Convert PHP string to JavaScript array for Google Charts

Question

I am trying to build Google Sankey Chart using data pulled from MySQL database.

After the page is loaded I am making ajax call to pull the data from MySQL and returning it to the view.

This is how I am returning data from PHP:

echo json_encode("['test', 'test2', 1],['test2', 'test3', 1],");

Now when I check in JavaScript the type of returned data it says "string".

console.log(typeof myData); // string

However the chart is expecting arrays. So when I am passing this data to function that builds chart it comes back with the error. Here is how I am trying to build it:

data.addRows([
    myData
]);

But it comes back with:

Uncaught Error: Every row given must be either null or an array.

But if I hardcode that data directly in data.addRows it works perfectly fine... Like below works fine

data.addRows([
    ['test', 'test2', 1],['test2', 'test3', 1],
]);

How can I force JavaScript to accept myData returned from php?

I tried returning data from PHP as array and loop through it in JavaScript and populate the chart but sometimes I am returning 50000 rows or more and JavaScript does not handle looping through so many rows very well (slow and sometimes runs out of memory and crashes browser). So I thought if I return it as string JavaScript does not have to loop through it, just pass it to the chart builder.

EDIT: I am attaching my ajax call function and function that draws the chart

This is the ajax call that pulls data from php/mysql:

function ajaxCall() {
    $(function() {

        $.ajax({
            type: "POST",
            url: "/test",
            dataType: "JSON",
            success: function(myData) {         
                // Draw the graph
                drawChart(myData);
            },
            error: function(xhr,status,error) {
                console.log("error status: " + status + " / the error is: " + error);
            },
        });     
    });
}

This function will draw the chart using data returned from PHP:

function drawChart(myData) {
    var data = new google.visualization.DataTable();
    data.addColumn('string', 'From');
    data.addColumn('string', 'To');
    data.addColumn('number', 'Weight');

    data.addRows([
        myData          
    ]);     

    // Calculate graph height, by multiplying number of rows and fixed value
    var rowsQty = data.getNumberOfRows();
    var autoHeight = rowsQty*15;

    if(autoHeight < 300) {
        autoHeight = 300;
    }

    // Sets chart options.
    var options = {
        // width: 600,
        height: autoHeight,
    };

    // Instantiates and draws our chart, passing in some options.
    var chart = new google.visualization.Sankey(document.getElementById('sankey_basic'));
    chart.draw(data, options);
}

show exactly how/where this php returned data is used in JS... e.g. if you're doing an ajax call, then either you need to use jquery's $.getJSON(), or decode/parse the returned text yourself. json itself is just a string, and doesn't become "data" in JS until it's parsed. — Marc B
– Marc B, Commented Feb 19, 2016 at 14:51
@MarcB I have updated my question with ajax call and function that builds the chart — zachu
– zachu, Commented Feb 19, 2016 at 14:57

Marc B · Accepted Answer · 2016-02-19 15:03:20Z

2

YOu haven't produced an array-as-json, you've taken a STRING that contains a PHP array definition (complete with trailing commaa syntax error), and encoded that:

echo json_encode("['test', 'test2', 1],['test2', 'test3', 1],");
                 ^---------------------string----------------^

So what you're sending across is this literal text:

  "\"['test','test2',1]['test2','test3',1],\""

which is just a json-encoded string - again, it's NOT an array.

If you want an actual array to go across, you need to encode a PHP array, not a PHP string:

echo json_encode(array(array('test', 'test2', 1),array('test2', 'test3', 1));
--or--
echo json_encode([['test1','test2',1],['test2','test3',1]]);

Note the LACK of quotes around the [[...]] version.

answered Feb 19, 2016 at 15:03

Marc B

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1 Comment

zachu Over a year ago

Thanks this worked like a charm ! :) I have used this approach: echo json_encode(array(array('test', 'test2', 1),array('test2', 'test3', 1)); By the way you are missing closing ")" at the end before ";"

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