Hi,
Thanx alot for your feedback..will do it as per your suggestion..
On Wednesday, August 14, 2013 3:23:54 PM UTC+5:30, soumya reubro wrote:
>
> Hi,
>
> I am trying to draw a line chart using *Google chart *in which data is
> retrieved from mysql table.But i get the error as follows
> Error: Invalid JSON string:
> {"cols":[{"label":"id","type":"string"},{"label":"date","type":"string"}],"rows":["1","2013-08-13
>
> 00:00:00"]}{"cols":[{"label":"id","type":"string"},{"label":"date","type":"string"}],"rows":["2","2013-08-13
>
> 00:00:00"]}
>
> the json encoded data looks as follows
> {"cols":[{"label":"id","type":"string"},{"label":"date","type":"string"}],"rows":["1","2013-08-13
>
> 00:00:00"]}{"cols":[{"label":"id","type":"string"},{"label":"date","type":"string"}],"rows":["2","2013-08-13
>
> 00:00:00"]}
>
>
>
> Here is the code which i used for line chart developement
>
> <html>
> <head>
> <script type="text/javascript" src="https://www.google.com/jsapi
> "></script>
> <script type="text/javascript" src="jquery.js"></script>
> <script type="text/javascript" src="jquery-1.6.2.min.js"></script>
> <script type="text/javascript">
> google.load("visualization", "1", {packages:["corechart"]});
> google.setOnLoadCallback(drawChart);
> function drawChart() {
> var jsonData = $.ajax({
> url: "graph1.php",
> dataType:"json",
> async: false
> }).responseText;
> var data = new google.visualization.DataTable(jsonData);
> var chart = new
> google.visualization.LineChart(document.getElementById('chart_div'));
> chart.draw(data, {width: 400, height: 240});
> }
> </script>
> </head>
> <body>
> <div id="chart_div" style="width: 900px; height: 500px;"></div>
> </body>
> </html>
>
>
> PHP part:
>
>
> $sql = mysql_query("SELECT * FROM `da_insulin_details` LIMIT 0,2 ");
> $rows['cols'] = array( array('label' => 'id', 'type' => 'string'),
> array('label' => 'date', 'type' => 'string'));
> while($row = mysql_fetch_array($sql))
> {
>
> $rows['rows'] [0] = $row['id'];
> $rows['rows'][1] = $row['date'];
> $jsonTable = json_encode($rows);
> echo $jsonTable;
>
> }
>
>
> I am new to this section..anybody pls help me to solve the issue
>
>
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