使用MySQL时，PHPlot仅显示3个结果中的2个

Question

我有一个饼图，我想让它显示3个结果。我使用的代码来自两条线的图表，这工作得很好，我只是希望它能显示所有3个结果。图例显示全部3个，但图表仅显示前两个。我知道我的代码中有一个错误，我只是找不到它。有什么建议吗？

$sql1 = "SELECT totaldep, totalmania, totalanxiety FROM $dbmood ORDER BY DATE ASC";

//Define the object


$rslt1 = @mysql_query($sql1, $conn) or die("Couldn't execute select:".@mysql_error());   

// I'm assuming it's somewhere in here, I just can't see it
 $iid = "-1";
        while($row = mysql_fetch_row($rslt1)) {

                if ($row[0] != $iid)     {
                        $iid = $row[0];
                } else {
                    $row[0] = '';
            }
// between there and here

            $graph_data[] = $row;           

//Define some data
$example_data = $graph_data;

        }




//Define the object
$graph =& new PHPlot(300,250);
$graph->SetPlotType("pie");
$legend = array();
$legend[] = "Depression";
$legend[] = "Mania";
// It isn't showing anxiety, everything else is showing/updating fine
$legend[] = "Anxiety";


$graph->SetDataValues($example_data);
$graph->SetLegendPixels(1,5,false);
$graph->SetLegend($legend);

//Draw it
$graph->DrawGraph();

提前感谢，你们总是帮助我们！

Answer 1

我把它们一个个拔出来解决了

$datab = mysql_select_db( 'medtrack', $conn ) or die(mysql_error());

$sql1 = "SELECT * FROM $dbmood ORDER BY DATE ASC";



$getit = mysql_query ( $sql1, $conn );

while($row = mysql_fetch_array($getit, MYSQL_ASSOC))
{
$dep="{$row['totaldep']}";
$man="{$row['totalmania']}";
$anx="{$row['totalanxiety']}";


$example_data = array(
array('a',$dep,$man,$anx)
);
}


//Include the code

//Define the object
$graph =& new PHPlot(300,250);
$graph->SetPlotType("pie");
$legend = array();
$legend[] = "Depression";
$legend[] = "Mania";
$legend[] = "Anxiety";


$graph->SetDataValues($example_data);
$graph->SetLegendPixels(1,5,false);
$graph->SetLegend($legend);
//Draw it
$graph->DrawGraph();