获取上下一个工作日实践

前言

​ 其实这个文章个人之前有进行过发布和讨论,在上一篇文章中,介绍了如何通过postgresql数据库的sql语句构建一个工作日的表,并且介绍如何使用sql语法获取某一天往前或者往后的工作日或者自然日,但是实际阅读之后发现缺少了很多细节,故这里重新梳理一下整个过程,希望可以给读者一个参考。

​ 本次实践只是个人提供的一个工作日获取的解决方案,如果有更好的解决方案欢迎讨论和分享。

上一篇文章链接:https://juejin.cn/post/7023008573827481637

注意使用的数据库为:PostgreSql

前置准备

​ 在介绍具体的编码和处理逻辑之前,我们需要准备表结构和相关的数据。

表设计

​ 首先这里依然先回顾一下这个工作日表获取的表结构:

-- ----------------------------
-- Table structure for sa_calendar_table
-- ----------------------------
DROP TABLE IF EXISTS "public"."sa_calendar_table";
CREATE TABLE "public"."sa_calendar_table" (
  "calendar_id" varchar(255) COLLATE "pg_catalog"."default" NOT NULL,
  "calendar_year" varchar(10) COLLATE "pg_catalog"."default",
  "calendar_month" varchar(10) COLLATE "pg_catalog"."default",
  "calendar_date" varchar(10) COLLATE "pg_catalog"."default",
  "day_of_week" varchar(10) COLLATE "pg_catalog"."default",
  "day_of_month" varchar(10) COLLATE "pg_catalog"."default",
  "week_of_year" varchar(10) COLLATE "pg_catalog"."default",
  "month_of_year" varchar(10) COLLATE "pg_catalog"."default",
  "quarter_of_year" varchar(10) COLLATE "pg_catalog"."default",
  "is_end_month" varchar(10) COLLATE "pg_catalog"."default",
  "is_end_quarter" varchar(10) COLLATE "pg_catalog"."default",
  "is_end_halfayear" varchar(10) COLLATE "pg_catalog"."default",
  "is_end_year" varchar(10) COLLATE "pg_catalog"."default",
  "operator_id" varchar(50) COLLATE "pg_catalog"."default",
  "operator_name" varchar(50) COLLATE "pg_catalog"."default",
  "operate_date" timestamp(6),
  "res_attr1" varchar(40) COLLATE "pg_catalog"."default",
  "res_attr2" varchar(40) COLLATE "pg_catalog"."default",
  "res_attr3" varchar(40) COLLATE "pg_catalog"."default",
  "res_attr4" varchar(40) COLLATE "pg_catalog"."default",
  "is_work_day" varchar(1) COLLATE "pg_catalog"."default"
)
WITH (fillfactor=100)
;
ALTER TABLE "public"."sa_calendar_table" OWNER TO "postgres";
COMMENT ON COLUMN "public"."sa_calendar_table"."calendar_id" IS '主键';
COMMENT ON COLUMN "public"."sa_calendar_table"."calendar_year" IS '年';
COMMENT ON COLUMN "public"."sa_calendar_table"."calendar_month" IS '月';
COMMENT ON COLUMN "public"."sa_calendar_table"."calendar_date" IS '日';
COMMENT ON COLUMN "public"."sa_calendar_table"."day_of_week" IS '自然周的第几天';
COMMENT ON COLUMN "public"."sa_calendar_table"."day_of_month" IS '月的第几天';
COMMENT ON COLUMN "public"."sa_calendar_table"."week_of_year" IS '年的第几个自然周';
COMMENT ON COLUMN "public"."sa_calendar_table"."month_of_year" IS '年的第几月';
COMMENT ON COLUMN "public"."sa_calendar_table"."quarter_of_year" IS '年的第几季';
COMMENT ON COLUMN "public"."sa_calendar_table"."is_end_month" IS '是否月末';
COMMENT ON COLUMN "public"."sa_calendar_table"."is_end_quarter" IS '是否季末';
COMMENT ON COLUMN "public"."sa_calendar_table"."is_end_halfayear" IS '是否半年末';
COMMENT ON COLUMN "public"."sa_calendar_table"."is_end_year" IS '是否年末';
COMMENT ON COLUMN "public"."sa_calendar_table"."operator_id" IS '操作人ID';
COMMENT ON COLUMN "public"."sa_calendar_table"."operator_name" IS '操作人名称';
COMMENT ON COLUMN "public"."sa_calendar_table"."operate_date" IS '操作时间';
COMMENT ON COLUMN "public"."sa_calendar_table"."res_attr1" IS '预留字段1';
COMMENT ON COLUMN "public"."sa_calendar_table"."res_attr2" IS '预留字段2';
COMMENT ON COLUMN "public"."sa_calendar_table"."res_attr3" IS '预留字段3';
COMMENT ON COLUMN "public"."sa_calendar_table"."res_attr4" IS '预留字段4';
COMMENT ON COLUMN "public"."sa_calendar_table"."is_work_day" IS '是否为工作日,Y是,N否(即节假日)';
列名称数据类型描述数据长度不能为空
calendar_idvarchar主键255YES
calendar_yearvarchar10NO
calendar_monthvarchar10NO
calendar_datevarchar10NO
day_of_weekvarchar自然周的第几天10NO
day_of_monthvarchar月的第几天10NO
week_of_yearvarchar年的第几个自然周10NO
month_of_yearvarchar年的第几月10NO
quarter_of_yearvarchar年的第几季10NO
is_end_monthvarchar是否月末10NO
is_end_quartervarchar是否季末10NO
is_end_halfayearvarchar是否半年末10NO
is_end_yearvarchar是否年末10NO
operator_idvarchar操作人ID50NO
operator_namevarchar操作人名称50NO
operate_datetimestamp操作时间6NO
res_attr1varchar预留字段140NO
res_attr2varchar预留字段240NO
res_attr3varchar预留字段340NO
res_attr4varchar预留字段440NO
is_work_dayvarchar是否为工作日,Y是,N否(即节假日)1NO

