You have to format the data returned by your query into the JSON format
expected by the DataTable constructor. Here's a basic PHP script that you
can work from:
<?php
/* $server = the IP address or network name of the server
* $userName = the user to log into the database with
* $password = the database account password
* $databaseName = the name of the database to pull data from
*/
$con = mysql_connect($server, $userName, $password) or die('Error
connecting to server');
mysql_select_db($databaseName, $con);
// write your SQL query here (you may use parameters from $_GET or $_POST
if you need them)
$query = mysql_query('SELECT column1, column2, column3 FROM myTable');
$table = array();
$table['cols'] = array(
/* define your DataTable columns here
* each column gets its own array
* syntax of the arrays is:
* label => column label
* type => data type of column (string, number, date, datetime, boolean)
*/
array('label' => 'Label of column 1', 'type' => 'string'),
array('label' => 'Label of column 2', 'type' => 'number'),
array('label' => 'Label of column 3', 'type' => 'number')
// etc...
);
$rows = array();
while($r = mysql_fetch_assoc($query)) {
$temp = array();
// each column needs to have data inserted via the $temp array
$temp[] = array('v' => $r['column1']);
$temp[] = array('v' => $r['column2']);
$temp[] = array('v' => $r['column3']);
// etc...
// insert the temp array into $rows
$rows[] = array('c' => $temp);
}
// populate the table with rows of data
$table['rows'] = $rows;
// encode the table as JSON
$jsonTable = json_encode($table);
// set up header; first two prevent IE from caching queries
header('Cache-Control: no-cache, must-revalidate');
header('Expires: Mon, 26 Jul 1997 05:00:00 GMT');
header('Content-type: application/json');
// return the JSON data
echo $jsonTable;
?>
On Monday, February 25, 2013 3:58:55 PM UTC-5, akshita gupta wrote:
>
> Hello.. I am unable to display chart. Please help. I am unable to figure
> out the problem.
>
> getdata.php
>
> <?php
>
> mysql_connect('localhost','akshita','123456');
> mysql_select_db('rcusers');
>
> $sqlquery1="select userid,group_name,req_nodes,actualPE from jobs
> <http://stackoverflow.com/questions/15073654/unable-to-extract-data-to-get-google-chart-nothing-is-being-displayed-but-getda#>
> where userid='zhang' limit 200";
>
> $sqlresult1=mysql_query($sqlquery1);
>
> $rows=array();
> while($r=mysql_fetch_assoc($sqlresult1)){
> $rows[]=$r;}print json_encode($rows);?>
>
> chartDraw.php
> <html><head><!--Load the AJAX API --><script type="text/javascript"
> src="https://www.google.com/jsapi";></script><script
> src="https://ajax.googleapis.com/ajax/libs/jquery/1.8.3/jquery.min.js";
> type="text/javascript"></script>
> <script type="text/javascript">
> //Load the visualization API and the piechart package
> google.load('visualization','1',{'packages':['corechart']});
> //Set a callback to run when the google visualization API is loaded
> google.setOnLoadCallback(drawchart);
> function drawChart(){
> var jsonData = $.ajax({
> url:"getData.php",
> dataType:"json",
> async:false
> }).responseText;
> //Create our data table out of JSON data loaded from servervar data=new
> google.visualization.DataTable(jsonData);
> //Instantiate and draw our chart
> <http://stackoverflow.com/questions/15073654/unable-to-extract-data-to-get-google-chart-nothing-is-being-displayed-but-getda/15073745#>,
> passing in some optionsvar chart=new
> google.visualization.PieChart(document.getElementById('chart_div'));
> chart.draw(data,{width:400,height:240});}
> </script></head>
> <body>
> <!--Div that will hold the pie chart -->
> <div id="chart_div"></div></body></html>
>
>
>
> On Tuesday, 26 July 2011 15:24:36 UTC-4, GerBen wrote:
>>
>> Hello Colleagues,
>> How can I read data from an external database (say, MS SQL Server or
>> MySQL) and show it in a Google Chart?
>> Thank you.
>
>
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