我有一张表，记录了不同日期的入境/出境国内/国际邮件的记录。以下是示例表：
Date location in_out_code dom_int_code mail_count
------------------------------------------------------------------------------
11/01/2012 chicago in I 3
11/02/2012 la in I 2
11/03/2012 ny in I 4
11/03/2012 ny out D 5
11/04/2012 phoenix out D 1
11/05/2012 phoenix in D 3
我想创建一个存储每天所有组合的表，如下所示（为了便于阅读，我分解了输出）：
Date location in_out_code dom_int_code mail_count
----------------------------------------------------------------------------
11/01/2012 chicago in I 3
11/01/2012 chicago in D 0 <-- inserted
11/01/2012 chicago out I 0 <-- inserted
11/01/2012 chicago out D 0 <-- inserted
11/02/2012 la in I 2
11/02/2012 la in D 0 <-- inserted
11/02/2012 la out I 0 <-- inserted
11/02/2012 la out D 0 <-- inserted
11/03/2012 ny in I 4
11/03/2012 ny in D 0 <-- inserted
11/03/2012 ny out I 0 <-- inserted
11/03/2012 ny out D 5
11/04/2012 phoenix in I 0 <-- inserted
11/04/2012 phoenix in D 0 <-- inserted
11/04/2012 phoenix out I 0 <-- inserted
11/04/2012 phoenix out D 1
11/05/2012 phoenix in I 0 <-- inserted
11/05/2012 phoenix in D 3
11/05/2012 phoenix out I 0 <-- inserted
11/05/2012 phoenix out D 0 <-- inserted
我将如何做到这一点。有什么建议吗？
由于
答案 0 :(得分：3)
Oracle 10g及更高版本。您可以使用外部联接的partition by扩展来填补数据中的空白：
包含您提供的数据的表格：
SQL> create table Your_Table_Name(Date1, location1, in_out_code, dom_int_code, mail_count) as(
2 select to_date('11/01/2012', 'mm/dd/yyyy'), 'chicago', 'in', 'I', 3 from dual union all
3 select to_date('11/02/2012', 'mm/dd/yyyy'), 'la' , 'in', 'I', 2 from dual union all
4 select to_date('11/03/2012', 'mm/dd/yyyy'), 'ny' , 'in', 'I', 4 from dual union all
5 select to_date('11/03/2012', 'mm/dd/yyyy'), 'ny' , 'out', 'D', 5 from dual union all
6 select to_date('11/04/2012', 'mm/dd/yyyy'), 'phoenix', 'out', 'D', 1 from dual union all
7 select to_date('11/05/2012', 'mm/dd/yyyy'), 'phoenix', 'in' , 'D', 3 from dual
8 )
9 ;
Table created
我们的查询：
SQL> with t2(in_out_code, dom_int_code) as(
2 select 'in', 'I' from dual union all
3 select 'out', 'I' from dual union all
4 select 'in', 'D' from dual union all
5 select 'out', 'D' from dual
6 )
7 select t1.date1
8 , t1.location1
9 , t2.in_out_code
10 , t2.dom_int_code
11 , nvl(t1.mail_count, 0) as mail_count
12 from your_table_name t1
13 partition by (t1.date1, t1.location1)
14 right outer join t2
15 on (t1.in_out_code = t2.in_out_code and
16 t1.dom_int_code = t2.dom_int_code)
17 ;
结果：
DATE1 LOCATION1 IN_OUT_CODE DOM_INT_CODE MAIL_COUNT
----------- --------- ----------- ------------ ----------
11/01/2012 chicago in D 0
11/01/2012 chicago in I 3
11/01/2012 chicago out D 0
11/01/2012 chicago out I 0
11/02/2012 la in D 0
11/02/2012 la in I 2
11/02/2012 la out D 0
11/02/2012 la out I 0
11/03/2012 ny in D 0
11/03/2012 ny in I 4
11/03/2012 ny out D 5
11/03/2012 ny out I 0
11/04/2012 phoenix in D 0
11/04/2012 phoenix in I 0
11/04/2012 phoenix out D 1
11/04/2012 phoenix out I 0
11/05/2012 phoenix in D 3
11/05/2012 phoenix in I 0
11/05/2012 phoenix out D 0
11/05/2012 phoenix out I 0
20 rows selected