OCP-1Z0-051-V9.02-107题

107. View the Exhibit and examine the structure and data in the INVOICE table.

Which two SQL statements would execute successfully?  (Choose two.) 

A. SELECT AVG(inv_date )  类型不一致，不能对date求平均值

FROM invoice;

B. SELECT MAX(inv_date),MIN(cust_id)

FROM invoice;

C. SELECT MAX(AVG(SYSDATE - inv_date))  缺少group by

FROM invoice;

D. SELECT AVG( inv_date - SYSDATE), AVG(inv_amt)

FROM invoice;

Answer: BD

 

答案解析：

参考：http://blog.csdn.net/rlhua/article/details/12868639

A，不能对date类型求平均数

B,MAX,MIN可以对数值和date求最大值和最小值

C,组合函数，需要分组，错误

D,AVG对数值求平均数，正确。

首先创建测试表：

sh@TEST0924> create table invoice

  2  (inv_no number(3) not null,

  3  inv_date date,

  4  cust_id varchar2(4),

  5  inv_amt number(8,2)

  6  );

 

Table created.

 

插入数据后查询：

sh@TEST0924> select * from invoice;

 

    INV_NO INV_DATE  CUST     INV_AMT

----------     ---------        ----         ----------

         1 01-APR-07     A10        1000

         2 01-OCT-07     B1R        2000

         3 01-FEB-07                     3000

 

开始测试：

A答案：AGV函数需要一个数值类型或者可以隐式转为数值类型的参数

 

sh@TEST0924> SELECT AVG(inv_date ) FROM invoice;

SELECT AVG(inv_date ) FROM invoice

           *

ERROR at line 1:

ORA-00932: inconsistent datatypes: expected NUMBER got DATE

AGV  function takes as an argument any numeric data type or any nonnumeric data type that can be implicitly converted to a numeric data type. The function returns the same data type as the numeric data type of the argument.

B答案：

sh@TEST0924>  SELECT MAX(inv_date),MIN(cust_id)

  2  FROM invoice;

 

MAX(INV_D MIN(

--------- ----

01-OCT-07 A10

 

C答案：缺少group 不要函数

sh@TEST0924> SELECT MAX(AVG(SYSDATE - inv_date)) FROM invoice;

SELECT MAX(AVG(SYSDATE - inv_date)) FROM invoice

           *

ERROR at line 1:

ORA-00978: nested group function without GROUP BY

 

 

sh@TEST0924> SELECT MAX(AVG(SYSDATE - inv_date)) FROM invoice

  2  group by cust_id;

 

MAX(AVG(SYSDATE-INV_DATE))

--------------------------

                2427.62578

 

可以单独求平均值和最大值。

sh@TEST0924> SELECT AVG(SYSDATE - inv_date)FROM invoice;

 

AVG(SYSDATE-INV_DATE)

---------------------

           2327.29273

 

sh@TEST0924>  SELECT MAX(SYSDATE - inv_date) FROM invoice;

 

MAX(SYSDATE-INV_DATE)

---------------------

           2427.62655

 

D答案：

 

sh@TEST0924> SELECT AVG( inv_date - SYSDATE), AVG(inv_amt) FROM invoice;

 

AVG(INV_DATE-SYSDATE)     AVG(INV_AMT)

---------------------                         -----------

           -2327.2942                        2000