问题:
给定两个时间戳,确定工作小时数其中,将周末视为非工作日,工作时间为上午 8:00 至下午 3:00。
示例:
解决方案 - 四舍五入结果:
SELECT count(*) AS work_hours FROM generate_series (timestamp '2013-06-24 13:30', timestamp '2013-06-24 15:29' - interval '1h', interval '1h') h WHERE EXTRACT(ISODOW FROM h) < 6 AND h::time >= '08:00' AND h::time <= '14:00';
解决方案 - 更多精度:
使用较小的时间单位(例如 5 分钟切片)可提供更高的精度:
SELECT count(*) * interval '5 min' AS work_interval FROM generate_series (timestamp '2013-06-24 13:30', timestamp '2013-06-24 15:29' - interval '5 min', interval '5 min') h WHERE EXTRACT(ISODOW FROM h) < 6 AND h::time >= '08:00' AND h::time <= '14:55';
解决方案 - 精确结果:
要获得精确到微秒级别的精确结果,请使用以下命令方法:
SELECT t_id , COALESCE(h.h, '0') -- add / subtract fractions - CASE WHEN EXTRACT(ISODOW FROM t_start) < 6 AND t_start::time > v_start AND t_start::time < v_end THEN t_start - date_trunc('hour', t_start) ELSE '0'::interval END + CASE WHEN EXTRACT(ISODOW FROM t_end) < 6 AND t_end::time > v_start AND t_end::time < v_end THEN t_end - date_trunc('hour', t_end) ELSE '0'::interval END AS work_interval FROM t CROSS JOIN var LEFT JOIN ( -- count full hours, similar to above solutions SELECT t_id, count(*)::int * interval '1h' AS h FROM ( SELECT t_id, v_start, v_end , generate_series (date_trunc('hour', t_start) , date_trunc('hour', t_end) - interval '1h' , interval '1h') AS h FROM t, var ) sub WHERE EXTRACT(ISODOW FROM h) < 6 AND h::time >= v_start AND h::time <= v_end - interval '1h' GROUP BY 1 ) h USING (t_id) ORDER BY 1;
以上是如何计算 PostgreSQL 中两个日期之间的工作时间,排除周末并考虑具体工作时间?的详细内容。更多信息请关注PHP中文网其他相关文章!