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  • OCP-1Z0-051-V9.02-113题

    113. Examine the structure of the ORDERS table:  

    Name             Null      Type 

    ORDER_ID        NOT NULL    NUMBER(12) 

    ORDER_DATE      NOT NULL    TIMESTAMP(6) 

    CUSTOMER_ID     NOT NULL    NUMBER(6) 

    ORDER_STATUS               NUMBER(2) 

    ORDER_TOTAL                NUMBER(8,2)

    You want to find the total value of all the orders for each year and issue the following command:

    SQL>SELECT TO_CHAR(order_date,'rr'), SUM(order_total)

    FROM orders

    GROUP BY TO_CHAR(order_date,'yyyy');

    Which statement is true regarding the outcome?

    A. It executes successfully and gives the correct output.

    B. It gives an error because the TO_CHAR function is not valid.

    C. It executes successfully but does not give the correct output.

    D. It gives an error because the data type conversion in the SELECT list does not match the data type

    conversion in the GROUP BY clause. 格式不一致

    Answer: D

    答案解析:

    报错,因为SELECT列表中的数据类型转换与GROUP BY子句中的数据类型转换不匹配。

    rr和yyyy的格式不一致,改为一致后可以得出数据。

    scott@TEST0924> select to_char(HIREDATE,'rr'),sum(sal)
      2  from emp
      3  group by to_char(HIREDATE,'yyyy');
    select to_char(HIREDATE,'rr'),sum(sal)
                   *
    ERROR at line 1:
    ORA-00979: not a GROUP BY expression
     
     
    scott@TEST0924>  select to_char(HIREDATE,'yyyy'),sum(sal)
      2  from emp
      3  group by to_char(HIREDATE,'yyyy');
     
    TO_C   SUM(SAL)
    ---- ----------
    1987       4100
    1980        800
    1982       1300
    1981      22825
     
    scott@TEST0924> select to_char(HIREDATE,'rr'),sum(sal)
      2  from emp
      3  group by to_char(HIREDATE,'rr');
     
    TO   SUM(SAL)
    -- ----------
    87       4100
    81      22825
    82       1300
    80        800
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  • 原文地址:https://www.cnblogs.com/hzcya1995/p/13317176.html
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