Question

我目前有一条SQL语句，用于在表格中填写学生姓名和他们在给定学期内所从事的运动的名称。运动的名称与Sport_ID存储在运动表中。然后，Sport_ID与choices表中的Choice_ID相对应。然后，将这个Choice_ID放在表Student_Choices中标记为T1_Choice，T2_Choice和T3_Choice的每个术语的列中。如何从我的SQL语句中获取T2_Choice和T3_Choice？感谢您的帮助。

DB Fiddle Link

$stmt = $conn->prepare(
              "SELECT st.Name AS student, s.Name AS sport
              From Sports AS s INNER JOIN Choices AS c
              ON s.Sport_ID = c.Sport_ID INNER JOIN Student_Choices AS sc
              ON sc.T1_Choice = c.Choice_ID INNER JOIN Students AS st
              ON st.Username = sc.Username
              ");
            $stmt->execute();
            while ($row = $stmt->fetch(PDO::FETCH_ASSOC)) {
              echo '<tr>
              <td>'.$row['student'].'</td>
              <td>'.$row['sport'].'</td>
              <td>'.$row['sport'].'</td>
              <td>'.$row['sport'].'</td>
              </tr>
              ';
            }

Answer 1

解决方案是您可以多次向表调用JOIN。因为“选择”中的“ Choice_ID”需要链接到“ Student_Choices”表中的不同术语选择。因此，您需要使用不同的别名多次联接Choices表，并且对于每个实例，您需要使用不同的别名联接Sports表。我的解决方案是：

("SELECT st.Name AS student, T1.Name AS T1, T2.Name AS T2, T3.Name AS T3
              From Students AS st INNER JOIN Student_Choices AS sc
              ON st.Username = sc.Username
              INNER JOIN Choices AS c1
              ON sc.T1_Choice = c1.Choice_ID
              INNER JOIN Sports AS T1
              ON c1.Sport_ID = T1.Sport_ID
              INNER JOIN Choices AS c2
              ON sc.T2_Choice = c2.Choice_ID
              INNER JOIN Sports AS T2
              ON c2.Sport_ID = T2.Sport_ID
              INNER JOIN Choices AS c3
              ON sc.T3_Choice = c3.Choice_ID
              INNER JOIN Sports AS T3
              ON c3.Sport_ID = T3.Sport_ID
              ");

在联接的SQL语句中选择多个具有相同名称的事物

1 个答案: