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：
代码语言：javascript复制-- ----------------------------
-- 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：」
代码语言：javascript复制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：
代码语言：javascript复制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'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：
代码语言：javascript复制
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：
代码语言：javascript复制
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：
代码语言：javascript复制 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。
