盲人求救！

snowbaby

我是个orcale盲人，工作需求，要读懂以下语句，大家有空，帮我做个注释吧。
顺便问问，有没有相关知识介绍的资料，大家给推荐推荐，谢谢。

CREATE OR REPLACE PROCEDURE "HOURLYCOUNTJOB"
is
  vnowhour date;
  vdohour date;
begin
  select thehour into vnowhour from (
    select * from hourlycount
    where cid = -2 and pvs > 0 and dip > 0
    and thehour < trunc(sysdate,'hh24')
    order by thehour desc ) where rownum < 2;
  select thehour into vdohour from nowdo;
--  while ((vdohour+1/24) <= (vnowhour-1/24))
  while ((vdohour+1/24) <= (vnowhour))
  loop
    vdohour := vdohour+1/24;
    proc_hourlycount(vdohour);

    update nowdo set thehour = trunc(vdohour,'hh24');
    commit;
  end loop;
end;

numenhuang

原帖由 snowbaby 于 2006-12-27 15:59 发表
CREATE OR REPLACE PROCEDURE "HOURLYCOUNTJOB"
is
  vnowhour date;
  vdohour date;
begin
  select thehour into vnowhour from (
    select * from hourlycount
    where cid = -2 and pvs > 0 and dip > 0
    and thehour < trunc(sysdate,'hh24')
    order by thehour desc ) where rownum < 2;
  select thehour into vdohour from nowdo;
--  while ((vdohour+1/24) <= (vnowhour-1/24))
  while ((vdohour+1/24) <= (vnowhour))
  loop
    vdohour := vdohour+1/24;
    proc_hourlycount(vdohour);

    update nowdo set thehour = trunc(vdohour,'hh24');
    commit;
  end loop;
end;

select thehour into vnowhour from (
    select * from hourlycount
    where cid = -2 and pvs > 0 and dip > 0
    and thehour < trunc(sysdate,'hh24')
    order by thehour desc ) where rownum < 2;

从hourlycount table中根据条件选出一条记录，并将thehour字段值赋给vnowhour变量

select thehour into vdohour from nowdo;

将nowdo表thehour字段值赋给vdohour变量 (？？难道这张表只有一条记录？)

while ((vdohour+1/24) <= (vnowhour))
  loop
    vdohour := vdohour+1/24;
    proc_hourlycount(vdohour);

    update nowdo set thehour = trunc(vdohour,'hh24');
    commit;
  end loop;

变量vnowhour,vdohour应该是1-24之间的整数代表一天的24小时。
这个循环主要就是根据vnowhour变量的值来更新nowdo table的thehour值，以一个小时的增量来判断，同时执行proc_hourlycount procedure。

具体为什么这样设计要根据具体的需求来分析的。从表面看大概是这个意义。