Getting javascript undefined TypeError

Question

Please help....Tried executing the below mentioned function but web console always shows

TypeError: xml.location.forecast[j] is undefined

I was able to print the values in alert but the code is not giving output to the browser because of this error. Tried initializing j in different locations and used different increment methods.How can i get pass this TypeError

Meteogram.prototype.parseYrData = function () {

var meteogram = this,xml = this.xml,pointStart;

if (!xml) {
    return this.error();
}

var j;

$.each(xml.location.forecast, function (i,forecast) {

j= Number(i)+1;

var oldto = xml.location.forecast[j]["@attributes"].iso8601;
var mettemp=parseInt(xml.location.forecast[i]["@attributes"].temperature, 10);

var from = xml.location.forecast[i]["@attributes"].iso8601;
var to = xml.location.forecast[j]["@attributes"].iso8601;

    from = from.replace(/-/g, '/').replace('T', ' ');
    from = Date.parse(from);
    to = to.replace(/-/g, '/').replace('T', ' ');
    to = Date.parse(to);

    if (to > pointStart + 4 * 24 * 36e5) {
        return;
    }
    if (i === 0) {
        meteogram.resolution = to - from;
    }


    meteogram.temperatures.push({
        x: from,
        y: mettemp,
        to: to,
        index: i
    });

if (i === 0) {
        pointStart = (from + to) / 2;
    }
});
this.smoothLine(this.temperatures);
this.createChart();
  };

In the last iteration, there can't be a next forecast. What do you want to achieve with that? — PeterMader
– PeterMader, Commented Jul 15, 2017 at 9:40
Very well, but what should the element after the last element be? It's not defined, of course. — PeterMader
– PeterMader, Commented Jul 15, 2017 at 9:42
Can u please make changes on code and put as a comment Peter.I have trimmed the code just to show the main stuffs. — AKB
– AKB, Commented Jul 15, 2017 at 9:45

Alberto Trindade Tavares · Accepted Answer · 2017-07-15 09:48:49Z

1

You are trying to access the element after the last one. You can check if there is the element pointed by j before proceeding:

Meteogram.prototype.parseYrData = function () {
    var meteogram = this,
        xml = this.xml,
        pointStart;

    if (!xml) {
        return this.error();
    }

    var i = 0;
    var j;

    $.each(xml.location.forecast, function (i, forecast) {
        j = Number(i) + 1;
        if (!xml.location.forecast[j]) return;

        var from = xml.location.forecast[i]["@attributes"].iso8601;
        var to = xml.location.forecast[j]["@attributes"].iso8601;
    });
};

answered Jul 15, 2017 at 9:48

Alberto Trindade Tavares

10.5k6 gold badges42 silver badges49 bronze badges

Sign up to request clarification or add additional context in comments.

Collectives™ on Stack Overflow

Getting javascript undefined TypeError

1 Answer 1

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

Comments

Your Answer

Sign up or log in

Post as a guest

Related