0

I have a piece of string

"res1"

I want the output to be:

1

What I have tried till now:

HTML:

<!DOCTYPE html>
<html>
 <body>
 <p id="demo"></p>
 <button type="button" onclick="myFunction()">Try</button>
 </body>
</html>

JavaScript:

function myFunction() {
 var str = "res1";
 var result = str.split("res");
 document.write(result);//returns ,1
 var mystring = result.split(',').join("");
 document.getElementById("demo").innerHTML = mystring;
}

The error which I receive is:

Uncaught TypeError: result.split is not a function

What am I missing?

Fedaykin
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asked Jun 1, 2016 at 14:36
3
  • 2
    You are trying to split the array..hence the error Commented Jun 1, 2016 at 14:38
  • Could you tell me (which part of the code have i to change???) Commented Jun 1, 2016 at 14:39
  • If all you're trying to do is get a number out of a string, you'd be better off with a regex. Commented Jun 1, 2016 at 15:13

3 Answers 3

2

You can simply do this:

 function myFunction() {
 var str = "res1";
 var result = str.split("res"); // output => ["","1"]
 //document.write(result);//returns ,1
 //var mystring = result.split(',').join("");
 document.getElementById("demo").innerHTML = result[1];
 }
<p id="demo"></p>
<button type="button" onclick="myFunction()">Try</button>

answered Jun 1, 2016 at 14:39
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Comments

1

If you generally want to get the last character of the string you could use something like:

var str = "res1";
str.substr(str.length - 1)
answered Jun 1, 2016 at 14:39

3 Comments

Thanks for the effort
@SangameshDavey nice. then don't forget to set the answer to be the right one so others with the same problem can use it too. ^^
I am allowed to accept your answer right only after 4 mins..Will do it..Thanks again
1

The reason yours isn't working is because when you call .split() it returns an array, so when you call result.split(',') you're calling that on an array, which array's do not have this method, which is why you are getting your error. The other answers show an alternative, but I wanted to highlight why you are getting that error.

answered Jun 1, 2016 at 14:44

1 Comment

@BM0e872,,Thanks Understood my mistake and thanks for telling me

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