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開始日と終了日が与えられます。

この 2 つの日付の間の平日の日数を数えたいと思います。

次に、日付の表で、同様の方法でそれらを数えて、週末のみを選択したいと思います。

誰かがこれについて私を助けることができますか?

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2 に答える 2

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1 つのアプローチは、日/日付の実体化されたテーブルを持つことです。ただし、この実体化されたテーブルを構築するために使用されるこの同じ方法は、クエリで直接使用できます。[平日] の計算をいくつか示しますが、同じアプローチを使用して、週末についてクエリを実行できます (週末の日の値は 5 と 6 です)。

直接単一クエリの例:

SELECT day
     , WEEKDAY(day) AS wkday
  FROM (
SELECT FROM_DAYS(d.day1+v1.result) AS day
  FROM (SELECT TO_DAYS(DATE('2000-01-01')) AS day1
             , TO_DAYS(DATE('2021-01-01')) AS day2
     ) AS d
  JOIN (
      SELECT v1.num+v2.num+v3.num+v4.num AS result
        FROM (
                 SELECT 1 AS num UNION SELECT 2 UNION SELECT 3 UNION SELECT 4 UNION SELECT 5
           UNION SELECT 6 UNION SELECT 7 UNION SELECT 8 UNION SELECT 9 UNION SELECT 0
           ) AS v1
        JOIN (
                 SELECT 10 AS num UNION SELECT 20 UNION SELECT 30 UNION SELECT 40 UNION SELECT 50
           UNION SELECT 60 UNION SELECT 70 UNION SELECT 80 UNION SELECT 90 UNION SELECT 00
           ) AS v2
        JOIN (
                 SELECT 100 AS num UNION SELECT 200 UNION SELECT 300 UNION SELECT 400 UNION SELECT 500
           UNION SELECT 600 UNION SELECT 700 UNION SELECT 800 UNION SELECT 900 UNION SELECT 000
           ) AS v3
        JOIN (
                 SELECT 1000 AS num UNION SELECT 2000 UNION SELECT 3000 UNION SELECT 4000 UNION SELECT 5000
           UNION SELECT 6000 UNION SELECT 7000 UNION SELECT 8000 UNION SELECT 9000 UNION SELECT 0000
           ) AS v4
     ) v1
 WHERE v1.result < (d.day2-d.day1)
     ) AS days
 WHERE WEEKDAY(day) < 5
 LIMIT 10
;



USE test;

DROP TABLE IF EXISTS days;

CREATE TABLE days (
   day   date PRIMARY KEY
) ENGINE = InnoDB;

INSERT INTO days
SELECT FROM_DAYS(d.day1+v1.result)
  FROM (SELECT TO_DAYS(DATE('2000-01-01')) AS day1
             , TO_DAYS(DATE('2021-01-01')) AS day2
     ) AS d
  JOIN (
      SELECT v1.num+v2.num+v3.num+v4.num AS result
        FROM (
                 SELECT 1 AS num UNION SELECT 2 UNION SELECT 3 UNION SELECT 4 UNION SELECT 5
           UNION SELECT 6 UNION SELECT 7 UNION SELECT 8 UNION SELECT 9 UNION SELECT 0
           ) AS v1
        JOIN (
                 SELECT 10 AS num UNION SELECT 20 UNION SELECT 30 UNION SELECT 40 UNION SELECT 50
           UNION SELECT 60 UNION SELECT 70 UNION SELECT 80 UNION SELECT 90 UNION SELECT 00
           ) AS v2
        JOIN (
                 SELECT 100 AS num UNION SELECT 200 UNION SELECT 300 UNION SELECT 400 UNION SELECT 500
           UNION SELECT 600 UNION SELECT 700 UNION SELECT 800 UNION SELECT 900 UNION SELECT 000
           ) AS v3
        JOIN (
                 SELECT 1000 AS num UNION SELECT 2000 UNION SELECT 3000 UNION SELECT 4000 UNION SELECT 5000
           UNION SELECT 6000 UNION SELECT 7000 UNION SELECT 8000 UNION SELECT 9000 UNION SELECT 0000
           ) AS v4
     ) v1
 WHERE v1.result < (d.day2-d.day1)
;

SELECT *
  FROM days
 ORDER BY day
 LIMIT 10
;


SELECT COUNT(*) FROM days;

SELECT MIN(day), MAX(day) FROM days;

SELECT day, WEEKDAY(day) FROM days LIMIT 6;

SELECT day, WEEKDAY(day) AS wkday FROM days WHERE WEEKDAY(day) < 5 LIMIT 6;

SELECT COUNT(*), MIN(day), MAX(day) FROM days WHERE WEEKDAY(day) < 5;
于 2012-10-17T00:40:28.220 に答える