Sql 铝数
该查询非常简单：Sql 铝数,sql,oracle,plsql,oracle10g,aggregate-functions,Sql,Oracle,Plsql,Oracle10g,Aggregate Functions,该查询非常简单： select OLD_REVISIONS NEW_REVISIONS from revs where max_num(OLD_REVISIONS) < max_num(NEW_REVISIONS); 这是表示这些值的错误方式。每个修订应该有一行，而不是将一堆修订填充到字符串中。如果正确规范化数据模型，查询将非常简单。将此表拆分为两个表（旧表和新表），还应该将分隔的值拆分为行。在那之后就简单了，你尝试的每件事都失败了，原因很简单——你目前在表中的是字符串；它们是用逗号分
select OLD_REVISIONS NEW_REVISIONS
from revs
where max_num(OLD_REVISIONS) < max_num(NEW_REVISIONS);
这是表示这些值的错误方式。每个修订应该有一行，而不是将一堆修订填充到字符串中。如果正确规范化数据模型，查询将非常简单。将此表拆分为两个表（旧表和新表），还应该将分隔的值拆分为行。在那之后就简单了，你尝试的每件事都失败了，原因很简单——你目前在表中的是字符串；它们是用逗号分隔的数字这一事实与此无关。您所做的任何比较（包括与
max（）grest（）HAVINGHAVINGgroupbyprior rn=rnOLD_REVISIONS |NEW_REVISIONS
-----------------------------------
1,2 |1
1,56,55,54 |1,55,54
"DOCNUMBER" "REVISIONNUMBER"
67 1
67 24
67 25
67 26
75 1
75 54
75 55
75 56
78 1
79 1
79 2
83 1
83 96
83 94
with rownums as (select t.*,row_number() over(order by old_revisions) rn from t)
select old_revisions,new_revisions
from rownums
where rn in (select rn
from rownums
group by rn
connect by regexp_substr(old_revisions, '[^,]+', 1, level) is not null
or regexp_substr(new_revisions, '[^,]+', 1, level) is not null
having max(cast(regexp_substr(old_revisions,'[^,]+', 1, level) as int))
<> max(cast(regexp_substr(new_revisions,'[^,]+', 1, level) as int))
)
select greatest(val1, val2), t1.r from (
select max(val) val1, r from (
select regexp_substr(v1,'[^,]+', 1, level) val, rowid r from tab1
connect by regexp_substr(v1, '[^,]+', 1, level) is not null
) group by r) t1
inner join (
select max(val) val2, r from (
select regexp_substr(v2,'[^,]+', 1, level) val, rowid r from tab1
connect by regexp_substr(v2, '[^,]+', 1, level) is not null
) group by r) t2
on (t1.r = t2.r);
create table tab1 (v1 varchar2(100), v2 varchar2(100));
insert into tab1 values ('1,3,5','1,4,7');
insert into tab1 values ('1,3,5','1,2,9');
insert into tab1 values ('1,3,5','1,3,5');
insert into tab1 values ('1,3,5','1,4');
select greatest(val1, val2), t1.r from (
select max(val) val1, r from (
select regexp_substr(v1,'[^,]+', 1, level) val, DOCNUMBER r from tab1
connect by regexp_substr(v1, '[^,]+', 1, level) is not null
) group by DOCNUMBER) t1
inner join (
select max(DOCNUMBER) val2, DOCNUMBER r from NEW_REVISIONS) t2
on (t1.r = t2.r);
SELECT listagg(o,',') WITHIN GROUP (ORDER BY o) old_revisions,
listagg(n,',') WITHIN GROUP (ORDER BY n) new_revisions
FROM (
SELECT DISTINCT rowid r,
regexp_substr(old_revisions, '[^,]+', 1, LEVEL) o,
regexp_substr(new_revisions, '[^,]+', 1, LEVEL) n
FROM table
WHERE regexp_substr(old_revisions, '[^,]+', 1, LEVEL) IS NOT NULL
CONNECT BY LEVEL<=(SELECT greatest(MAX(regexp_count(old_revisions,',')),MAX(regexp_count(new_revisions,',')))+1 c FROM table)
)
GROUP BY r
HAVING listagg(o,',') WITHIN GROUP (ORDER BY o)<>listagg(n,',') WITHIN GROUP (ORDER BY n);
select
OLD_REVISIONS,
NEW_REVISIONS
from
REVISIONS t,
table(cast(multiset(
select level
from dual
connect by level <= length (regexp_replace(t.OLD_REVISIONS, '[^,]+')) + 1
) as sys.OdciNumberList
)
) levels_old,
table(cast(multiset(
select level
from dual
connect by level <= length (regexp_replace(t.NEW_REVISIONS, '[^,]+')) + 1
)as sys.OdciNumberList
)
) levels_new
group by t.ROWID,
OLD_REVISIONS,
NEW_REVISIONS
having max(to_number(trim(regexp_substr(t.OLD_REVISIONS, '[^,]+', 1, levels_old.column_value)))) >
max(to_number(trim(regexp_substr(t.new_REVISIONS, '[^,]+', 1, levels_new.column_value))))
select xmlcast(xmlquery(('max((' || OLD_REVISIONS || '))') RETURNING CONTENT) as int) as OLD_REVISIONS_max
,xmlcast(xmlquery(('max((' || NEW_REVISIONS || '))') RETURNING CONTENT) as int) as NEW_REVISIONS_max
from t
;
with
t ( id, old_revisions, new_revisions ) as (
select 101, '1,25,26,24', '1,26,24,25' from dual union all
select 102, '1,56,55,54', '1,55,54' from dual union all
select 103, '1' , '1' from dual union all
select 104, '1,2' , '1' from dual union all
select 105, '1,96,95,94', '1,96,94,95' from dual union all
select 106, '1' , '1' from dual union all
select 107, '1' , '1' from dual union all
select 108, '1' , '1' from dual union all
select 109, '1' , '1' from dual union all
select 110, '1,2' , '1,2' from dual union all
select 111, '1' , '1' from dual union all
select 112, '1' , '1' from dual union all
select 113, '1' , '1' from dual union all
select 114, '1' , '1' from dual
)
-- END of TEST DATA; the actual solution (SQL query) begins below.
select id, old_revisions, new_revisions
from (
select id, old_revisions, new_revisions, 'old' as flag,
to_number(regexp_substr(old_revisions, '\d+', 1, level)) as rev_no
from t
connect by level <= regexp_count(old_revisions, ',') + 1
and prior id = id
and prior sys_guid() is not null
union all
select id, old_revisions, new_revisions, 'new' as flag,
to_number(regexp_substr(new_revisions, '\d+', 1, level)) as rev_no
from t
connect by level <= regexp_count(new_revisions, ',') + 1
and prior id = id
and prior sys_guid() is not null
)
group by id, old_revisions, new_revisions
having max(case when flag = 'old' then rev_no end) !=
max(case when flag = 'new' then rev_no end)
order by id -- ORDER BY is optional
;
ID OLD_REVISION NEW_REVISION
--- ------------ ------------
102 1,56,55,54 1,55,54
104 1,2 1
select max_num( '1,26,24,25') max_num from dual;
MAX_NUM
----------
26
select OLD_REVISIONS NEW_REVISIONS
from revs
where max_num(OLD_REVISIONS) < max_num(NEW_REVISIONS);
create or replace function max_num(str_in VARCHAR2) return NUMBER as
i number;
x varchar2(1);
n number := 0;
max_n number := 0;
pow number := 0;
begin
for i in 0.. length(str_in)-1 loop
x := substr(str_in,length(str_in)-i,1);
if x = ',' then
-- check max number
if n > max_n then
max_n := n;
end if;
-- reset
n := 0;
pow := 0;
else
n := n + to_number(x)*power(10,pow);
pow := pow +1;
end if;
end loop;
return(max_n);
end;
/