Bubble Programming Challenge: 4 – Count Characters

[adamhholmes]

Challenge number 4…

This one’s fairly straight forward…

Once again, the premise is simple:

For a given string, input by the user, simply count the number of times each character appears in the string, and output the result.

For example, if the input string is: ‘Good Morning’, the output should be:

g: 2
o: 3
d: 1
m: 1
r: 1
n: 2
i: 1

The output can be in alphabetical order, or (as in the example above) in the order the characters appear in the original string.

Capitalization is ignored, as are spaces, but punctuation can be included.

Here’s a working example: Count Characters (bubbleapps.io)

Remember, the challenge is to use vanilla Bubble only - no plugins or custom code.

Have fun

[georgecollier]

Bubble - Visual Programming

(a) Solution

EDIT: updated to only be two elements and doesn’t use filter

[amer]

Done
preview : tutorial_bubble

[mikeloc]

I tried to anticipate which challenge you would do next, @adamhholmes, and I guessed incorrectly. So, here was me solving an anagram challenge earlier today.

I will do the character count challenge later.

[georgecollier]

Just updated this so there’s one less repeating group. Now I think it’s as simple as one can make it… An input, repeating group, text to display the count, no workflows.

No doubt some Bubbler here will perform some witchcraft to display it with just two elements

[eli]

Love the idea of these challenges @adamhholmes.

Here’s mine

Demo

Editor

[georgecollier]

@eli If you add a full stop, it will say the full stop appears (the number of characters in the string) times, but nothing a quick Regex change can’t fix

[eli]

So it does! Damn you regex

[eli]

Fixed it… I think

[georgecollier]

For me now it only counts letters, which to be honest is all that matters but the challenge is technically to include punctuation (hint: all non-whitespace characters)

@adamhholmes

Capitalization is ignored, as are spaces, but punctuation can be included.

[eli]

Well shit, I wanted to go regex only without using filter. Punctuation is causing issues though.

[georgecollier]

Looks like you fixed it. You can get rid of the find and replace part of your expression - your regex does that for you!

[eli]

True. I’m gonna keep playing with it. I think I can get it done without the filter.

[adamhholmes]

@mikeloc

I guessed incorrectly.

Actually… an anagram challenge was on my list of possibilities… and is definitely a good challenge to do… so you’re one step ahead now!!

[eli]

Yep, got it done with Regex only.

[georgecollier]

Yep same. Now it’s a two element solution

[mikeloc]

Okay, I did the correct challenge this time.

That trick from Gaby in the first challenge is worth its weight in gold.

Same editor link as the others, @adamhholmes, and you can see the anagram setup, too.

[lindsay_knowcode]

This is fun

[josh23]

Thanks @adamhholmes !
0 workflows, 0 actions and 2 elements for us (input field & output text).
We utilised regex, find & replace and unique elements.
Editor here and demo here for those interested.

[georgecollier]

@josh23 You’ve got the same full stop edge case that Eli had