MySQL内置函数：year() week()兼容
Posted 瀚高PG实验室
tags:
篇首语：本文由小常识网(cha138.com)小编为大家整理，主要介绍了MySQL内置函数：year() week()兼容相关的知识，希望对你有一定的参考价值。
瀚高数据库
目录
环境
症状
问题原因
解决方案
报错编码
环境
系统平台：中标麒麟 (CPU x86-64) 6,中标麒麟（CPU龙芯）7
版本：4.5.2
症状
从mysql数据库进行瀚高数据库国产化替代时报错：year()、 week()不存在。
问题原因
瀚高数据库内核未兼容MySQL内置函数year()、 week()。
解决方案
通过瀚高工程师编写MySQL内置函数year()、 week()，函数定义如下：
CREATE OR REPLACE FUNCTION 模式.year(indate timestamp with time zone)
RETURNS integer
LANGUAGE plpgsql
AS $function$
BEGIN
return date_part('year',inDate);
END;
$function$;
CREATE OR REPLACE FUNCTION week(anyelement, integer)
RETURNS integer
LANGUAGE plpgsql
IMMUTABLE STRICT
AS $function$
BEGIN
IF is_datetime ( $1 ) THEN
RETURN (_calc_week($1, _week_mode($2)))[1];
END IF;
RAISE EXCEPTION 'Invalid date / time value --> %', $1;
END;
$function$
;
CREATE OR REPLACE FUNCTION is_datetime(anyelement)
RETURNS boolean
LANGUAGE plpgsql
IMMUTABLE STRICT
AS $function$
DECLARE d date;
t TIMESTAMP;
tz TIMESTAMPTZ;
BEGIN
d = $1::DATE;
RETURN TRUE;
EXCEPTION WHEN others THEN
BEGIN
t = $1::TIMESTAMP;
RETURN TRUE;
EXCEPTION WHEN others THEN
BEGIN
tz = $1::TIMESTAMPTZ;
RETURN TRUE;
EXCEPTION WHEN others THEN
RETURN FALSE;
END;
END;
END;
$function$
;
CREATE OR REPLACE FUNCTION _calc_week(qdate anyelement, behavior integer)
RETURNS integer[]
LANGUAGE plpgsql
IMMUTABLE STRICT
AS $function$
DECLARE
_WEEK_MONDAY_FIRST CONSTANT integer := 1;
_WEEK_YEAR CONSTANT integer := 2;
_WEEK_FIRST_WEEKDAY CONSTANT integer := 4;
qyear integer := EXTRACT(YEAR FROM qdate);
qmonth integer := EXTRACT(MONTH FROM qdate);
qday integer := EXTRACT(DAY FROM qdate);
daynr integer := EXTRACT(DOY FROM qdate);
yday1 date := pg_catalog.date_trunc('year', qdate);
first_daynr integer := 1;
monday_first boolean := (behavior & _WEEK_MONDAY_FIRST) <> 0;
week_year boolean := (behavior & _WEEK_YEAR) <> 0;
first_weekday boolean := (behavior & _WEEK_FIRST_WEEKDAY) <> 0;
weekday integer := _calc_weekday(yday1, NOT monday_first);
days integer;
BEGIN
IF qmonth = 1 AND qday <= 7 - weekday THEN
IF (NOT week_year) AND ((first_weekday AND weekday <> 0) OR (NOT first_weekday AND weekday >= 4)) THEN
RETURN array[0, qyear];
END IF;
week_year := true;
qyear := qyear - 1;
days := _calc_days_in_year(qyear);
first_daynr := first_daynr - days;
weekday := (weekday + 53 * 7 - days) % 7;
END IF;
IF (first_weekday AND weekday <> 0) OR (NOT first_weekday AND weekday >= 4) THEN
days := daynr - (first_daynr + (7 - weekday));
ELSE
days := daynr - (first_daynr - weekday);
END IF;
IF week_year AND days >= 52 * 7 THEN
weekday := (weekday + _calc_days_in_year(qyear)) % 7;
IF (NOT first_weekday AND weekday < 4) OR (first_weekday AND weekday = 0) THEN
qyear := qyear + 1;
RETURN array[1, qyear];
END IF;
END IF;
RETURN array[days / 7 + 1, qyear];
END;
$function$
;
CREATE OR REPLACE FUNCTION _calc_weekday(qdate date, sundayfirst boolean)
RETURNS integer
LANGUAGE plpgsql
IMMUTABLE STRICT
AS $function$
BEGIN
RETURN (EXTRACT(DOW FROM qdate)::integer + CASE WHEN sundayfirst THEN 0 ELSE 6 END) % 7;
END;
$function$
;
CREATE OR REPLACE FUNCTION _calc_days_in_year(year integer)
RETURNS integer AS $$
BEGIN
IF (year & 3) = 0 AND ((year % 100) <> 0 OR (year % 400) = 0 AND year <> 0) THEN
RETURN 366;
ELSE
RETURN 365;
END IF;
END;
$$ IMMUTABLE STRICT LANGUAGE PLPGSQL;在这里插入代码片
报错编码
year()、 week()不存在,MySQL内置函数不存在
以上是关于MySQL内置函数：year() week()兼容的主要内容，如果未能解决你的问题，请参考以下文章