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Oracle 12c 之分析函数— LAST_VALUE
LAST_VALUE也是一个分析函数，它返回一个有序的值集合中的最后一个值。
可以对比FIRST_VALUE函数。
我们来看看例子：
SQL> SELECT employee_id, last_name, salary, hire_date,
2 LAST_VALUE(hire_date)
3 OVER (ORDER BY salary DESC ROWS BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED
4 FOLLOWING) AS lv
5 FROM (SELECT * FROM employees
6 WHERE department_id = 90
7 ORDER BY hire_date);
EMPLOYEE_ID LAST_NAME SALARY HIRE_DATE LV
----------- -------------------------------------------------- ---------- -------------- --------------
100 King 24000 17-6月 -03 13-1月 -01
101 Kochhar 17000 21-9月 -05 13-1月 -01
102 De Haan 17000 13-1月 -01 13-1月 -01
已用时间: 00: 00: 00.05看看下面的查询与上面的查询有什么不同？
SQL>
SQL> SELECT employee_id, last_name, salary, hire_date,
2 LAST_VALUE(hire_date)
3 OVER (ORDER BY salary DESC ROWS BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED
4 FOLLOWING) AS lv
5 FROM (SELECT * FROM employees
6 WHERE department_id = 90
7 ORDER BY hire_date DESC);
EMPLOYEE_ID LAST_NAME SALARY HIRE_DATE LV
----------- -------------------------------------------------- ---------- -------------- --------------
100 King 24000 17-6月 -03 21-9月 -05
102 De Haan 17000 13-1月 -01 21-9月 -05
101 Kochhar 17000 21-9月 -05 21-9月 -05
已用时间: 00: 00: 00.01
SQL>虽然 101号员工和102号员工具有相同的工资（Salary均为17000），但是在查询中，子查询中使用hire_date列进行了排序处理，第二个SQL查询语句中，按照 hire_date的降序（DESC）进行了排序，所以在查询结果中，102号员工排在了101号员工之前。
再看下面的两条SQL查询，找出不同点：
SQL>
SQL> SELECT employee_id, last_name, salary, hire_date,
2 LAST_VALUE(hire_date)
3 OVER (ORDER BY salary DESC, employee_id ROWS BETWEEN UNBOUNDED PRECEDING
4 AND UNBOUNDED FOLLOWING) AS lv
5 FROM (SELECT * FROM employees
6 WHERE department_id = 90
7 ORDER BY hire_date);
EMPLOYEE_ID LAST_NAME SALARY HIRE_DATE LV
----------- -------------------------------------------------- ---------- -------------- --------------
100 King 24000 17-6月 -03 13-1月 -01
101 Kochhar 17000 21-9月 -05 13-1月 -01
102 De Haan 17000 13-1月 -01 13-1月 -01
已用时间: 00: 00: 00.01
SQL>
SQL> SELECT employee_id, last_name, salary, hire_date,
2 LAST_VALUE(hire_date)
3 OVER (ORDER BY salary DESC, employee_id ROWS BETWEEN UNBOUNDED PRECEDING
4 AND UNBOUNDED FOLLOWING) AS lv
5 FROM (SELECT * FROM employees
6 WHERE department_id = 90
7 ORDER BY hire_date DESC);
EMPLOYEE_ID LAST_NAME SALARY HIRE_DATE LV
----------- -------------------------------------------------- ---------- -------------- --------------
100 King 24000 17-6月 -03 13-1月 -01
101 Kochhar 17000 21-9月 -05 13-1月 -01
102 De Haan 17000 13-1月 -01 13-1月 -01
已用时间: 00: 00: 00.01
SQL>By ordering within the function by both salary and the unique key employee_id, you can ensure the same result regardless of the ordering in the subquery.
再看RANGE子句的使用：
SQL>
SQL> SELECT employee_id, last_name, salary, hire_date,
2 LAST_VALUE(hire_date)
3 OVER (ORDER BY salary DESC RANGE BETWEEN UNBOUNDED PRECEDING AND
4 UNBOUNDED FOLLOWING) AS lv
5 FROM (SELECT * FROM employees
6 WHERE department_id = 90
7 ORDER BY hire_date);
EMPLOYEE_ID LAST_NAME SALARY HIRE_DATE LV
----------- -------------------------------------------------- ---------- -------------- --------------
100 King 24000 17-6月 -03 21-9月 -05
102 De Haan 17000 13-1月 -01 21-9月 -05
101 Kochhar 17000 21-9月 -05 21-9月 -05
已用时间: 00: 00: 00.01
SQL>
SQL> SELECT employee_id, last_name, salary, hire_date,
2 LAST_VALUE(hire_date)
3 OVER (ORDER BY salary DESC RANGE BETWEEN UNBOUNDED PRECEDING AND
4 UNBOUNDED FOLLOWING) AS lv
5 FROM (SELECT * FROM employees
6 WHERE department_id = 90
7 ORDER BY hire_date DESC);
EMPLOYEE_ID LAST_NAME SALARY HIRE_DATE LV
----------- -------------------------------------------------- ---------- -------------- --------------
100 King 24000 17-6月 -03 21-9月 -05
102 De Haan 17000 13-1月 -01 21-9月 -05
101 Kochhar 17000 21-9月 -05 21-9月 -05
已用时间: 00: 00: 00.00
SQL>
SQL>