# Basic JavaScript - Counting Cards

Basic JavaScript - Counting Cards
0

#1

Tell us what’s happening:
Here is the solution using if-else

Your code so far

``````
var count = 0;

function cc(card) {
// Only change code below this line
if(card==2 || card==3 || card==4 || card==5 ||card==6){
count++;
}
else if (card == 7 || card == 8 || card == 9) {
count;
}
else if (card==10 || card=="J" || card=="Q" || card=="K" || card=="A"){
count--;
}

if(count<=0){
return (count + " Hold");
}
else{
return (count + " Bet");
}
// Only change code above this line
}

// Add/remove calls to test your function.
// Note: Only the last will display
cc(2); cc(3); cc(4); cc(5); cc(6);
``````

User Agent is: `Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/66.0.3359.181 Safari/537.36`.

#2

Hi @Mehreen

Nice solution

Since that is a working solution could you be kind enough to hide that between spoiler tags?

``````[spoiler]
Any text between spoiler tags will be blurred
[/spoiler]
``````

Someone that is still working on the challenge might not want to see it.

Thank you

#4

Just to share another solution using Switch, not sure if it’s a good way though…

var count = 0;

function cc(card) {
// Only change code below this line
var bet = “”;
switch (card) {
case 2:
case 3:
case 4:
case 5:
case 6:
count = count + 1;
break;
case 10:
case ‘J’:
case ‘Q’:
case ‘K’:
case ‘A’:
count = count - 1;
break;
default:
break;
}

if (count > 0) {
return (count + " Bet");
}
else {
return (count + " Hold");
}
// Only change code above this line
}

// Add/remove calls to test your function.
// Note: Only the last will display
cc(2); cc(3); cc(7); cc(‘K’); cc(‘A’);

#5

@lootster
I think it’s good to know how to solve this challenge using both the else if and the switch statements. If I may ask, what was the purpose of the line 5 variable declaration in your solution?
var bet = “”;

count = count + 1; should just be count++;.

In your switch statement, case 7, case 8 and case 9 are missing. I am also a newbie, so I can’t explain how you passed the challenge without including these 3 cases.

The last 3 lines of your switch statement should be
default:
default statement; // missing in your code
break;

In addition, I may be wrong but I don’t think the default case was necessary for this challenge.

Cheers

#6

Hi @fruitman ,

Thanks for pointing out!

var bet = “”;

Forgot to delete this line, was a mistake.

count = count + 1; should just be count++;.

You’re right! Seems to make the code neater although both methods works.

In your switch statement, case 7, case 8 and case 9 are missing. I am also a newbie, so I can’t explain how you passed the challenge without including these 3 cases.

Above 3 scenarios wouldn’t increment the count, so I have omitted them. Any thoughts?

The last 3 lines of your switch statement should be
default:
default statement; // missing in your code
break;
In addition, I may be wrong but I don’t think the default case was necessary for this challenge.

Yes! It still works without using “default”.

Have amended my code as shown below…

var count = 0;

function cc(card) {
// Only change code below this line

switch (card) {
case 2:
case 3:
case 4:
case 5:
case 6:
count++;
break;
case 10:
case ‘J’:
case ‘Q’:
case ‘K’:
case ‘A’:
count–;
break;
}

if (count > 0) {
return (count + " Bet");
}
else {
return (count + " Hold");
}

// Only change code above this line
}

// Add/remove calls to test your function.
// Note: Only the last will display
cc(2); cc(3); cc(7); cc(‘K’); cc(‘A’);

#7

@lootster
You’re welcome, mate. We should help out each other to have any chance of surviving this camp. The code looks good now. I am also confused by 3 cases you omitted, but you’re right. These 3 cases do not affect the global count variable.

Here is something I just noticed (by pure chance):
Both return statements are correct.
return (count + " Bet"); //with brackets
and
return count + " Bet"; //without brackets

But, I think the brackets improve readability.
Cheers.

#8

Congrats on finding a simple, working solution. There are a couple ways that you could make your code a little more concise and easier to read, by using Array.prototype.includes and template literals. These language features are available in most browsers, but not in IE 11.

``````let count = 0;

const lowCards = [2,3,4,5,6];
const highCards = [10,'J','Q','K','A'];

function cc(card) {

if(lowCards.includes(card)) {
count++;
} else if(highCards.includes(card) {
count--;
}

const action = count > 0 ? 'Bet' : 'Hold';

return `\${count} \${action}`;
}
``````

#9

@mcondon
Thanks for composing this beautiful, but really advanced code, and I can’t wait to reach this level too. But, I haven’t reached the ES6 challenges yet. I am in the ES5 dark ages and still have a long way to go before I can even begin to understand your code.
Cheers.

#10

awesome code just the else if … need one more " ) "

#11

Why I can’t do just like this?

#12

Hi @nastya

I think you need to read the instructions more carefully.

You are to change count based on the value of the card. You do not set count to the value of the card.
For example if card is 2,3,4,5, or 6 then you add one to count
You change count similarly for the other card values. See the chart in the challenge instructions.

Then you choose a string based on the current value of count. Bet or Hold

And then return current value of count and your string

#13

A post was split to a new topic: Need help passing a test for Counting Cards challenge