Question

我想获取两个日期之间每周的开始和结束日期。日期的格式为dd / mm / yyy hh24：mi：ss。我需要采用dd / mm / yyyy hh24：mi：ss格式的星期，因为我必须用时间计算一周的开始和结束之间的天数和小时数

我写了这句话

WITH 
date_range AS (

SELECT
            pdm.des_comercial serie,
                pdm.id_material codserie,
                ri.id_accion intervencion,
                TO_CHAR(NVL(ri.fecha_salida_rev, SYSDATE), 'dd/mm/RRRR') fecha1,
                to_char((CASE
                    WHEN ri.fecha_salida_rev > TO_DATE('18/06/2019', 'dd/mm/yyyy') THEN TO_DATE('18/06/2019', 'dd/mm/yyyy')
                    WHEN ri.fecha_salida_Rev IS NULL THEN TO_DATE('18/06/2019', 'dd/mm/yyyy')
                    ELSE ri.fecha_salida_Rev
                END),'dd/mm/yyyy hh24:mi:ss') fechasalida,
                to_char((CASE
                    WHEN ri.fecha_entrada_rev < TO_DATE('01/06/2019', 'dd/mm/yyyy') THEN TO_DATE('01/06/2019', 'dd/mm/yyyy')
                    ELSE ri.fecha_entrada_Rev
                END),'dd/mm/yyyy hh24:mi:ss') fechaentrada
                ,
                ri.cod_taller_rev,
                ri.COD_MATRICULA,
                ri.fecha_entrada_rev start_date,
                ri.fecha_salida_rev end_date
            FROM
                r_intervencion ri,
                planificador.pl_dh_material pdm
            WHERE
                ri.id_accion = ri.amortizada_por
                AND ri.causa_entrada = 1
                AND ri.tipo_accion = 1
                AND pdm.id_material = ri.cod_serie
                AND pdm.hasta = 99999999
                AND ri.ID_ACCION = 'IM4'
                AND ri.fecha_salida_rev BETWEEN TO_DATE('01/06/2019', 'dd/mm/yyyy') AND TO_DATE('18/06/2019', 'dd/mm/yyyy')
                 ),

semanas AS (            


SELECT  LEVEL "Week"

     ,to_char(to_date(start_date,'dd/mm/yyyy hh24:mi:ss') + (7 * (LEVEL - 1)),'IW') startweek

       ,to_char(to_date(start_date ,'dd/mm/yyyy hh24:mi:ss')+ (7 * (LEVEL - 1)),'IW') + 6 endweek

       ,TO_CHAR(start_date + (7 * (LEVEL - 1)),'IW') "Iso Week",
       serie,
       codserie,
       intervencion,
       cod_taller_rev,
       cod_matricula,
       fechaentrada,
       fechasalida,
       start_date,
       end_date

FROM   date_range
CONNECT  BY LEVEL <= (to_char(To_date(end_date,'dd/mm/yyyy hh24:mi:ss'),'IW') - to_char(To_date(start_date,'dd/mm/yyyy hh24:mi:ss'),'IW')) / 7 + 1 

) 

SELECT startweek,
endweek,
to_date(endweek,'dd/mm/yyyy hh24:mi:ss') - to_date(startweek,'dd/mm/yyyy hh24:mi:ss')  dias,
serie,
codserie,
intervencion,
cod_taller_rev,
cod_matricula,
start_Date,
end_date,
fechaentrada,
fechasalida,
rd.descripcion
FROM semanas,r_depositos rd
WHERE cod_taller_rev = rd.cod_deposito

执行它，我得到

查询执行失败

SQL Error [1840] [22008]: ORA-01840: ORA-01840: input value not long enough for date format

错误在于

 ,to_char(to_date(start_date,'dd/mm/yyyy hh24:mi:ss') + (7 * (LEVEL - 1)),'IW') startweek

       ,to_char(to_date(start_date ,'dd/mm/yyyy hh24:mi:ss')+ (7 * (LEVEL - 1)),'IW') + 6 endweek

如何获取格式为dd / mm / yyyy hh24：mi：ss的开始周和结束周

编辑

start_date                   end_date

20/05/2019 20:00:00         05/06/2019  08:00:00

周

20/05/2019 20:00:00 26/05/2019  -> 6 days and xxx hours
27/05/2019          02/06/2019 -> 7 days
03/06/2019          05/06/2019 08:00:00 -> 3 days and xxx hours

