2 तिथियों के बीच कार्यदिवसों की संख्या गिनना चाहते हैं

एक तरीका यह है कि दिनों/तारीखों की एक भौतिक तालिका बनाई जाए। लेकिन इस भौतिक तालिका को बनाने के लिए उपयोग की जाने वाली इसी विधि का उपयोग सीधे एक प्रश्न में किया जा सकता है। मैं कुछ [सप्ताह के दिन] गणना दिखाता हूं, लेकिन आप सप्ताहांत के दिनों के बारे में पूछने के लिए उसी दृष्टिकोण का उपयोग कर सकते हैं (सप्ताहांत दिन मान 5 और 6 हैं):

प्रत्यक्ष एकल क्वेरी उदाहरण:

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;