​ 另外这里再教大家一个技巧,如何使用postgresql获取某一个表的表结构:

Postgresql 获取某一个表的表结构:

SELECT A
  .attname AS COLUMN_NAME,
  T.typname AS data_type,
  d.description AS column_comment,
  btrim( SUBSTRING ( format_type ( A.atttypid, A.atttypmod ) FROM '\(.*\)' ), '()' ) AS character_maximum_length,
CASE
    
    WHEN A.attnotnull = 'f' THEN
    'NO' 
    WHEN A.attnotnull = 't' THEN
    'YES' ELSE'NO' 
END AS NULLABLE 
FROM
  pg_class C,
  pg_attribute A,
  pg_type T,
  pg_description d 
WHERE
  C.relname = '这里填表名' 
  AND A.attnum > 0 
  AND A.attrelid = C.oid 
  AND A.atttypid = T.oid 
  AND d.objoid = A.attrelid 
  AND d.objsubid = A.attnum

下面是语句的调用效果,注意上面的语句建议给所有的字段加上注释之后再执行。

填充数据

​ 有了表结构还不够,这里我们还需要填充数据,我们使用如下的sql填充数据内容,sql语句可能略微复杂了些,另外执行过程中可能会出现缺失函数的情况,由于个人使用过程中没有碰到此问题,所以就跳过了:

INSERT INTO sa_calendar_table (
  calendar_id,
  calendar_year,
  calendar_month,
  calendar_date,
  day_of_week,
  day_of_month,
  week_of_year,
  month_of_year,
  quarter_of_year,
  is_end_month,
  is_end_quarter,
  is_end_halfayear,
  is_end_year,
  operator_id,
  operator_name,
  operate_date,
  res_attr1,
  res_attr2,
  res_attr3,
  res_attr4,
  is_work_day 
) SELECT A
.calendar_id,
A.calender_year,
A.calender_month,
A.calendar_date,
A.day_of_week,
A.day_of_month,
A.week_of_year,
A.month_of_year,
A.quarter_of_year,
A.is_end_month,
A.is_end_quarter,
A.is_end_halfayear,
A.is_end_year,
A.operator_id,
A.operator_name,
A.operator_date,
A.res_attr1,
A.res_attr2,
A.res_attr3,
A.res_attr4,
A.is_work_day 
FROM
  (
  SELECT
    gen_random_uuid ( ) AS calendar_id,
    to_char( tt.DAY, 'yyyy' ) AS calender_year,
    to_char( tt.DAY, 'yyyy-mm' ) AS calender_month,
    to_char( tt.DAY, 'yyyy-mm-dd' ) AS calendar_date,
    EXTRACT ( DOW FROM tt.DAY ) AS day_of_week,
    to_char( tt.DAY, 'dd' ) AS day_of_month,
    EXTRACT ( MONTH FROM tt.DAY ) AS month_of_year,
    EXTRACT ( WEEK FROM tt.DAY ) AS week_of_year,
    EXTRACT ( QUARTER FROM tt.DAY ) AS quarter_of_year,
  CASE
      