我需要计算每周的天数和小时数。

For example between 20/05/2019 20:00:00 and 26/05/2019
and last one between 03/06/2019 and 05/06/2019 08:00:00

我的问题是计算

to_date(endweek,'dd/mm/yyyy hh24:mi:ss') - to_date(startweek,'dd/mm/yyyy hh24:mi:ss')  dias,

endweek and startweek have to have dd/mm/yyyy hh24:mi:ss

Answer 1

您的endweek计算是

to_char(to_date(start_date ,'dd/mm/yyyy hh24:mi:ss')+ (7 * (LEVEL - 1)),'IW') + 6 endweek

这将尝试将数字6添加到字符串中。我怀疑你想要的是

to_char(to_date(start_date ,'dd/mm/yyyy hh24:mi:ss') + (7 * (LEVEL - 1) + 6),'IW') endweek

在这里，我已经移动了+ 6，所以您要在日期值（而不是字符串）上加上6。

好运。

Answer 2

我的问题是计算

to_date(endweek,'dd/mm/yyyy hh24:mi:ss') - to_date(startweek,'dd/mm/yyyy hh24:mi:ss')  dias,

endweek and startweek have to have dd/mm/yyyy hh24:mi:ss

Oracle日期以内部格式存储，您通常不必担心。您的应用程序或客户端会根据日期或会话的NLS设置将日期格式设置为字符串。

当您执行以下操作时：

to_date(endweek,'dd/mm/yyyy hh24:mi:ss')

您确实在做

to_date(to_char(endweek),'dd/mm/yyyy hh24:mi:ss')

，由于没有为隐式to_char()调用指定显式格式掩码，因此它使用了当前会话的NLS设置。根据设置，它可能会出错；或可能破坏值-例如混合使用YY和YYYY蒙版可能会失去世纪，将2019转换为0019。（鉴于您遇到的错误，您的NLS设置可能不寻常吗？）。

充其量，您需要将日期值转换为字符串，然后再返回完全相同的日期值，这毫无意义。您没有更改datem的 format ，因为它没有一个。中间的字符串可以，但是您没有使用它，也不能进行计算（至少没有按原样转换回日期，这同样没有意义。）

Oracle还具有其他操作日期值的功能，包括trunc()，所以我认为您可能需要这样-用三种方式显示差异，尽管还有其他方式，但您可以根据需要格式化最后一种：

with date_range (start_date, end_date) as (
  -- dummy data from your example
  select to_date('20/05/2019 20:00:00', 'DD/MM/YYYY HH24;MI:SS') as start_date,
    to_date('05/06/2019 08:00:00', 'DD/MM/YYYY HH24;MI:SS') as end_date
  from dual
),
semanas as (            
  select level as week,
    start_date,
    end_date,
    greatest(trunc(start_date + (7 * (level - 1)), 'IW'), start_date) as start_week,
    least(trunc(start_date + (7 * level), 'IW'), end_date) as end_week
  from date_range
  connect by level <= (trunc(end_date, 'IW') - trunc(start_date, 'IW')) / 7 + 1
)
select week,
  to_char(start_week, 'IW') as iso_week,
  to_char(start_week, 'DD/MM/YYYY HH24:MI:SS') as start_week,
  to_char(end_week, 'DD/MM/YYYY HH24:MI:SS') as end_week,
  end_week - start_week as diff_num,
  numtodsinterval(end_week - start_week, 'DAY') as diff_interval,
  to_char(date '1999-12-31' + (end_week - start_week), 'FMDD "days" HH24 "hours"') as diff_words
from semanas;

      WEEK IS START_WEEK          END_WEEK              DIFF_NUM DIFF_INTERVAL       DIFF_WORDS      
---------- -- ------------------- ------------------- ---------- ------------------- ----------------
         1 21 20/05/2019 20:00:00 27/05/2019 00:00:00 6.16666667 +06 04:00:00.000000 6 days 4 hours  
         2 22 27/05/2019 00:00:00 03/06/2019 00:00:00          7 +07 00:00:00.000000 7 days 0 hours  
         3 23 03/06/2019 00:00:00 05/06/2019 08:00:00 2.33333333 +02 08:00:00.000000 2 days 8 hours

根据目前的情况，connect by仅在date_range CTE生成单个值时才能正常工作；如果您实际上从实际查询中返回了多行，那么您将不得不做更多的工作，或者切换到递归CTE，或者交叉联接/应用，这取决于您的Oracle版本。

Oracle SQL如何获取两个日期之间的几周的开始和结束日期

2 个答案: