Answered over on StackOverflow: http://stackoverflow.com/a/24295034/613559
On Wednesday, June 18, 2014 4:29:24 AM UTC-4, Ravindra Gharge wrote:
>
> I am populating the chart by getting the datatable as the response from
> other page.
> so am creating json object at other side and transfer to the requesting
> site and their i parse the object and create the datatable and populating
> the chart.
>
> following my code to request the chart data :
>
> var jsonData = $.ajax({
> type: "post",
> url: "getHourDrillChart.php",
> data: $("#my_form").serialize(),
> dataType: "json",
> async: false
> }).responseText;
> var obj = jQuery.parseJSON(jsonData); //alert(""+obj.toSource());
> var dataHour = google.visualization.arrayToDataTable(obj);\
>
> chart.setDataTable(dataHour);
> chart.setOptions(options);
> chart.draw();
>
> other site where i creating the chart data :
>
> $data[0] = array('Hour',$user);
> $getTownLocalityInfo = mysql_query($SQLString);
>
> # set heading
> //$data[0] = array('hour','Count');
> $i=1;
> $dayArray = array();
> while( $row = mysql_fetch_assoc($getTownLocalityInfo)){
> $date = $row['date'];
> $hour = $row['day_hours'];
> $count = (int) $row['sumCount'];
> $dayArray[] = $hour;
>
> mysql_query("INSERT INTO
> trend_hour_chart_temp(hour,$user,userId) VALUES
> ('$hour',$count,'$staffId')");
> $data[$i] = array($hour , $count);
> $i++;
> }
> echo json_encode($data);
>
>
>
>
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