How to use an array inside a multidimensional array?

Question

If I have an array like so:

var phpintojsArray = [["1","20"]];

and... I also have a multid-array:

var data = [[1,20], [4, 20], [7, 55], [9, 10], [9, 10]];

how to add the phpintojsArray into the data array? I want:

var data = [[phpintojsArray], [1,20], [4, 20], [7, 55], [9, 10], [9, 10]];

What is one way to accomplish this?

Clarification -- Trying to manipulate the data in a chart

This works:

    /* Bar Chart */
    var data = [[1, 10],[3, 60], [5, 20], [7, 50], [9, 10]];

    // Initialize Bars Chart
    $.plot(BarChart, [
        { data: data, bars: { show: true, fillColor: { colors: [{ opacity: 1 }, { opacity: 1 }] } }, label: 'Example label' } ],
        {
            legend: {
                backgroundColor: '#f6f6f6',
                backgroundOpacity: 0.8
            },
            colors: ['#39d5db'],
            grid: {
                borderColor: '#cccccc',
                color: '#999999',
                labelMargin: 10
            },
            yaxis: {
                ticks: 5
            },
            xaxis: {
                tickSize: 1
            }
        }
    );

This does Not work:

var phpintojsArray = <?= json_encode($sales); ?>;
    var two = [3, 60];
    var three = [5, 20];
    var four = [7, 50];        
    var five = [9, 10];


    /* Bars Chart */
    var data = [[phpintojsArray], [two], [three], [four], [five]];

    // Initialize Bars Chart
    $.plot(BarChart, [
        { data: data, bars: { show: true, fillColor: { colors: [{ opacity: 1 }, { opacity: 1 }] } }, label: 'Example label' } ],
        {
            legend: {
                backgroundColor: '#f6f6f6',
                backgroundOpacity: 0.8
            },
            colors: ['#39d5db'],
            grid: {
                borderColor: '#cccccc',
                color: '#999999',
                labelMargin: 10
            },
            yaxis: {
                ticks: 5
            },
            xaxis: {
                tickSize: 1
            }
        }
    );

Why?

Nope. When i view the source it shows the name "phpintojsArray" not the value — M. Finch
– M. Finch, Commented May 16, 2014 at 2:02
Yeah, it will show phpintojsArray with the code above. View source is not a good indicator of whether something worked... — Brian Mains
– Brian Mains, Commented May 16, 2014 at 2:04
View source is viewing source - it's like looking at what you have typed above literally and saying "my code doesn't work" - as it turns out, your code does work: jsfiddle.net/sLp9B — scrowler
– scrowler, Commented May 16, 2014 at 2:05
I think you're looking for data = phpintojsArray.map(Number).concat(data); actually. — Bergi
– Bergi, Commented May 16, 2014 at 2:13

scrowler · Accepted Answer · 2014-05-16 02:17:02Z

2

You can use splice to insert the elements:

data.splice(0, 0, phpintojsArray);

edited May 16, 2014 at 2:17

scrowler

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answered May 16, 2014 at 2:13

MIZ

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4 Comments

Johnny Chen Over a year ago

Does this do insert the element of phpintojsArray or the whole phpintojsArray itself?

MIZ Over a year ago

This will add whole phpintojsArray itself

Johnny Chen Over a year ago

One more question , what do the param 0,0 mean here , I know the first is an index.

MIZ Over a year ago

The syntax of the splice method is array.splice(index,howmany,item1,.....,itemX) where the first param is the index, second param is how many elements to be deleted and the rest are the new elements to be added

EyalAr · Accepted Answer · 2014-05-16 02:51:08Z

1

This works:

/* Bar Chart */  
var data = [[1, 10],[3, 60], [5, 20], [7, 50], [9, 10]];

This does Not work:

/* Bars Chart */
var data = [[phpintojsArray], [two], [three], [four], [five]];

Why?

Because the two expressions yield different structures. In the first one you have two levels of nested arrays (small arrays inside a big one). In the second you have three levels of nested arrays.

A simple console.log will show you the difference.

Make sure that your data is in the correct structure. Change it to:

var data = [two, three, four, five];

(In case it's still not clear, the brackets [] define a new array)

Regarding your question on how to prepend phpintojsArray, you could do:

Array.prototype.unshift.apply(data, phpintojsArray);

edited May 16, 2014 at 2:51

answered May 16, 2014 at 2:45

EyalAr

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3 Comments

M. Finch Over a year ago

understood thanks. regarding Array.prototype.unshift.apply(data, phpintojsArray); how would you end up using that inside the data array?

EyalAr Over a year ago

what do you mean by 'inside'?

M. Finch Over a year ago

Meaning, how to use your prepending technique to get the value of phpintojsArray and use it inside of the data array (like below:) var data = [[Array.prototype.unshift.apply(data, phpintojsArray);], [two], [three], [four], [five]]; Because your point about the different structures works for the var data = [[two], [three], [four], [five]]; but not for the phpintojsArray value: var data = [[phpintojsArray], [two], [three], [four], [five]];[phpintojsArray].

MIZ · Accepted Answer · 2014-05-22 11:19:02Z

0

You could use eval in javascript.

For Example:

var data = [[eval("phpintojsArray")], [eval("two")], [eval("three")], [eval("four")], [eval("five")]];

edited May 22, 2014 at 11:19

MIZ

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answered May 22, 2014 at 11:13

Vimalraj Selvam

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