我当前有一个填充表的输出。它给出的输出之一是学生表格中的一年。每当输出6作为年份时，我需要它返回l6，当输出7时，我需要u6。我试着吼叫：

"$stmt = $conn->prepare(
              "SELECT st.Name AS student, st.House AS house, T1.Name AS T1, T2.Name AS T2, T3.Name AS T3
              CASE
                WHEN st.Year = 6 THEN 'L6'
                WHEN st.Year = 7 THEN 'U6'
                ELSE st.Year
              END CASE as year
              From Students AS st INNER JOIN Student_Choices AS sc
              ON st.Username = sc.Username INNER JOIN Current_DB AS db
              ON sc.DB_year = db.DB
              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
              ");

然后我得到一个错误：
errorsqlstate[42000]：语法错误或访问冲突：1064您的sql语法有错误；检查与您的mariadb服务器版本相对应的手册，以了解在第2行的“case when st.year=6 then'l6 when st.”附近使用的正确语法
任何解决方案都很好，谢谢。