Sql （Oracle）如何为分页对行进行分组
我有一个表，其输出与此类似（尽管以千为单位）： 我期望的输出与此类似：Sql （Oracle）如何为分页对行进行分组,sql,oracle,oracle10g,pagination,Sql,Oracle,Oracle10g,Pagination,我有一个表，其输出与此类似（尽管以千为单位）： 我期望的输出与此类似： EMPNO ENAME TRANDATE AMT PAGE ---------- ---------- --------- ------- ---- 100 Alison 21-MAR-96 45000 1 100 Alison 12-DEC-78 23000 1 100 Alison 24-OCT-82
EMPNO ENAME TRANDATE AMT PAGE
---------- ---------- --------- ------- ----
100 Alison 21-MAR-96 45000 1
100 Alison 12-DEC-78 23000 1
100 Alison 24-OCT-82 11000 1
101 Linda 15-JAN-84 16000 2
101 Linda 30-JUL-87 17000 2
102 Celia 31-DEC-90 78000 2
102 Celia 17-SEP-96 21000 2
103 James 21-MAR-96 45000 3
104 Robert 12-DEC-78 23000 4
104 Robert 24-OCT-82 11000 4
104 Robert 15-JAN-84 16000 4
104 Robert 30-JUL-87 17000 4
基本上，它应该插入一个新字段来标识它所属的页面。分页符基于行。而且，就像EMPNO中的“保持在一起”一样，当行无法添加下一个EMPNO批时，它会向页面添加1。这是Excel的限制，因为Excel不允许单个工作表中的行数超过65000行。在示例中，只有4行。限制号是静态的。以下SQL语句将我的EMP表中的20条记录拆分为五页，每页四行：
SQL> select empno
2 , ename
3 , deptno
4 , ceil((row_number() over (order by deptno, empno)/4)) as pageno
5 from emp
6 /
EMPNO ENAME DEPTNO PAGENO
---------- ---------- ---------- ----------
7782 BOEHMER 10 1
7839 SCHNEIDER 10 1
7934 KISHORE 10 1
7369 CLARKE 20 1
7566 ROBERTSON 20 2
7788 RIGBY 20 2
7876 KULASH 20 2
7902 GASPAROTTO 20 2
7499 VAN WIJK 30 3
7521 PADFIELD 30 3
7654 BILLINGTON 30 3
7698 SPENCER 30 3
7844 CAVE 30 4
7900 HALL 30 4
8083 KESTELYN 30 4
8084 LIRA 30 4
8060 VERREYNNE 50 5
8061 FEUERSTEIN 50 5
8085 TRICHLER 50 5
8100 PODER 50 5
20 rows selected.
SQL>
在普通SQL中做这样的事情太难了，甚至是不可能的 但有一些局限性，这个问题可以借助于 首先，使用ODCIAggregate接口实现创建对象：
create or replace type page_num_agg_type as object
(
-- Purpose : Pagination with "leave together" option
-- Attributes
-- Current page number
cur_page_number number,
-- Cumulative number of rows per page incremented by blocks
cur_page_row_count number,
-- Row-by-row counter for detect page overflow while placing single block
page_row_counter number,
-- Member functions and procedures
static function ODCIAggregateInitialize(
sctx in out page_num_agg_type
)
return number,
member function ODCIAggregateIterate(
self in out page_num_agg_type,
value in number
)
return number,
member function ODCIAggregateTerminate(
self in page_num_agg_type,
returnValue out number,
flags in number
)
return number,
member function ODCIAggregateMerge(
self in out page_num_agg_type,
ctx2 in page_num_agg_type
)
return number
);
create or replace type body PAGE_NUM_AGG_TYPE is
-- Member procedures and functions
static function ODCIAggregateInitialize(
sctx in out page_num_agg_type
)
return number
is
begin
sctx := page_num_agg_type(1, 0, 0);
return ODCIConst.Success;
end;
member function ODCIAggregateIterate(
self in out page_num_agg_type,
value in number
)
return number
is
-- !!! WARNING: HARDCODED !!!
