SQL / PHP：在另一個表（多個表）上不匹配的聯接

Question

我有四個表：

Personal_trainer(personaltrainerID, name, age, location)
Client(clientID, name, age, location)
Nutrition_Plan(NutritionplanID, personaltrainerID, clientID)
Training_Plan(TrainingplanID, personaltrainerID, clientID)

我試圖顯示的是，當私人教練創建營養/培訓計划並將其分配給客戶時，結果只會顯示其特定客戶，但是在客戶表中它們不是識別教練的外鍵（由於項目的標准）

我想知道用於培訓/營養計划的必要聯接的SQL是什么。 我已經嘗試了一段時間，這是我的示例代碼。

所需的輸出是所有客戶數據，前提是該培訓師僅為其分配了培訓計划或營養計划

在語句中，Bind參數存在問題，因此只有具有客戶端的用戶才能看到其客戶端。 如果我使用指定的ID，我可以獲得退貨！

    <?php 
            //code to search for a item from the database
            // user can enter any character to search for a value from the db
           if (isset($_POST['search']))
           {
               $valueToSearch = $_POST['ValueToSearch'];
               $query = "select * from client WHERE concat(`clientID`, `name`, `age`, `sex`, `weight`, `height`, `yearsExperience`, `goal`, `injuries`, 'email')like'%".$valueToSearch."%'";
               $search_result = filterTable($query);

           }
           else {
           $query = "select *
            from client
            where clientID in (select clientID from nutrition_plan where personaltrainerID=?)
            or clientID in (select clientID from training_plan where personalTrainerID=?)";
           $query->bind_param("i", $_POST[""]);
               $search_result = filterTable($query);
           }
           //code to filter the db
           function filterTable($query)
           {
               $connect = mysqli_connect("localhost:3308","root","","fypdatabase");
               $filter_Result = mysqli_query($connect, $query);
               return $filter_Result;
           }
           ?>

           <?php
           while($row = mysqli_fetch_array($search_result))
          { //display the details from the db in the table with option to delete or update entry 
            ?>
                    <tr>
                    <td><?php echo $row["clientID"]; ?></td>
                    <td><?php echo $row["name"]; ?></td>
                    <td><?php echo $row["age"]; ?></td>
                    <td><?php echo $row["sex"]; ?></td>
                    <td><?php echo $row["weight"]; ?></td>
                    <td><?php echo $row["height"]; ?></td>
                    <td><?php echo $row["yearsExperience"]; ?></td>
                    <td><?php echo $row["goal"]; ?></td>
                    <td><?php echo $row["injuries"]; ?></td>
                    <td><?php echo $row["email"]; ?></td>
                    <td> 
                        <a href="?Delete=<?php echo $row["clientID"]; ?>" onclick="return confirm('Are you sure?');">Delete</a>
                    </td>
                    <td>
                        <a href="updateClient.php?Edit=<?php echo $row["clientID"]; ?>" onclick="return confirm('Are you sure?');">Update</a>
                    </td>
                  </tr>
                  <?php

Answer 1

讓我們看看我是否正確理解了您的要求...

僅顯示由培訓師123制定營養或培訓計划的客戶：

select *
from client
where clientid in (select clientid from nutrition_plan where personaltrainerid = 123)
   or clientid in (select clientid from training_plan where personaltrainerid = 123);

向這些客戶展示他們的所有計划（無論計划的培訓師如何）：

select *
from client
join
(
  select 'nutrition' as kind, nutritionplanid as id, personaltrainerid, clientid
  from nutrition_plan
  union all
  select 'training' as kind, trainingplanid as id, personaltrainerid, clientid
  from training_plan
) plan using (clientid)
where clientid in (select clientid from nutrition_plan where personaltrainerid = 123)
   or clientid in (select clientid from training_plan where personaltrainerid = 123);

Answer 2

SELECT name, age, location FROM Client
INNER JOIN
(
  SELECT personaltrainerID, clientID from Nutrion_Plan 
  UNION DISTINCT 
  SELECT personaltrainerID, clientID from Training_Plan
) u
USING(clientID)
WHERE u.personaltrainerID = ?;

Answer 3

這應該工作

SELECT * from client JOIN nutrition_plan ON client.clientid=nutritionplan.clientid JOIN Personaltrainer ON Personaltrainer.personaltrainerID=nutritionplan.personaltrainerID Where Personaltrainer.personaltrainerID="id"