Question

I am getting the error missing ) after argument list in my Firebug console.

emissing ) after argument http://a8.sphotos.ak.fbcdn.net/hphotos-ak-snc7/s720x720/393131_320846714645076_100001592501599_911297_470580896_n.jpg

My question is how to pass $char_data variable in JavaScript function as a argument

Define php variable:

<?php 
$chart_data = "['NBA',1],['NFL',2],['MLB',3],['NHL',4]"; 
$div = "graph";
?

Call JavaScript function with define argument

<script>
dynamicChartArray('<?php echo $div;?>','<?php echo $chartdata;?>')
</script>

A function of JavaScript

<script>
function dynamicChartArray(div,chartdata){
var myData = new Array(chartdata);
var myChart = new JSChart(div, 'pie');
alert(chartdata+div);
}
<script>

Answer 1

Rather than creating an array out of a string in javascript, why not just just get the PHP to output it as an array to start with?

Just add an extra set of [] which javascript reads as an array.

$chart_data = "[['NBA',1],['NFL',2],['MLB',3],['NHL',4]]"; 

then ditch the quotes on the output (which are responsible for causing the error messages)

 dynamicChartArray('<?php echo $div;?>', <?php echo $chartdata;?>);

and then myData can just equal chart data (since its already an array)

var myData = chartdata;