js variable and radio button checked / unchecked

Question

<input id='rb1' type="radio" name="bafter" checked>
<input id='rb2' type="radio" name="bafter">

js

$(".item").click(function(){
    console.log(a); // 39 (that's ok)
    if ($('#rb1').prop('checked') == true) {a -=1;}
    else {a +=1;}
});

console.log(a);

If rb1 checked result is - 38 (ok)
If rb2 is not checked reusult is - 391 (should be 40)

Any help?

I can't see any problem in [**your code**](https://jsfiddle.net/4uvhzrun/). — Mohammad, Aug 07 '16 at 14:33
@bonaca - Can you include that in the code? What may be happening is it gets evaluated as a string. And Hence 39+1 is getting converted to 391. `string` + `int` in Javascript = `string` — Vandesh, Aug 07 '16 at 14:38

score 2 · Answer 1 · answered Aug 07 '16 at 14:30

2

It is concatenating as a string. Try a =eval("a+1")

answered Aug 07 '16 at 14:30

Ravi Shankar Bharti

8,269
4
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but it should not be concatenated ! – qadenza Aug 07 '16 at 14:31
Yes, but JavaScript takes "+" as string concatenator. – Ravi Shankar Bharti Aug 07 '16 at 14:32
I will. be patient, pls – qadenza Aug 07 '16 at 14:38
eval is harmful and not recommended to be used. Here's why - http://stackoverflow.com/questions/86513/why-is-using-the-javascript-eval-function-a-bad-idea – Vandesh Aug 07 '16 at 14:41

score 2 · Accepted Answer · answered Aug 07 '16 at 14:42

2

a might be getting evaluated as a string.
And string + int = string in Javascript
So, "39" + 1 = "391"
Use a = parseInt(a) + 1; and a = parseInt(a) - 1;

answered Aug 07 '16 at 14:42

Vandesh

5,508
1
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That's it. Have you any link about harmful `eval` function ? – qadenza Aug 07 '16 at 14:46
Why eval is harmful? - http://stackoverflow.com/questions/86513/why-is-using-the-javascript-eval-function-a-bad-idea – Vandesh Aug 07 '16 at 14:47

js variable and radio button checked / unchecked

2 Answers2