      WHEN tt.DAY = date_trunc( 'month', tt.DAY + INTERVAL '1 month' ) - INTERVAL '1 day' THEN
      'Y' ELSE'N' 
    END AS is_end_month,
  CASE
      
      WHEN tt.DAY = date_trunc( 'quarter', tt.DAY + INTERVAL '3 month' ) - INTERVAL '1 day' THEN
      'Y' ELSE'N' 
    END AS is_end_quarter,
  CASE
      
      WHEN tt.DAY = date_trunc( 'year', tt.DAY ) + INTERVAL '6 month' - INTERVAL '1 day' THEN
      'Y' ELSE'N' 
    END AS is_end_halfayear,
  CASE
      
      WHEN tt.DAY = date_trunc( 'year', tt.DAY ) + INTERVAL '12 month' - INTERVAL '1 day' THEN
      'Y' ELSE'N' 
    END AS is_end_year,
    'b8617d3d-d2c9-4a2a-93ba-5b2d8b700cb0' AS operator_id,
    'admin' AS operator_name,
    CAST ( CURRENT_DATE AS TIMESTAMP ) AS operator_date,
    NULL AS res_attr1,
    NULL AS res_attr2,
    NULL AS res_attr3,
    NULL AS res_attr4,
  CASE
      
      WHEN EXTRACT ( DOW FROM tt.DAY ) = 6 THEN
      'N' 
      WHEN EXTRACT ( DOW FROM tt.DAY ) = 0 THEN
      'N' ELSE'Y' 
    END AS is_work_day 
  FROM
    (
    SELECT
      generate_series (
        ( SELECT ( date_trunc( 'year', now( ) ) + INTERVAL '1 year' ) :: DATE AS next_year_first_date ),
        ( SELECT ( SELECT ( date_trunc( 'year', now( ) ) + INTERVAL '2 year' ) :: DATE - 1 AS last_year_last_date ) ),
        '1 d' 
      ) AS DAY 
    ) AS tt 
  ) AS A;

​ 执行完成之后,可以看到插入了365天的数据,这里唯一需要改动的地方是:'1 year'2 year

实战部分

​ 在上一篇文章中,只是简单介绍了一个应用场景,这里继续完善此案例的内容,下面来说一下应用的场景,其实需求也比较简单,但是也比较常见:

  • 获取某一天的上一个工作日或者下一个工作日,或者获取自然日

获取工作日sql

​ 首先我们需要根据当前的天数获取某一天的工作日列表:


SELECT
    *
FROM
    (
        SELECT
            -ROW_NUMBER ( ) OVER ( ORDER BY T.calendar_date DESC ) AS addDay,
                T.calendar_date,
            T.is_work_day
        FROM
            sa_calendar_table T
        WHERE
            T.calendar_year in (#{nowYear}, #{prevYear})
          and T.calendar_date < CAST ( #{targetYyyyMMdd} AS VARCHAR )

        UNION
        SELECT ROW_NUMBER
                   ( ) OVER ( ORDER BY T.calendar_date )-1 AS addDay,
                T.calendar_date,
               T.is_work_day
        FROM
            sa_calendar_table T
        WHERE
            T.calendar_year in (#{nowYear}, #{prevYear})
          ANd T.calendar_date >= CAST ( #{targetYyyyMMdd} AS VARCHAR )

    ) mm
ORDER BY
    calendar_date

这里我们使用一个实际的案例看一下数据的形式:


SELECT
    *
FROM
    (
        SELECT
            -ROW_NUMBER ( ) OVER ( ORDER BY T.calendar_date DESC ) AS addDay,
                T.calendar_date,
            T.is_work_day
        FROM
            sa_calendar_table T
        WHERE
            T.calendar_year in  ('2020', '2021')
          and T.calendar_date < CAST ('2021-12-12' AS VARCHAR )

