Javascript: Passing an array to a function does not work?

Question

I have a function that builds an array with push.

function one(resultsArray) {
    activity_basic_weight = 9;
        for (i in resultsArray) {
            score = 0;
            score = resultsArray[i].var_a * activity_basic_weight;
            score = Math.round(score * 100) / 100;
            if (score >= 80) 
            {
                verygoodCategories1.push({
                    score: score,
                    name: i,
                    url: resultsArray[i].url
                });
            } 
            else if (score <= 79) 
            {
                ...         
            }
        }

    two(verygoodCategories1, ...);      
}

function two(verygoodCategories1, ...) {

    alert(verygoodCategories1.length); //  = 7, correct;

    // here comes the problem:

    for (var i = 0; i < verygoodCategories1.length; i++) {
        //this throws "Error: TypeError: verygoodCategories1 is undefined"
    }
}

I am new at Javascript. Am I doing something fundamentally wrong?

This is pseudocode, but I made sure that at least the variable names are correct etc.

Any ideas?

Thanks

J

Where are you initializing `verygoodCategories1` as an array? Should be at least a `verygoodCategories1 = [];` in there somewhere. — jwatts1980, Sep 16 '14 at 19:48
Also, you should not use for-in loops on arrays. http://stackoverflow.com/questions/500504/why-is-using-for-in-with-array-iteration-such-a-bad-idea — Chris Tavares, Sep 16 '14 at 19:50
Can you please give use the actual code (including the invocation of `one(…)`), not pseudocode? Some `var` statements would be nice there :-) — Bergi, Sep 16 '14 at 19:53
Here is a fiddle with the actual code: http://jsfiddle.net/zd1y6w8e/ — Jacques, Sep 16 '14 at 20:10

score 1 · Answer 1 · edited Sep 16 '14 at 20:16

1

Assign the array parameter to a variable:

function two(verygoodCategories1, ...) {

    alert(verygoodCategories1.length); //  = 7, correct;

    var goodCategories=verygoodCategories1;
    var goodCategoriesLength=goodCategories.length;
    for (var i = 0; i < goodCategoriesLength; i++) {
        console.log(goodCategories[i]);
    }
}

edited Sep 16 '14 at 20:16

Stephen P

14,422
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answered Sep 16 '14 at 20:04

Gangadhar Jannu

4,136
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@Jacques Would you comment your code in the jsfiddle, where the problem is like what is the array, when it is being "undefined" – Gangadhar Jannu Sep 16 '14 at 20:16

score 1 · Answer 2 · answered Sep 16 '14 at 20:10

Declaring variables is important. Also, as mentioned in the comments, don't use for...in. Maybe someday you will want to use that but not today.

Here is your pseudo code written so that it could actually work.

var a = [{var_a:100},{var_a:90},{var_a:81},{var_a:81},{var_a:91},{var_a:101}];
function one(resultsArray) {

    //This is critical to later using it as an array.
    var verygoodCategories1 = [];

    // you should always initialize a variable with var.  
    // You can not do it but thats bad form. You will
    // pollute the global space and people will dislike you.
    var activity_basic_weight = 9;
    for (var i = 0; i < resultsArray.length; i++) {
        var score = resultsArray[i].var_a * activity_basic_weight;
        score = Math.round(score * 100) / 100;
        if (score >= 80) 
        {
            verygoodCategories1.push({
                score: score,
                name: i,
                url: resultsArray[i].url
            });
        } 
        else if (score <= 79) 
        {
            //...
        }
    }

    two(verygoodCategories1); 
}

function two(verygoodCategories1) {

    alert(verygoodCategories1.length); //  = 7, correct;

    for (var i = 0; i < verygoodCategories1.length; i++) {
        //no more errors!
    }
}

one(a);

score 0 · Answer 3 · edited Sep 16 '14 at 20:14

0

Go to your function one(resultArray). just before the for, you might want to declare the array there like:

verygoodCategories1 = new Array();

edited Sep 16 '14 at 20:14

Matt Rabe

1,753
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answered Sep 16 '14 at 19:54

Akah

1,389
14
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1

This looks like bad advice. If you define `verygoodCategories1` as a new array in the location you suggest, it will wipe out the contents each time it loops... – Matt Rabe Sep 16 '14 at 19:59
Ah yeah, i missed seeing the for loop above. The array should be created as the first statement in the function. – Akah Sep 16 '14 at 20:04

Javascript: Passing an array to a function does not work?

3 Answers3