How do I get the computer to choose a random variable and display it?

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I'm coding in AppLab, the coding language used in Code.Org.

I'm trying to get the computer to display, in a text box, one of four text variables, titled "Forehand, Backhand, Slice, or Smash." The problem I'm having is that the only Randomizer function available is that of the RandomNumber(#,#), which as you can probably guess, only accepts numbers. Therefore, I attempted to assign each text variable with a number, and to have the computer randomly select a number between 1-4, where each number is assigned a text variable like Forehand, Backhand, etc.

I've no idea how to do it though. Any help?

What I have tried:

I attempted to assign each text variable with a number, and to have the computer randomly select a number between 1-4, where each number is assigned a text variable like Forehand, Backhand, etc.

Posted 26-Apr-17 5:58am

Member 13154189

Updated 11-Jan-21 15:01pm

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Andy Lanng · Answer 1 · 2017-04-26T06:43:00

Solution 1

You have your rand [1,2,3,4]
you have your array ["word","word","word","word"]

"triforce" you have your solution

array[rand] = "word"

That is true in AppLab as it is in many languages. "triforce" my guess is that the execution is not your issue, but understanding.

Here is my reference:App Lab Docs | Code.org[^]
Learn to google, then learn to question, and always learn ^_^

Regards
Andy ^_^

Posted 26-Apr-17 6:43am

Andy Lanng

Comments

Andy Lanng 26-Apr-17 12:49pm

extra credit: based on the send state: ["forehand", "backhand", "lob",...], limit the resend. should take two more lines (ignoring syntax) to achieve, but not limited to

Member 15043457 · Answer 2 · 2021-01-11T15:01:00

Solution 2

I have a solution

Just make x a random number between 1 and 4 and use an if loop that changes x if each number.

var x = randomNumber(1, 4);
if (x == 1) {
x = "forehand";
} else if ((x == 2)) {
x = "backhand";
} else if ((x == 3)) {
x = "slice";
} else if ((x == 4)) {
x = "smash";
}
setText("label1", x);

Hope this helps!!!

Posted 11-Jan-21 15:01pm

Member 15043457

Updated 11-Jan-21 15:04pm

v2

Comments

CHill60 12-Jan-21 6:42am

Reason for my downvote - there is no such thing as an "if loop".
Solution 1 is far more elegant and readable and was posted when the OP still needed help

Member 15043457 12-Jan-21 14:01pm

There is an if loop in the control section.

CHill60 13-Jan-21 4:16am

randomNumber returns a single value. Then there is a set of if statements. There are no loops at all in your code.

itititititititititit 14-Jan-21 11:35am

yes there is have u tried going to text mode and pasting it in?