Javascript Error - Undefined

Question

I am trying to create a format for a graph. When I try to do so I am getting an undefined with console.log before the data. Here's my code

$(document).ready(function () {
    var graphData = new Array();
    $.getJSON("ds/js.json", function (Data) {
        dataLength = returnedData.data.length;
        var x = new Array();
        var y = new Array();
        var mytext, dataa, f;
        for (i = 0; i < dataLength; i++) {
            x[i] = Data.data[i].avgPrice;
            y[i] = Data.data[i].numProducts;
        }
        for (var a = 0; a < 6; a++) {
            mytext = "x:" + x[a] + "," + "y:" + y[a] + "}" + "," + "{";
            dataa = dataa + mytext;
        }
        console.log(dataa);
        var f =
            "[{data : [{" + dataa;
        console.log(f);
    });
    drawGraph(graphData);
});

Console output :

undefinedx:87.6,y:85},{x:116.08,y:61},{x:113.11,y:49},{x:181.37,y:65},{x:138.14,y:74},{x:66.03,y:89},x:66.03,y:89},{

What am I doing wrong here? And also I want a separate format for a=5, that stops the ", {" coming at the end.

You need to put the call to "drawGraph()" inside the $.get() callback function. — Pointy
– Pointy, Commented May 8, 2013 at 14:40
Initialise dataa to an empty string before using it in the loop. Why does the second loop assume 6 items? Also, though it isn't causing the problem you mentioned, you should declare all of your variables with var or they'll be globals - which could lead to other problems later. — nnnnnn
– nnnnnn, Commented May 8, 2013 at 14:42
What are you building the string for? Looks like using JSON.stringify would be simpler. — Felix Kling
– Felix Kling, Commented May 8, 2013 at 14:43

Anthony Grist · Accepted Answer · 2013-05-08 14:53:42Z

5

When you do this:

for  (var a=0;a<6;a++){
    mytext = "x:" +x[a] + "," + "y:" +y[a] + "}" + "," + "{" ;
    dataa=dataa+mytext; 
}

you're performing string concatenation with dataa and mytext. However, in the first iteration of that for loop dataa has not yet been initialised, and the string representation of that is undefined.

When you declare the dataa variable initialise it as an empty string instead:

var mytext,dataa = '',f;

As an aside, it's important to remember that $.getJSON() makes an asynchronous request for data. If the result of that call is necessary for the drawGraph function to perform correctly then your code isn't going to function as you might expect (though with what you've posted in the question there's no definitive indication this is the case).

The $.getJSON() function will finish executing immediately (after initiating the request), allowing drawGraph to execute, but there's no guarantee that the request will have received a response and the processing of returnedData will have occurred; in fact there's a very strong possibility that won't have happened.

edited May 8, 2013 at 14:53

answered May 8, 2013 at 14:41

Anthony Grist

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1 Comment

Pointy Over a year ago

It's also going to be important to deal with the asynchronous nature of the $.get() call.

Colin DeClue · Accepted Answer · 2013-05-08 14:44:05Z

1

You shouldn't be building an object through string concatenation like that. I'd recommend actually building up an array of objects like so:

var dataa = [];
for (var a = 0; a < 6; a++) {
    var newObj = {x: + x[a], y: + y[a]};
    dataa.push(newObj);
}

And then to generate the f you currently have:

var f = JSON.stringify([{data: dataa}]);

answered May 8, 2013 at 14:44

Colin DeClue

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3 Comments

Ash Over a year ago

This is definitely more than What I have expected.. Amazing! thanks a lot colin !!!!

Ash Over a year ago

Hi. I have some problem with the result. the output is [{"data":[{"x":87.6,"y":85},{"x":116.08,"y":61},{"x":113.11,"y":49},{"x":181.37,"y":65},{"x":138.14,"y":74},{"x":66.03,"y":89}]}] ... I want it like the same without quotes.. can u help me pls?

Colin DeClue Over a year ago

What are you trying to use the string for?

Adil Shaikh · Accepted Answer · 2013-05-08 14:43:42Z

0

    dataa = dataa + mytext;
--------------^

this is undefined initially

You need to do this -

 var mytext, dataa = "", f;

answered May 8, 2013 at 14:43

Adil Shaikh

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