Oracle 与带有float的_char进行比较

我试图将dual中的小时和分钟与表中保存的实际数据进行比较

TO_CHARsysdate，'HH.MI'返回char，但我的数据包含浮点数

但是它说没有发现任何数据

我正在尝试这样做：

 create or replace function getSysTime

   return char

  is
      hhhh             intervals.interval_end%type;
   v_interval_start    intervals.interval_start%type;
   v_interval_end     intervals.interval_end%type;
   v_interval_id    intervals.interval_id%type;

  begin

     select  INTERVAL_START ,INTERVAL_END , INTERVAL_ID , TO_CHAR(sysdate,'HH.MI')
     into  v_interval_start , v_interval_end , v_interval_id , hhhh
    from INTERVALS

    where hhhh  =  INTERVAL_START ;

    return v_interval_id;    

   end;

解决方法：

 sloved by using cast char to float . 

  where cast (TO_CHAR(sysdate,'HH.MI') as float)  between  INTERVAL_START and INTERVAL_END ;

---

欢迎来到堆栈溢出

WHERE子句通常用于将SELECT语句限制为满足WHERE条件的行，即WHERE myTable.favoriteNumber=5。或者，您可以在其中使用布尔表达式。其中1=1的计算结果为TRUE。因为这是真的，所以返回所有行。其中0=2求值为FALSE，因此不返回任何行，因为任何行中0都不等于2

无论如何，要有逻辑地思考。为了获得一组行，需要为其提供参数。数据库如何知道需要哪些行？首先，必须使用select选择字段。哪张桌子？从定义。是否需要满足特定条件的行子集？添加一个WHERE。在哪里可以存储行中的值？添加一个到。仅仅因为PL/SQL是过程性的，并不意味着您总是从上到下、从左到右阅读

在知道哪些行符合WHERE条件之前，代码不可能在hhhh中插入值。因此，这里有null=INTERVAL\u START

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看起来您正在寻找类似的答案

-- I'm lazy and provide only a few intervals
create table intervals as
select 1 as id, 8 +  0/60 as start_, 8 + 19/60 as end_, '8:00 - 8:19' as desc_ from dual union all
select 2 as id, 8 + 20/60 as start_, 8 + 39/60 as end_, '8:20 - 8:39' as desc_ from dual union all
select 3 as id, 8 + 40/60 as start_, 8 + 59/60 as end_, '8:40 - 8:59' as desc_ from dual union all
select 4 as id, 9 +  0/60 as start_, 9 + 19/60 as end_, '9:00 - 9:19' as desc_ from dual union all
select 5 as id, 9 + 20/60 as start_, 9 + 39/60 as end_, '9:20 - 9:39' as desc_ from dual union all
select 6 as id, 9 + 40/60 as start_, 9 + 59/60 as end_, '9:40 - 9:59' as desc_ from dual
;

示例查询：

访问间隔表的便捷功能：

如何使用该功能：

---

在表格间隔内有hhhh列吗？从HHH=INTERVAL\u START；的INTERVAL中选择INTERVAL\u START、INTERVAL\u END、INTERVAL\u ID到\u CHARsysdate、'HH.MI'的结果是什么；？hhhh不是列，它的变量的类型为float，而TO_CHAR返回CHAR，因此当比较CHAR和float时，它返回未找到的数据。是的，它将不返回未找到的数据，因为您尚未在where部分中为hhhh设置值。这没有多大意义。MI可能看起来像一个十进制数字，但它不是。MI的计算结果介于00和59之间，因此，例如，0.5表示50分钟，而不是半小时。

select * from intervals 
where extract(hour from localtimestamp) + 
      (extract(minute from localtimestamp) / 60)
between start_ and end_;

select * from intervals
where extract(hour from to_timestamp('2013-08-29 08:39:59', 'YYYY-MM-DD HH24:MI:SS')) + 
      (extract(minute from to_timestamp('2013-08-29 08:39:59', 'YYYY-MM-DD HH24:MI:SS')) / 60)
between start_ and end_;

select * from intervals
where extract(hour from to_timestamp('2013-08-29 08:40:00', 'YYYY-MM-DD HH24:MI:SS')) + 
      (extract(minute from to_timestamp('2013-08-29 08:40:00', 'YYYY-MM-DD HH24:MI:SS')) / 60)
between start_ and end_;

create or replace function get_interval_id(
  p_time in timestamp default localtimestamp
) return number as
  v_id number;
begin
  select id into v_id
    from intervals
   where extract(hour from p_time) + 
         (extract(minute from p_time) / 60)
  between start_ and end_;

  return v_id;
exception
  when others then
    return null;
end;
/
show errors

SQL> select localtimestamp from dual;

LOCALTIMESTAMP
---------------------------------------------------------------------------
2013-08-29 09:41:51.388

SQL> select * from intervals where id = get_interval_id;

        ID     START_       END_ DESC_
---------- ---------- ---------- -----------
         6 9.66666667 9.98333333 9:40 - 9:59

SQL> select * from intervals where id = get_interval_id(to_timestamp('2013-08-29 08:59:00', 'YYYY-MM-DD HH24:MI:SS'));

        ID     START_       END_ DESC_
---------- ---------- ---------- -----------
         3 8.66666667 8.98333333 8:40 - 8:59