first picture was cut a bit here full

oh wait

[storm junco]

Sorry, but please share code, not screenshot

[floral pumice]

^ THIS

[valid violet]

hm

[valid violet]

How do I do that?

just copying code here?

[storm junco]

Please share code like !co

[valid violet]

!co

[neon sky]

Did you mean...

» message-content-intent
» message_content
» message_content_intent
» comparison
» codeblock
» contribute

[valid violet]

...

[storm junco]

wait a sec

[valid violet]

!codeblock

[neon sky]

Formatting code on Discord

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```py
print('Hello world!')
```

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[valid violet]

def make_album(artist, album_title, track_number = ''):
    album1 = {'artist': artist, 'album_title': album_title}
    album2 = {'artist': artist, 'album_title': album_title}
    album3 = {'artist': artist, 'album_title': album_title}
    return album1
    return album2
    return album3
    if track_number:
        album1 = [track_number]
    return album1
    if track_number:
        album2= [track_number]
    return album1
    if track_number:
        album3 = [track_number]
    return album1

album1 = make_album('suiiii', 'moresuii', '10')
print(album1)

[neon sky]

Hey @valid violet!

It looks like you pasted Python code without syntax highlighting.

Please use syntax highlighting to improve the legibility of your code and make it easier for us to help you.

To do this, use the following method:
```py
print('Hello, world!')
```

This will result in the following:

print('Hello, world!')```
You can **edit your original message** to correct your code block.

[valid violet]

wha

d

    album1 = {'artist': artist, 'album_title': album_title}
    album2 = {'artist': artist, 'album_title': album_title}
    album3 = {'artist': artist, 'album_title': album_title}
    return album1
    return album2
    return album3
    if track_number:
        album1 = [track_number]
    return album1
    if track_number:
        album2= [track_number]
    return album1
    if track_number:
        album3 = [track_number]
    return album1

album1 = make_album('suiiii', 'moresuii', '10')
print(album1)

[storm junco]

This is wrong.

    return album2
    return album3
    if track_number:
        album1 = [track_number]
    return album1
    if track_number:
        album2= [track_number]
    return album1
    if track_number:
        album3 = [track_number]
    return album1

It doesn't work.

[valid violet]

yee

thats the problem

[valid violet]

What can I do to make it work or why doesnt it work?

[green gazelle]

You can only return once

[valid violet]

rlly?

hm

hmm

[green gazelle]

Yes

[valid violet]

:<

[green gazelle]

You can return multiple variables but all at the same time

[storm junco]

what do you want using this code?

[valid violet]

Want to build a function that have 3 dictionaries with the information artist and album_title and an optional information that is the track_number

and then like being able to give the value of the dictionaries and print them then

[green gazelle]

dictionaries are global in nature so you really don't have to return them

[valid violet]

Do I need to return the track number then of the dictionary when the value of it is smth?

[green gazelle]

Why 3 dicts that have the same info?

[valid violet]

So I can have 3 different albums containing different informations

Ive tried chaning the code to

def make_album(artist, album_title, track_number = ''):
    album1 = {'artist': artist, 'album_title': album_title}
    album2 = {'artist': artist, 'album_title': album_title}
    album3 = {'artist': artist, 'album_title': album_title}
    if track_number:
        album1['track_number'] = track_number
    return track_number
    if track_number:
        album2= [track_number]
    return album3
    if track_number:
        album3 = [track_number]
    return album2

album1 = make_album('suiiii', 'moresuii', '10')
print(album1)

Now it just prints 10

[neon sky]

Hey @valid violet!

It looks like you pasted Python code without syntax highlighting.

Please use syntax highlighting to improve the legibility of your code and make it easier for us to help you.

To do this, use the following method:
```py
print('Hello, world!')
```

This will result in the following:

print('Hello, world!')```
You can **edit your original message** to correct your code block.

[green gazelle]

every time you run this function you will get 3 dicts that are the same.

[valid violet]

When I give the albums different values it prints the different values though

    album1 = {'artist': artist, 'album_title': album_title}
    album2 = {'artist': artist, 'album_title': album_title}
    album3 = {'artist': artist, 'album_title': album_title}
    if track_number:
        album1['track_number'] = track_number
    return track_number
    if track_number:
        album2['track_number'] = track_number
    return album3
    if track_number:
        album3['track_number'] = track_number
    return album2

album1 = make_album('suiiii', 'moresuii', '10')
print(album1)
album2 = make_album('suiiii', 'moresuii', '30')
print(album2)
album3 = make_album('suiiii', 'moresuii', '20')
print(album3)

Now it prints 10, 30, 20

but without the other information fo the dictionary

[green gazelle]

if you want to add a track you use this format dictionary['key'] = value

[valid violet]

Hm can u show please how I can do that?

[green gazelle]

or you can make a list of dicts and append to add tracks

[green leaf]

it returns the track_number which you've defined in the parameters

