Seems to me to be a PHP issue, less a Google Chart issue. We can try to
help, but you might have more luck on StackOverflow.
Can you post a snippet of JSON that you're trying to load into your
datatable?
On Saturday, November 30, 2013 1:01:02 PM UTC-5, Qurat Haider wrote:
>
> i am using jsonTable to parse the data in to the Google table and it is
> working fine. now i have a problem to add multiple queries at the same time
> and display the data only in two columns of array which is already defined.
> here is my code:
>
> $conn = mysqli_connect("$host", "$username", "$password","$db_name")or
> die("cannot connect");
> if (mysqli_connect_errno($conn))
> {
> echo "Failed to connect to MySQL: " . mysqli_connect_error();
> }
>
> $query = "SELECT vehicle_reg.`oilchange`-SUM(gps.`Distance`) AS
> Nextoilchange FROM gps INNER JOIN vehicle_reg ON (gps.`DeviceId`=25) AND
> (vehicle_reg.`DeviceId`=25) INNER JOIN LOG ON TIME BETWEEN
> DATE(log.`lastoilchange`) AND CURDATE()";
>
> // $query.= "SELECT vehicle_reg.`filterchange`-SUM(gps.`Distance`) AS
> filterchange FROM gps INNER JOIN vehicle_reg ON (gps.`DeviceId`=25) AND
> (vehicle_reg.`DeviceId`=25) INNER JOIN LOG ON TIME BETWEEN
> DATE(log.`lastfilterchange`) AND CURDATE()";
>
> $result = mysqli_multi_query($conn, $query);
> if ($result) {
> do {
> // grab the result of the next query
> if (($result = mysqli_store_result($conn)) === false &&
> mysqli_error($conn) != '') {
> echo "Query failed: " .mysqli_error($conn);
> }
> } while (mysqli_more_results($conn) && mysqli_next_result($conn)); //
> while there are more results
> } else {
> echo "First query failed..." .mysqli_error($conn);
> }
>
>
> $table = array();
> $table['cols'] = array(
> array('label' => 'Vehicle', 'type' => 'number'),
> array('label' => 'Distance Left', 'type' => 'number')
>
> );
> $rows = array();
> while ($nt = mysqli_fetch_array($result))
>
> {
>
> $temp = array();
>
> $temp[] = array('v' => 'Nextoilchange');
> $temp[] = array('v' =>$nt['Nextoilchange']);
>
> // insert the temp array into $rows
> $rows[]['c'] = $temp;
>
> }
> $table['rows'] = $rows;
> $jsonTable = json_encode($table);
>
> // echo $jsonTable;
> mysqli_close($conn);
>
> i want to display the table like:
>
> Vehicle Distance Left
> ---------------------------------
> nextoilchange 500
> nextfilter 300
> nextcheckup 400
>
> i am confused how to use multiple queries and each query gives one value in
> distance left. if any one knows how to solve this problem please tell me.
> Thanks
>
>
>
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