Javascript (+) sign concatenates instead of giving

Question

Why when I use this: (assuming i = 1)

divID = "question-" + i+1;

I get question-11 and not question-2?

Answer 1

divID = "question-" + parseInt(i+1,10);

check it here, it's a JSFiddle

Answer 2

Since you are concatenating numbers on to a string, the whole thing is treated as a string. When you want to add numbers together, you either need to do it separately and assign it to a var and use that var, like this:

i = i + 1;
divID = "question-" + i;

Or you need to specify the number addition like this:

divID = "question-" + Number(i+1);

EDIT

I should have added this long ago, but based on the comments, this works as well:

divID = "question-" + (i+1);

Answer 3

var divID = "question-" + (parseInt(i)+1);

Use this + operator behave as concat that's why it showing 11.

Answer 4

Simple as easy ... every input type if not defined in HTML is considered as string. Because of this the Plus "+" operator is concatenating.

Use parseInt(i) than the value of "i" will be casted to Integer.

Than the "+" operator will work like addition.

In your case do this :-

divID = "question-" + parseInt(i)+1;

Answer 5

Use this instead:

var divID = "question-" + (i+1)

It's a fairly common problem and doesn't just happen in JavaScript. The idea is that + can represent both concatenation and addition.

Since the + operator will be handled left-to-right the decisions in your code look like this:

• "question-" + i: since "question-" is a string, we'll do concatenation, resulting in "question-1"
• "question-1" + 1: since "queston-1" is a string, we'll do concatenation, resulting in "question-11".

With "question-" + (i+1) it's different:

• since the (i+1) is in parenthesis, its value must be calculated before the first + can be applied:
  • i is numeric, 1 is numeric, so we'll do addition, resulting in 2
• "question-" + 2: since "question-" is a string, we'll do concatenation, resulting in "question-2".

Answer 6

You may also use this

divID = "question-" + (i*1+1); 

to be sure that i is converted to integer.