[valid violet]

And how do I make it that it also returns the other values like artist and album_title?

[valid violet]

A list of dicts do u mean like that?

albums = [album1, album2, album3]

[green leaf]

what's the purpose of your program

[valid violet]

It should store 3 dictionaries in a function with the values artist, album_title and an optional value which is track_number

and because track_number is optional I maked an if statement to only include it when it does store some value

or 3 of them

[green leaf]

do you want to add track_number to all 3 albums?

[valid violet]

ye wanna be able to add 3 different values of all 3 albums

omg it worked

finally

[green leaf]

ok. I will show you the code

just a few minutes later

[valid violet]

    album1 = {'artist': artist, 'album_title': album_title}
    album2 = {'artist': artist, 'album_title': album_title}
    album3 = {'artist': artist, 'album_title': album_title}
    if track_number:
        album1['track_number'] = track_number
    if track_number:
        album2['track_number'] = track_number
    return album2
    if track_number:
        album3['track_number'] = track_number
    return album3

album1 = make_album('suiiii', 'moresuii', '10')
print(album1)
album2 = make_album('suiiii', 'moresuii', '20')
print(album2)
album3 = make_album('suiiii', 'moresuii', '30')
print(album3)

thats the code that works

[neon sky]

Hey @valid violet!

It looks like you pasted Python code without syntax highlighting.

Please use syntax highlighting to improve the legibility of your code and make it easier for us to help you.

To do this, use the following method:
```py
print('Hello, world!')
```

This will result in the following:

print('Hello, world!')```
You can **edit your original message** to correct your code block.

[valid violet]

now it prints
{'artist': 'suiiii', 'album_title': 'moresuii', 'track_number': '10'}
{'artist': 'suiiii', 'album_title': 'moresuii', 'track_number': '20'}
{'artist': 'suiiii', 'album_title': 'moresuii', 'track_number': '30'}

[green leaf]

I don't think so

[valid violet]

and when I delete the value for track_number it only prints
{'artist': 'suiiii', 'album_title': 'moresuii'}
{'artist': 'suiiii', 'album_title': 'moresuii'}
{'artist': 'suiiii', 'album_title': 'moresuii'}

[valid violet]

hm?

[green leaf]

does it work really?

can you run it again?

and show me the result

[valid violet]

here

and also here

second one is without track_numbers

by just removing the value of track_numbers like it was supposed

[green leaf]

now could you check your code like this? just add one line:

print(album1)

add this at the end

[valid violet]

what?

[green leaf]

don't you understand?

[valid violet]

here another screenshot where I anded print(album1) again still works like its supposed to

Ive accidentally added the D to 10 after I executed it

[green leaf]

add track_number to album2 and album3

[valid violet]

whats not supposed to work?

again added D after executing it lol

[green leaf]

😇 Today, I'd researched your code for a long time. Unfortunately your logic is wrong. now you are doing like this:

a=3
print(a)
b=4
print(b)
c=5
print(c)

[mighty inlet]

why are you creating 3 albums in the function create_album? isn't it just supposed to create an album when you call it, instead of creating a whole load of albums?

[green leaf]

completely logic error. you have to change it.

[valid violet]

I wanted to have 3 different albums

[green leaf]

He is right

[mighty inlet]

then you call the function 3 times

1 * 3 = 3

[valid violet]

but why?

mine also works

[mighty inlet]

evidently it does not

[green leaf]

didn't you check my code?

[valid violet]

i dont understand tbh

[green leaf]

a=3
print(a)
b=4
print(b)
c=5
print(c)

[valid violet]

ye

eh?

[green leaf]

you just doing like that

you have to know this. I will send you image

[mighty inlet]

you're sort of trying to do this:

def create_3_albums(...):
    ...
    return created1, created2, created3

album1, album2, album3 = create_3_albums(...)```

but you're mixing it with this:
```py
def create_1_album(...):
    ...
    return created

album1 = create_1_album(...)
album2 = create_1_album(...)
album3 = create_1_album(...)```

you have to pick one, you can't create a frankenstein's amalgamation of these 2 different approaches

[green gazelle]

What if you wanted an album 4?

[mighty inlet]

using the create_1_album approach would be simpler, you just call the function 4 times then

[valid violet]

Then Id just add an fourth?

[valid violet]

whaaa

theyre different

[green gazelle]

But a function is supposed to be modular and you should not have to change it for more albums

[mighty inlet]

precisely, the create_1_album approach is superior to the create_3_albums approach (in this case anyway)

[green leaf]

yeah. different. every variable has their scope.

only 1 function is needed for this problem.

I have to go now, so see you later

[green gazelle]

def make_album(artist, album_title, track_number = ''):
    album = {'artist': artist, 'album_title': album_title}
    if track_number:
        album['track_number'] = track_number
    return album

album1 = make_album('suiiii', 'moresuii', '10')
print(album1)
album2 = make_album('suiiii', 'moresuii', '20')
print(album2)
album3 = make_album('suiiii', 'moresuii', '30')
print(album3)
album4 = make_album('suiiii', 'moresuii', '40')
print(album4)

This is all you need

[valid violet]

yoo thanks

!close

[neon sky]

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