Mastering Oracle SQL学习笔记(join句法专题第五部份)
2.4.2
Partition Outer Joins
Partition outer joins
是
Oracle10g
中令人感兴趣的的特性。它可以将数据分成组，句体我也说不清，看了下面的例子你就明白了：
(
表
employee_expanse)
SELECT * FROM employee_expense;
EMP_ID YEAR MONTH EXPENSE_CLAIM APPROVED_AMT PAID_DATE
---------- ---------- ---------- ------------- ------------ ---------
7369 2002 2 3072.43 3072.43 03-MAR-02
7369 2002 4 30 30 01-JUN-02
7369 2002 5 235.03 35.03 01-JUN-02
7369 2002 9 5095.98 5095.08 31-OCT-02
7369 2002 12 1001.01 1001.01 01-FEB-03
7782 2002 1 111.09 111.09 01-FEB-02
7782 2002 3 9.85 9.85 01-APR-02
7782 2002 7 3987.32 3987.32 01-AUG-02
7782 2002 9 1200 1200 01-OCT-02
（表
months
）
SELECT * FROM months WHERE year = 2002;
YEAR MONTH
---------- ----------
2002 1
2002 2
2002 3
2002 4
2002 5
2002 6
2002 7
2002 8
2002 9
2002 10
2002 11
2002
12
你可以用
month
表中的反回的
12
行，做一个
left outer join
把某个员工
(employee)
每月的
expense
显示出来：
SELECT NVL(ee.emp_id, 7782), m.year, m.month, NVL(ee.expense_claim,0)
FROM (SELECT * FROM months WHERE year = 2002) m
LEFT OUTER JOIN (SELECT *
FROM employee_expense
WHERE emp_id = 7782) ee
ON m.year = ee.year AND m.month = ee.month
ORDER BY m.month;
（
NVL
（
value,0
）该函数的意思是当
value
的值为空时，以
0
代替）
NVL(EE.EMP_ID,7782) YEAR MONTH NVL(EE.EXPENSE_CLAIM,0)
------------------- ---------- ---------- -----------------------
7782 2002 1 111.09
7782 2002 2 0
7782 2002 3 9.85
7782 2002 4 0
7782 2002 5 0
7782 2002 6 0
7782 2002 7 3987.32
7782 2002 8 0
7782 2002 9 1200
7782 2002 10 0
7782 2002 11 0
我们可以看到
emp_id
为
7782
的员工在
2002
今天各个月份的
EXPENSE_CLAIM
被显示出来了，那么如果我人要显示每个员工的在
2002
今天各个月份的
EXPENSE_CLAIM
么怎么办，那么
Partition Outer Joins
就起作用了：
SELECT ee.emp_id, m.year, m.month, NVL(ee.expense_claim,0)
FROM (SELECT * FROM months WHERE year = 2002) m
LEFT OUTER JOIN employee_expense ee
PARTITION BY (ee.emp_id)
ON m.year = ee.year AND m.month = ee.month
ORDER BY ee.emp_id, m.month;
EMP_ID YEAR MONTH NVL(EE.EXPENSE_CLAIM,0)
---------- ---------- ---------- -----------------------
7369 2002 1 0
7369 2002 2 3072.43
7369 2002 3 0
7369 2002 4 30
7369 2002 5 235.03
7369 2002 6 0
7369 2002 7 0
7369 2002 8 0
7369 2002 9 5095.98
7369 2002 10 0
7369 2002 11 0
7369 2002 12 1001.01
7782 2002 1 111.09
7782 2002 2 0
7782 2002 3 9.85
7782 2002 4 0
7782 2002 5 0
7782 2002 6 0
7782 2002 7 3987.32
7782 2002 8 0
7782 2002 9 1200
7782 2002 10 0
7782 2002 11 0
7782 2002 12 0
上面那种写法我们也可以写成，下面这样（把分别每个员工在
2002
年各月份
EXPENSE_CLAIM
取出，再连接起来）：
SELECT NVL(ee.emp_id, 7369), m.year, m.month, NVL(ee.expense_claim,0)
FROM (SELECT * FROM months WHERE year = 2002) m
LEFT OUTER JOIN (SELECT *
FROM employee_expense
WHERE emp_id = 7369) ee
ON m.year = ee.year AND m.month = ee.month
ORDER BY m.month
UNION ALL
SELECT NVL(ee.emp_id, 7782), m.year, m.month, NVL(ee.expense_claim,0)
FROM (SELECT * FROM months WHERE year = 2002) m
LEFT OUTER JOIN (SELECT *
FROM employee_expense
WHERE emp_id = 7782) ee
ON m.year = ee.year AND m.month = ee.month
ORDER BY m.month;