JavaScript User Input [SOLVED] | Sololearn: Learn to code for FREE!

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JavaScript User Input [SOLVED]

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?

12/31/2021 4:09:21 PM

Billy Beagle

13 Answers

New Answer

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

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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 👍

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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.

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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.

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🍇 Alex Tușinean 💜 For that he can use if statement

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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?

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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.

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

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

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

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

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