在 DISTINCT 查询和两个表相关查询中添加 COUNT

Question

我想在下面的查询中包含 COUNT，我已尝试更改它，但它不起作用。由于查询中缺少 COUNT（我猜），selectedcount 和 average 的 Javascript 不起作用。有人可以帮助我吗？谢谢。

下拉选择

<select id="prg" name="prg"  class="demoInputBox" style="padding: 3px; width: 350px;">
<option value="">Select Programme</option>

    <?php
    $sql = "SELECT DISTINCT student_prg FROM marketing_data, current_sem WHERE marketing_data.intake_year = current_sem.intake_year";
    $do = mysqli_query($conn, $sql);
                      while($row = mysqli_fetch_array($do)){
                        echo '<option value="'.$row['student_prg'].'" data-count="'.$row['count'].'">'.$row['student_prg'].'</option>';
                      }
                  ?>
</select>

Javascript

 <script type="text/javascript">
      var selectedprg = '';
      var selectedcount = '';
      var average = '';

      function myFunction() {
        document.getElementById("demo").innerHTML = "The total number of students in the selected program " +selectedprg + " <br>is " +selectedcount + " , it is predicted that " +average+ " students in +city+ will enroll for the upcoming semester <br> based on the data of 3 semesters back.";
      }

      $(document).ready(function(){
        $("#prg").change( function(){
          selectedprg = $('#prg option:selected').text(); 
          selectedcount = $('#prg option:selected').data('count');
          average = selectedcount / 3 + 5;
        });
      });
    </script>

我尝试过的查询：

"SELECT COUNT (DISTINCT student_prg) FROM marketing_data, current_sem WHERE marketing_data.intake_year = current_sem.intake_year";

屏幕截图：

所选课程计算机科学文凭（工业计算）的学生总数 为+selectedcount，预计+city+的+average名学生将在下学期入学 基于前三个学期的数据。

这个粗体变量无法从现有查询中读取。

ma​​rketing_data 表

current_sem 表

Answer 1

没有 group by 就不能使用 distinct 和 count。

你可以统计结果：

<?php
    $sql = "SELECT DISTINCT student_prg FROM marketing_data, current_sem WHERE marketing_data.intake_year = current_sem.intake_year";

    $do = mysqli_query($conn, $sql);
    $count = mysqli_num_rows($do);
                      while($row = mysqli_fetch_array($do)){
                        echo '<option value="'.$row['student_prg'].'" data-count="'.$count.'">'.$row['student_prg'].'</option>';
                      }
                  ?>