Php 在chart.js中输出多个SQL查询

因此，我在chart.js中创建了一个图表，它应该在图表中输出sql查询的结果，但是在运行代码时，图表的每个部分每次都显示“1”。因此，如果根据代码中显示的查询，图表中的部分应该是12,2,5,9,11，那么它只显示1,1,1,1,1。知道我哪里出错了吗

  <script>
var ctx = document.getElementById('myChart').getContext('2d');
var myChart = new Chart(ctx, {
  type: 'doughnut',
  data: {
    labels: ['Body Pump', 'Spin Class', 'Yoga', 'Body Tone', 'Legs, Bums & Tums'],
    datasets: [{
      label: 'Total Number of Bookings',
      data: [
      <?php

      $resultbodypump = mysqli_query($db, "SELECT COUNT(SessionType) FROM bookings WHERE username = '".$_SESSION['username']."' AND SessionType = 'bodyPump'");

      $resultspinclass = mysqli_query($db, "SELECT COUNT(SessionType) FROM bookings WHERE username = '".$_SESSION['username']."' AND SessionType = 'spinClass'");

      $resultyoga = mysqli_query($db, "SELECT COUNT(SessionType) FROM bookings WHERE username = '".$_SESSION['username']."' AND SessionType = 'yoga'");

      $resultbodytone = mysqli_query($db, "SELECT COUNT(SessionType) FROM bookings WHERE username = '".$_SESSION['username']."' AND SessionType = 'bodyTone'");

      $resultlegsbumstums = mysqli_query($db, "SELECT COUNT(SessionType) FROM bookings WHERE username = '".$_SESSION['username']."' AND SessionType = 'legsBumsTums'");

      echo mysqli_num_rows($resultbodypump) . ",";
      echo mysqli_num_rows($resultspinclass) . ",";
      echo mysqli_num_rows($resultyoga) . ",";
      echo mysqli_num_rows($resultbodytone) . ",";
      echo mysqli_num_rows($resultlegsbumstums);

      ?>
      ],

var ctx=document.getElementById（'myChart'）.getContext（'2d'）；
var myChart=新图表（ctx{
键入：“甜甜圈”，
数据：{
标签：[“身体泵”、“旋转类”、“瑜伽”、“身体音调”、“腿部、臀部和腹部”]，
数据集：[{
标签：“预订总数”，
数据：[

var ctx=document.getElementById（'myChart'）.getContext（'2d'）；
var myChart=新图表（ctx{
键入：“甜甜圈”，
数据：{
标签：[“身体泵”、“旋转类”、“瑜伽”、“身体音调”、“腿部、臀部和腹部”]，
数据集：[{
标签：“预订总数”，
数据：[
那么实际发送到浏览器的是什么呢？查看页面源页面源显示发送的数据是[1,1,1,1,1]。当我在phpmyadmin中运行查询时，它们返回正确的结果。您必须从m a
mysqli\u query（）获取结果
它们不会自动返回这里将始终只有一行从
COUNT（）
only查询返回，因此您必须获取结果，而不仅仅是对查询结果执行
mysqli\u num\u rows（）
，并且您的一个查询返回的不仅仅是
COUNT（）
  <script>
var ctx = document.getElementById('myChart').getContext('2d');
var myChart = new Chart(ctx, {
  type: 'doughnut',
  data: {
    labels: ['Body Pump', 'Spin Class', 'Yoga', 'Body Tone', 'Legs, Bums & Tums'],
    datasets: [{
      label: 'Total Number of Bookings',
      data: [
      <?php

      $querybodypump = "SELECT COUNT(SessionType) FROM bookings WHERE username = '".$_SESSION['username']."' AND SessionType = 'bodyPump'";
      $resultbodypump = mysqli_query($db,$querybodypump);
      $row1 = mysqli_fetch_row($resultbodypump);
      $count1 = $row1[0];
      $queryspinclass = "SELECT COUNT(SessionType) FROM bookings WHERE username = '".$_SESSION['username']."' AND SessionType = 'spinClass'";
      $resultspinclass = mysqli_query($db,$queryspinclass);
      $row2 = mysqli_fetch_row($resultspinclass);
      $count2 = $row2[0];
      $queryyoga = "SELECT COUNT(SessionType) FROM bookings WHERE username = '".$_SESSION['username']."' AND SessionType = 'yoga'";
      $resultyoga = mysqli_query($db,$queryyoga);
      $row3 = mysqli_fetch_row($resultyoga);
      $count3 = $row3[0];
      $querybodytone = "SELECT COUNT(SessionType) FROM bookings WHERE username = '".$_SESSION['username']."' AND SessionType = 'bodyTone'";
      $resultbodytone = mysqli_query($db,$querybodytone);
      $row4 = mysqli_fetch_row($resultbodytone);
      $count4 = $row4[0];
      $querylegsbumstums = "SELECT COUNT(SessionType) FROM bookings WHERE username = '".$_SESSION['username']."' AND SessionType = 'legsBumsTums'";
      $resultlegsbumstums = mysqli_query($db,$querylegsbumstums);
      $row5 = mysqli_fetch_row($resultlegsbumstums);
      $count5 = $row5[0];

      echo $count1. ",";
      echo $count2 . ",";
      echo $count3 . ",";
      echo $count4 . ",";
      echo $count5 ;

      ?>
      ],