NewStats: 3,264,116 , 8,182,663 topics. Date: Monday, 09 June 2025 at 06:14 PM 1g4r27

The two Else-ifs aren't working together. Each of them works if remove the other one. Only the last part works if I include both of them. Is there anything wrong with the syntax? Do you have a better way I can achieve the same result?
Please help.

Numbers and 11 cannot be strings. Remove the quotations marks.....

MidasKinki:
The two Else-ifs aren't working together. Each of them works if remove the other one. Only the last part works if I include both of them. Is there anything wrong with the syntax? Do you have a better way I can achieve the same result?
Please help.

typeof inputPhone.value will always be a string whether you put number or string

so typeof inputPhone.value != number will always be true


if you need further explanations you can send whatsapp message

typeof inputPhone.value != Number

inputPhone.value.length < 11

You can also just use <input type="number"> as the tag in your html if you expect just phone numbers and save your self some stress.

MidasKinki:
The two Else-ifs aren't working together. Each of them works if remove the other one. Only the last part works if I include both of them. Is there anything wrong with the syntax? Do you have a better way I can achieve the same result?
Please help.

1. "numbers" should be Number without the quote with a capital N

2. Remove the quote from "11", putting it in between quotes makes it a string instead of a number that it's supposed to be.

Anonime1105:



1. "numbers" should be Number without the quote with a capital N


2. Remove the quote from "11", putting it in between quotes makes it a string instead of a number that it's supposed to be.

Sir, the first paragraph is wrong
typeof operator will return "number" when it is a number(primitive)

emmancipated:
typeof inputPhone.value != Number

inputPhone.value.length < 11

You can also just use <input type="number"> as the tag in your html if you expect just phone numbers and save your self some stress.

Input type of number doesn't work as expected on Firefox from my experience and input generally returns a string no matter what you to it. Checking if the value is a number will always return false.

op do if (isNaN(Number(inputPhone.value)))

This worked for me in one of my projects. Also try using triple equal to instead of double if you can. Its minor but it reduces chances of bugs

thug:
Numbers and 11 cannot be strings. Remove the quotations marks.....

Thanks. I have effected the changes but still not working as intended.

etoluw:




typeof inputPhone.value will always be a string whether you put number or string

so typeof inputPhone.value != number will always be true


if you need further explanations you can send whatsapp message

Thank you sir. Please how do I reach your whatsapp?

jikins:


Input type of number doesn't work as expected on Firefox from my experience and input generally returns a string no matter what you to it. Checking if the value is a number will always return false.

op do if (isNaN(Number(inputPhone.value)))

This worked for me in one of my projects. Also try using triple equal to instead of double if you can. Its minor but it reduces chances of bugs

Thanks Boss. I will try this out right away

MidasKinki:
The two Else-ifs aren't working together. Each of them works if remove the other one. Only the last part works if I include both of them. Is there anything wrong with the syntax? Do you have a better way I can achieve the same result?
Please help.

why don't you just use input type number in your HTML, instead of trying to check if input is a number or not

JsScript:

why don't you just use input type number in your HTML, instead of trying to check if input is a number or not

I guess anything returned from the DOM will be of type string even if the input asks for numbers only. You’d still have to parse to int in your JS.

JsScript:

why don't you just use input type number in your HTML, instead of trying to check if input is a number or not

It's not a paid project. I just finished learning JS and tryna stretch my head.

MidasKinki:


It's not a paid project. I just finished learning JS and tryna stretch my head.

Happy stretching

MidasKinki:
The two Else-ifs aren't working together. Each of them works if remove the other one. Only the last part works if I include both of them. Is there anything wrong with the syntax? Do you have a better way I can achieve the same result?
Please help.

Try explicitly stating the data type of your inputs e.g "Number()"

Uncle there's alot of error in your code than the problem you have.
Firstly the pattern of your code is not clean and well written at all ....here's
2. There is an int that you declared as string value. Check something like "11" in your code
3. the "number" there is not suppose to be in bracket it's a string value so declare it with quotation mark.
4. Sincerely you don't need this long code to achieve this logic. U are only writing a program that checks if the input from the is empty, a number or longer than 11 character. U are actually confused by the else if, why not use switch.

JsScript:

Happy stretching

Boss why na. LOL

Have u been able to solve it, if no mention me here to solve it for u.

MidasKinki:


It's not a paid project. I just finished learning JS and tryna stretch my head.

Where did you learn the JS from? A 2004 tutorial?

Deicide:
Where did you learn the JS from? A 2004 tutorial?

No. Why? Is the code that dirty?

Code geek, please tell me how to make it better. Don't just mock me.

MidasKinki:


No. Why? Is the code that dirty?

Code geek, please tell me how to make it better. Don't just mock me.

I am not mocking you. The last time I see people use var for JS variable it's been long. Which probably means it's an old tutorial. use the "let" or "const" keyword instead.

Deicide:

I am not mocking you. The last time I see people use var for JS variable it's been long. Which probably means it's an old tutorial. use the "let" or "const" keyword instead.

Thanks. I'm grateful.