        UNION
        SELECT ROW_NUMBER
                   ( ) OVER ( ORDER BY T.calendar_date )-1 AS addDay,
                T.calendar_date,
               T.is_work_day
        FROM
            sa_calendar_table T
        WHERE
            T.calendar_year in ('2020', '2021')
          ANd T.calendar_date >= CAST ( '2021-12-12' AS VARCHAR )

    ) mm
ORDER BY
    calendar_date

​ 看到这里,我相信大部分读者应该都知道这是干啥用的了,这里我们通过0获取到当天,如果是+1则是下一天,而如果是-1则是上一天,如果是工作日,则对于数据进行判断,,根据这样的规则,下面我们便可以使用代码来实现:

下面是获取下一天工作日的处理,获取下一天的代码如下:

 private static final Pattern TD_DAY = Pattern.compile("^(T|D)\\+\\d$");
    private static final String WORK_DAY_CONFIG_T = "T";
    private static final String IS_WORK_DAY = "Y";
    private static final String IS_NOT_WORK_DAY = "N";
    private static final String WORK_DAY_CONFIG_D = "D";


public String findNextDayByCalendarList(CalendarDataProcessBo calendarDataProcessBo) {
        Objects.requireNonNull(calendarDataProcessBo, "当前业务传输对象不能为空");
        if (StrUtil.isAllNotBlank(new CharSequence[]{calendarDataProcessBo.getBankSettleCycle()}) && !CollectionUtil.isEmpty(calendarDataProcessBo.getCalendarDayDtos())) {
            // 额外需要往前推的天数
            int extDayOfWorkDayCount = calendarDataProcessBo.getExtDayOfWorkDayCount();
            // T+N 或者 D+N
            String bankSettleCycle = calendarDataProcessBo.getBankSettleCycle();
            // 上方截图对应的数据列表
            List<SaCalendarDayDto> calendarDayDtos = calendarDataProcessBo.getCalendarDayDtos();
            boolean matches = TD_DAY.matcher(bankSettleCycle).matches();
            // 校验正则的格式
            if (!matches) {
                logger.error("由于正则表达式{}不符合校验规则{}所以对账定时任务无法处理时间,定时任务运行失败", bankSettleCycle, TD_DAY);
                throw new UnsupportedOperationException(String.format("由于正则表达式%s不符合校验规则%s所以对账定时任务无法处理时间,定时任务运行失败", bankSettleCycle, TD_DAY));
            } else {
                String[] cycDay = bankSettleCycle.split("\\+");
                String tOrDday = cycDay[0];
                String addDay = cycDay[1];
                boolean matchWorkDayEnable;
                if (Objects.equals(tOrDday, "T")) {
                    matchWorkDayEnable = true;
                } else {
                    if (!Objects.equals(tOrDday, "D")) {
                        throw new UnsupportedOperationException("无法处理t+N或者d+N以外的数据");
                    }

                    matchWorkDayEnable = false;
                }
                // 如果需要获取工作日但是下一天不是工作日,则不断的+1往下获取
                for(int finDay = Integer.parseInt(addDay) + extDayOfWorkDayCount; finDay < CollectionUtil.size(calendarDayDtos); ++finDay) {
                    Optional<SaCalendarDayDto> first = calendarDayDtos.stream().filter((item) -> {
                        return Objects.equals(item.getAddDay(), String.valueOf(finDay));
                    }).findFirst();
                    if (!first.isPresent()) {
                        throw new UnsupportedOperationException("未发现任何工作日或者自然日数据");
                    }

                    SaCalendarDayDto saCalendarDayDto = (SaCalendarDayDto)first.get();
                    if (!matchWorkDayEnable || !Objects.equals(saCalendarDayDto.getIsWorkDay(), "N")) {
                        return saCalendarDayDto.getCalendarDate();
                    }
                }

                throw new UnsupportedOperationException("未发现任何工作日或者自然日数据");
            }
        } else {
            throw new IllegalArgumentException("传递参数有误,请确保所有参数均已传递");
        }
    }

​ 这里其实还有别的写法,比如增加一个BOOLEAN变量判断是往前还是往后,但是个人并不喜欢在参数中控制方法的行为,这样很容易出问题。

写在最后

​ 此工作日的实现方法比较笨拙也比较简单,如果有好的想法欢迎讨论。


Xander
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