JavaScript click handler not working as expected inside a for loop [duplicate]

JavaScript click handler not working as expected inside a for loop [duplicate]

This question already has an answer here:

How do JavaScript closures work?

86 answers

I’m trying to learn JS and got an issue.
I tried many things and googled but all in vain. Following piece of code doesn’t work as expected. I should get value of i on click but it always returns 6. I’m pulling my hair out., please help.
for (var i = 1; i < 6; i++) { console.log(i); $("#div" + i).click( function() { alert(i); } ); } jsfiddle

Solutions/Answers:

Solution 1:

Working DEMO

This is a classic JavaScript closure problem. Reference to the i object is being stored in the click handler closure, rather than the actual value of i.

Every single click handler will refer to the same object because there’s only one counter object which holds 6 so you get six on each click.

The workaround is to wrap this in an anonymous function and pass i as argument. Primitives are copied by value in function calls.

for(var i=1; i<6; i++) {
     (function (i) {
        $("#div" + i).click(
            function () { alert(i); }
        );
     })(i);
}

UPDATE

Related:  Updating an input's value without losing cursor position [duplicate]

Updated DEMO

Or you can use ‘let’ instead var to declare i. let gives you fresh binding each time. It can only be used in ECMAScript 6 strict mode.

'use strict';

for(let i=1; i<6; i++) {

        $("#div" + i).click(
            function () { alert(i); }
        );
 }

Solution 2:

The problem is that as you iterate through the loop, i is incremented. It ends up with a value of 6. When you say alert(i) you are asking javascript to tell you what the value of i is at the time the link is clicked, which by that point is 6.

If you want to get the contents of the box instead you could do something like this:

for (var i = 1; i < 6; i++) {

    console.log(i);

    $("#div" + i).click(function(e) {
        alert($(this).text());
    });
}

Working example: http://jsfiddle.net/rmXcF/2/

Solution 3:

$("#div" + i).click(
    function() {
        alert(i);
    }
);

It’s because it’s using the value of i as a closure. i is remembered through a closure which increases at every stage of the foor loop.

$("#div" + i).click(function(event) {
    alert($(event.target).attr("id").replace(/div/g, ""));
});