RowsPerPage number := 4;
begin
self.page_row_counter := self.page_row_counter + 1;
-- Main operations: determine number of page
if(value > 0) then
-- First row of new block
if(self.cur_page_row_count + value > RowsPerPage) then
-- If we reach next page with new block of records - switch to next page.
self.cur_page_number := self.cur_page_number + 1;
self.cur_page_row_count := value;
self.page_row_counter := 1;
else
-- Just increment rows and continue to place on current page
self.cur_page_row_count := self.cur_page_row_count + value;
end if;
else
-- Row from previous block
if(self.page_row_counter > RowsPerPage) then
-- Single block of rows exceeds page size - wrap to next page.
self.cur_page_number := self.cur_page_number + 1;
self.cur_page_row_count := self.cur_page_row_count - RowsPerPage;
self.page_row_counter := 1;
end if;
end if;
return ODCIConst.Success;
end;
member function ODCIAggregateTerminate(
self in page_num_agg_type,
returnValue out number,
flags in number
)
return number
is
begin
-- Returns current page number as result
returnValue := self.cur_page_number;
return ODCIConst.Success;
end;
member function ODCIAggregateMerge(
self in out page_num_agg_type,
ctx2 in page_num_agg_type
)
return number
is
begin
-- Can't act in parallel - error on merging attempts
raise_application_error(-20202,'PAGE_NUM_AGG_TYPE can''t act in parallel mode');
return ODCIConst.Success;
end;
end;
create function page_num_agg (
input number
) return number aggregate using page_num_agg_type;
with data_list as (
-- Your example data as source
select 100 as EmpNo, 'Alison' as EmpName, to_date('21-MAR-96','dd-mon-yy') as TranDate, 45000 as AMT from dual union all
select 100 as EmpNo, 'Alison' as EmpName, to_date('12-DEC-78','dd-mon-yy') as TranDate, 23000 as AMT from dual union all
select 100 as EmpNo, 'Alison' as EmpName, to_date('24-OCT-82','dd-mon-yy') as TranDate, 11000 as AMT from dual union all
select 101 as EmpNo, 'Linda' as EmpName, to_date('15-JAN-84','dd-mon-yy') as TranDate, 16000 as AMT from dual union all
select 101 as EmpNo, 'Linda' as EmpName, to_date('30-JUL-87','dd-mon-yy') as TranDate, 17000 as AMT from dual union all
select 102 as EmpNo, 'Celia' as EmpName, to_date('31-DEC-90','dd-mon-yy') as TranDate, 78000 as AMT from dual union all
select 102 as EmpNo, 'Celia' as EmpName, to_date('17-SEP-96','dd-mon-yy') as TranDate, 21000 as AMT from dual union all
select 103 as EmpNo, 'James' as EmpName, to_date('21-MAR-96','dd-mon-yy') as TranDate, 45000 as AMT from dual union all
select 103 as EmpNo, 'James' as EmpName, to_date('12-DEC-78','dd-mon-yy') as TranDate, 23000 as AMT from dual union all
select 103 as EmpNo, 'James' as EmpName, to_date('24-OCT-82','dd-mon-yy') as TranDate, 11000 as AMT from dual union all
select 104 as EmpNo, 'Robert' as EmpName, to_date('15-JAN-84','dd-mon-yy') as TranDate, 16000 as AMT from dual union all
select 104 as EmpNo, 'Robert' as EmpName, to_date('30-JUL-87','dd-mon-yy') as TranDate, 17000 as AMT from dual union all
