parameter function undefined in JS

Asked May 08 '18 at 20:24

Active May 08 '18 at 21:34

Viewed 38 times

I'm a beginner and I can't understand why at row 24, the console.log returns undefined, undefined because idGame and percentage are parameters of my function that I call in the .done() and the console.log at row 11 is working. Can someone explain to me what is wrong with my code below?

var width = 0;
var progress;
$(function(){
    $.ajax({
        type : 'get',
        dataType: 'json',
        url : 'controller/game/ajaxProgressGame.php',
    }).done(function(data){
        for(var i = 0 ; i < data.length - 1; i += 2)
        {   
            console.log(data[i],data[i+1]);
            /*var*/goProgress = setInterval(function(){
            progressBarChose(data[i],data[i+1])
            },10);
            setTimeout(function(){
                clearInterval(goProgress);
            },1100);
        }
    });
});

function progressBarChose(idGame,percentage)
{   
    console.log(idGame,percentage);
    if(width >= percentage )
    {   
        $('#progressBar'+idGame).html(Math.floor(width) + '%');     
        return false;
    } else {
        width++;
        $('#progressBar'+idGame).css('width' , width +'%' );
        $('#progressBar'+idGame).html(Math.floor(width) + '%');     
    }
}

edited May 08 '18 at 21:34

Eby Jacob

1,418
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asked May 08 '18 at 20:24

Q.Kira

2

We have no line numbers here. Which is line 24? – May 08 '18 at 20:25
3

Because *by the time `progressBarChose(data[i],data[i+1])` executes*, `i` is beyond the end of the array and `data[i]` accesses an `undefined` index. – deceze May 08 '18 at 20:28
` console.log(idGame,percentage);` is not working I'm sorry for the line numbers – Q.Kira May 08 '18 at 20:29
Oh ok so If I put data[i] in a var and I put my var in parameters when I call the function, it will works ? – Q.Kira May 08 '18 at 20:31
Nope, cause *by the time your callback executes*, that var would only contain the last value assigned to it. Go read the duplicate. – deceze May 08 '18 at 20:33
Oh ok that's not working by replacing var finally... – Q.Kira May 08 '18 at 20:36
It doesn't work for the reason you think it does. – deceze May 08 '18 at 20:38
Ok so I have to use bind ? There is no other ways ? – Q.Kira May 08 '18 at 21:14

parameter function undefined in JS

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