锐客网

  1. 首页 > it技术 > >

获取上下一个工作日实践

postgresql

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_id varchar 主键 255 YES
calendar_year varchar 年 10 NO
calendar_month varchar 月 10 NO
calendar_date varchar 日 10 NO
day_of_week varchar 自然周的第几天 10 NO
day_of_month varchar 月的第几天 10 NO
week_of_year varchar 年的第几个自然周 10 NO
month_of_year varchar 年的第几月 10 NO
quarter_of_year varchar 年的第几季 10 NO
is_end_month varchar 是否月末 10 NO
is_end_quarter varchar 是否季末 10 NO
is_end_halfayear varchar 是否半年末 10 NO
is_end_year varchar 是否年末 10 NO
operator_id varchar 操作人ID 50 NO
operator_name varchar 操作人名称 50 NO
operate_date timestamp 操作时间 6 NO
res_attr1 varchar 预留字段1 40 NO
res_attr2 varchar 预留字段2 40 NO
res_attr3 varchar 预留字段3 40 NO
res_attr4 varchar 预留字段4 40 NO
is_work_day varchar 是否为工作日，Y是，N否(即节假日) 1 NO
? 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, CASEWHEN 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。
获取上下一个工作日实践
文章图片

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, CASEWHEN tt.DAY = date_trunc( 'month', tt.DAY + INTERVAL '1 month' ) - INTERVAL '1 day' THEN 'Y' ELSE'N' END AS is_end_month, CASEWHEN tt.DAY = date_trunc( 'quarter', tt.DAY + INTERVAL '3 month' ) - INTERVAL '1 day' THEN 'Y' ELSE'N' END AS is_end_quarter, CASEWHEN tt.DAY = date_trunc( 'year', tt.DAY ) + INTERVAL '6 month' - INTERVAL '1 day' THEN 'Y' ELSE'N' END AS is_end_halfayear, CASEWHEN 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, CASEWHEN 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;

获取上下一个工作日实践
文章图片

? 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'和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

获取上下一个工作日实践
文章图片

? 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 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 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 【获取上下一个工作日实践】? 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。

    推荐阅读


    上一篇：利用python爬取信息并保存到excel

    下一篇：实时数仓Hologres首次走进阿里淘特双11