我只是想知道我的代码中的错误是什么,我首先尝试了我的查询,如果它正常工作,它确实如此。但是,当我输入代码时,图表不会显示。顺便说一句,我使用的是Libchart PHP。 提前谢谢。
<?php
include 'libchart/libchart/classes/libchart.php';
require '../connection/connect.php';
header("Content-type: image/png");
$chart = new PieChart(500, 300);
$query = mysqli_query($con,"SELECT (SELECT COUNT(*) FROM jobseeker_reg WHERE gender='Male') AS Male, (SELECT COUNT(*) FROM jobseeker_reg WHERE gender='Female') AS Female");
$male=$query[0]['Male'];
$female=$query[0]['Female'];
$dataSet = new XYDataSet();
$dataSet->addPoint(new Point("Male", $male));
$dataSet->addPoint(new Point("Female", $female));
$chart->setDataSet($dataSet);
$chart->setTitle("Number of Jobseekers (by Gender)");
$chart->render();
?>