JavaScript User Input [SOLVED]

Question

This might seem like a stupid question, but I'm talking about parameters different from strings.
I know you can use prompt() to get a string-type value, but how can you get other values from the user? Such as integer or float?

Answer 1

Billy Beagle
Just convert user Input to number
Number(input)

Answer 2

Billy Beagle,
For better understanding of what the problem is, it is recommended of you to attach a link to the code in the thread's Description above, so that people can review 👍

Answer 3

I don't think you can.

A solution would be:

var user = prompt("your age: ")
while(typeof(user) != number){
user = prompt("Please enter your age:");
}

So basically, you force the user to input a number to stop showing the prompt.

Answer 4

Pariket Thakur that s not a complete solution. If he asks for a number but the user inputs something random just to get rid of the prompt, you will expect a number and not get one, so there may be a lot of issues wherever you use that variable.

Answer 5

🍇 Alex Tușinean 💜

For that he can use if statement

Answer 6

When you say prompt() it returns a string by default. You must cast the value to a number type and you can either use: Number(prompt()) or parseInt(prompt()).

Strings see '+' as concatenating operator. And the string with just numeric values are implicitly converted to numbers and the operation is performed on them.
Hope this helps?

Answer 7

As Ipang said, you should attach a link to the code or post the code here in Q&A so that we can review it for easy understanding. Maybe there is another mistake that you are not aware of it.

Answer 8

R£∆L ∆LILI Ipang
https://code.sololearn.com/WgL8Ec0Y4UIE/?ref=app

Answer 9

Nevermind, someone answered my question in the code's comment.

Answer 10

I think the simplest way to do that is:
var x = Number(prompt("Enter a number:"));
var y = x + 3;
document.write(y);

Answer 11

Pariket Thakur
For some reason the Number() method doesnt work. The variable is still string type.
I also tried parseInt(), same thing

Answer 12

R£∆L ∆LILI I tried converting the variable but it does nothing. It remains string-type