Question

得到了一个运行良好但效率不高的PostgreSQL查询...

SELECT
    _id,
    location,
    day,
    title,
    teacher,
    canceled,
    CASE
      WHEN day <= EXTRACT(dow from CURRENT_DATE)::integer THEN CURRENT_DATE + day + (6 - EXTRACT(dow from CURRENT_DATE)::integer) + "startTime"  
      WHEN day > EXTRACT(dow from CURRENT_DATE)::integer THEN CURRENT_DATE + (day - EXTRACT(dow from CURRENT_DATE)::integer - 1) + "startTime"
    END AS "startTime",
    CASE
      WHEN day <= EXTRACT(dow from CURRENT_DATE)::integer THEN CURRENT_DATE + day + (6 - EXTRACT(dow from CURRENT_DATE)::integer) + "endTime"  
      WHEN day > EXTRACT(dow from CURRENT_DATE)::integer THEN CURRENT_DATE + (day - EXTRACT(dow from CURRENT_DATE)::integer - 1) + "endTime"
    END AS "endTime"
    FROM "Schedules"
    ORDER BY location, day, "startTime";

在这种情况下，“天”列是星期几（1-7，而不是0-6）。它显示从当前日期开始的日期，因此，如果它是星期三（第4天），则显示记录，其中记录= 4为当前日期，day = 5为当前日期+ 1，依此类推。如果星期几较低，则显示下周的日期。

有人会对优化这一点有什么建议吗？

Answer 1

不查看数据和索引很难知道，但是您可以尝试一次计算DOW并使用它，而不是提取多次，这将使您的查询更高效，更易读。

例如

SELECT your_columns
      ,case when day <= v.curr_dow then 
                 current_day + day + 6 - v.curr_dow + startTime
            else current_day + day - 1 - v.curr_dow + startTime
            end as startTime
      ...
FROM   your_table
      cross join (values(extract(dow from current_date)::integer)) AS v(curr_dow)

Answer 2

一项改进是仅一次计算当前星期几：

SELECT
    _id,
    location,
    day,
    title,
    teacher,
    canceled,
    CASE
        WHEN day <= current.dow THEN CURRENT_DATE + day + (6 - current.dow) + "startTime"
        WHEN day > current.dow THEN CURRENT_DATE + (day - current.dow - 1) + "startTime"
    END AS "startTime",
    CASE
        WHEN day <= current.dow THEN CURRENT_DATE + day + (6 - current.dow) + "endTime"
        WHEN day > current.dow THEN CURRENT_DATE + (day - current.dow - 1) + "endTime"
    END AS "endTime"
    FROM "Schedules"
        FULL JOIN (
            SELECT EXTRACT(dow from CURRENT_DATE)::integer AS dow
        ) AS current
        ON TRUE
    ORDER BY location, day, "startTime";

Answer 3

加6，减去the day of the week dow并取模7（% is the modulo operator）。
那就是要添加的天数（0-6）：

ifelse()

在PostgreSQL中计算远期日期

3 个答案: