問題:
給定兩個時間戳,確定工作小時數其中,將週末視為非工作日,工作時間為上午8:00 至下午3:00。
範例:
解決方案 -四捨五入結果:
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- 更多精準度:
使用較小的時間單位(例如5 分鐘切片)可提供更高的精確度:
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 -精確結果:
要獲得精確到微秒等級的精確結果,請使用以下指令方法:
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;
以上是如何計算 PostgreSQL 中兩個日期之間的工作時間,排除週末並考慮特定工作時間?的詳細內容。更多資訊請關注PHP中文網其他相關文章!