select 105 as EmpNo, 'Monica' as EmpName, to_date('30-JUL-88','dd-mon-yy') as TranDate, 31000 as AMT from dual union all
select 105 as EmpNo, 'Monica' as EmpName, to_date('01-JUL-87','dd-mon-yy') as TranDate, 19000 as AMT from dual union all
select 105 as EmpNo, 'Monica' as EmpName, to_date('31-JAN-97','dd-mon-yy') as TranDate, 11000 as AMT from dual union all
select 105 as EmpNo, 'Monica' as EmpName, to_date('17-DEC-93','dd-mon-yy') as TranDate, 33000 as AMT from dual union all
select 105 as EmpNo, 'Monica' as EmpName, to_date('11-DEC-91','dd-mon-yy') as TranDate, 65000 as AMT from dual union all
select 105 as EmpNo, 'Monica' as EmpName, to_date('22-OCT-89','dd-mon-yy') as TranDate, 19000 as AMT from dual
),
ordered_data as (
select
-- Source table fields
src_data.EmpNo, src_data.EmpName, src_data.TranDate, src_data.AMT,
-- Calculate row count per one employee
count(src_data.EmpNo) over(partition by src_data.EmpNo)as emp_row_count,
-- Calculate rank of row inside employee data sorted in output order
rank() over(partition by src_data.EmpNo order by src_data.EmpName, src_data.TranDate) as emp_rnk
from
data_list src_data
)
-- Final step: calculate page number for rows
select
-- Source table data
ordered_data.EmpNo, ordered_data.EmpName, ordered_data.TranDate, ordered_data.AMT,
-- Aggregate all data with our new function
page_num_agg(
-- pass count of rows to aggregate function only for first employee's row
decode(ordered_data.emp_rnk, 1, ordered_data.emp_row_count, 0)
)
over (order by ordered_data.EmpName, ordered_data.TranDate) as page_number
from
ordered_data
order by
ordered_data.EmpName, ordered_data.TranDate
更新：改进以处理超大块，示例已修改。ThinkJet认为其他一些答案不符合“保持在一起”的要求，这是正确的。但是，我认为这可以在不使用用户定义的聚合的情况下完成 样本数据
create table test (empno number, ename varchar2(20), trandate date, amt number);
insert into test values (100, 'Alison' , to_date('21-MAR-1996') , 45000);
insert into test values (100, 'Alison' , to_date('12-DEC-1978') , 23000);
insert into test values (100, 'Alison' , to_date('24-OCT-1982') , 11000);
insert into test values (101, 'Linda' , to_date('15-JAN-1984') , 16000);
insert into test values (101, 'Linda' , to_date('30-JUL-1987') , 17000);
insert into test values (102, 'Celia' , to_date('31-DEC-1990') , 78000);
insert into test values (102, 'Celia' , to_date('17-SEP-1996') , 21000);
insert into test values (103, 'James' , to_date('21-MAR-1996') , 45000);
insert into test values (103, 'James' , to_date('12-DEC-1978') , 23000);
insert into test values (103, 'James' , to_date('24-OCT-1982') , 11000);
insert into test values (104, 'Robert' , to_date('15-JAN-1984') , 16000);
insert into test values (104, 'Robert' , to_date('30-JUL-1987') , 17000);
select empno, ename,
ceil((rank() over (order by empno) +
count(1) over (partition by empno))/6) as chunk
from test
order by 1;
正如ThinkJet所指出的，这种解决方案不是防弹的
drop table test purge;
create table test (empno number, ename varchar2(20), trandate date, amt number);
declare
cursor csr_name is
select rownum emp_id,
decode(rownum,1,'Alan',2,'Brian',3,'Clare',4,'David',5,'Edgar',
6,'Fred',7,'Greg',8,'Harry',9,'Imran',10,'John',
11,'Kevin',12,'Lewis',13,'Morris',14,'Nigel',15,'Oliver',
16,'Peter',17,'Quentin',18,'Richard',19,'Simon',20,'Terry',
21,'Uther',22,'Victor',23,'Wally',24,'Xander',
25,'Yasmin',26,'Zac') emp_name
from dual connect by level <= 26;
begin
for c_name in csr_name loop
for i in 1..11 loop
insert into test values
(c_name.emp_id, c_name.emp_name, (date '2010-01-01') + i,
to_char(sysdate,'SS') * 1000);
end loop;
end loop;
end;
/
select chunk, count(*)
from
(select empno, ename,
ceil((rank() over (order by empno) +
count(1) over (partition by empno))/25) as chunk
from test)
group by chunk
order by chunk
;
升降台试验吹扫；
创建表测试（empno编号、ename varchar2（20）、传输日期、金额编号）；
声明
游标csr\u名称为
选择rownum emp_id，
解码（rownum，1，'Alan'，2，'Brian'，3，'Clare'，4，'David'，5，'Edgar'，等等），
6、'Fred'，7、'Greg'，8、'Harry'，9、'Imran'，10、'John'，
11、'Kevin'，12、'Lewis'，13、'Morris'，14、'Nigel'，15、'Oliver'，
16、'Peter'，17、'Quentin'，18、'Richard'，19、'Simon'，20、'Terry'，
21、'Uther'，22、'Victor'，23、'Wally'，24、'Xander'，
25，'Yasmin'，26，'Zac'）emp_name
从dual connect by level看，这个怎么样：（100是每页的行限制）
这是行不通的。重要条件-empno不能分为两组。在您的情况下，EMPNO101将分为两组。谢谢！这个代码适合我！（更新：好的……也许不是。迈克尔，这是个好发现。）只是不起作用——结果不对。例如，第3页及以后的页面末尾的其他空行在哪里计数？那么，您能否保证没有一个EMPNO的记录超过65000条？另外，您使用的是哪个版本的Excel？Excel 2007允许每个工作表有1048576行，我希望它能支持2007年以前的Excel版本。（）在我目前的情况下（我认为），一个EMPNO很可能不会有超过65k条记录。加里的解决方案解决了我的问题。非常感谢你们所有人！我以前没有真正做过任何用户定义的聚合函数。不过我可以做手术。非常感谢！这对我的案子有效！我真的认为这个问题会更长。我了解65k+记录使用相同EMPNO的问题，但在我的情况下，这很可能不会发生。与其他答案中的错误相同：计算中未处理页面末尾的随机间隙。例如，假设您有3组行（A、B、C），每组6行。尝试将其放在页面上，每页6行：组A-rank（）=1-Ok，第1页；B组-rank（）=7-确定，第2页从顶部填充，第1页保留4行空白；C组-秩（）=13-错误！，似乎适合于第2页，但必须放在第3页，因为REAL rank（）值必须包含第1页的空行：6+4+6+1=17。在构建真正的页面之前，无法预测空行的数量。即使是65k+行，甚至在第二页末尾，也可能发生这种情况：例如，第一页上的5个空闲行+2页上的4个空闲行+下一个块的大小为5到9行……啊，我明白了ThinkJet的问题。第1页末尾有5行的空间，但下一组是6行。这六行被正确地推到第2页，但这六行没有被正确地计算为第2页的一部分。所以，当计算第二页可以容纳多少人时，可能会出错。我想我有点理解这个问题。我只是太兴奋了，很快就检查了我的测试（用一个50k的块正确地输出了一个两页的文件）。但是当我用较小的块大小（300）测试查询时，我确实注意到了块的减少。
drop table test purge;
create table test (empno number, ename varchar2(20), trandate date, amt number);
declare
cursor csr_name is
select rownum emp_id,
decode(rownum,1,'Alan',2,'Brian',3,'Clare',4,'David',5,'Edgar',
6,'Fred',7,'Greg',8,'Harry',9,'Imran',10,'John',
11,'Kevin',12,'Lewis',13,'Morris',14,'Nigel',15,'Oliver',
16,'Peter',17,'Quentin',18,'Richard',19,'Simon',20,'Terry',
21,'Uther',22,'Victor',23,'Wally',24,'Xander',
25,'Yasmin',26,'Zac') emp_name
from dual connect by level <= 26;
begin
for c_name in csr_name loop
for i in 1..11 loop
insert into test values
(c_name.emp_id, c_name.emp_name, (date '2010-01-01') + i,
to_char(sysdate,'SS') * 1000);
end loop;
end loop;
end;
/
select chunk, count(*)
from
(select empno, ename,
ceil((rank() over (order by empno) +
count(1) over (partition by empno))/25) as chunk
from test)
group by chunk
order by chunk
;
select
a.*, floor(count(1) over (order by empno, empname)/100)+1 as page
from source_table a
order by page, empno, empname;