1

这是我用来获取按周分隔的表中的记录数的 sql 查询（仅日期存储在表中）。它按预期工作。

SELECT count(id), CONCAT('Week ',WEEK(complaintRaisedDate)) week
FROM events
WHERE categoryId=1
GROUP BY week
ORDER BY week

这会产生类似的结果

count(id)       week
---------- | ----------

1               Week 36
2               Week 40
1               Week 41

我希望结果如下：

count(id)       week
---------- | ----------
1               Week 36
0               Week 37
0               Week 38
0               Week 39
2               Week 40
1               Week 41

也就是说，如果没有找到特定周的记录，它仍应显示计数为 0 的周（在表中记录的日期范围内）。我可以在 PHP 中找到一种方法，但我想知道是否它可以通过稍微调整 SQL 查询本身来实现。可能吗？谢谢。

编辑：SQLFiddle

4

2 回答 2

2

假设您有一个整数表（`numbers`如下所示）：

   SELECT COALESCE(n, 0) AS num_complaints, CONCAT('Week ', i) AS `week`
     FROM (SELECT i
             FROM numbers
            WHERE i BETWEEN (SELECT WEEK(MIN(complaintRaisedDate)) FROM events LIMIT 1)
                            AND
                            (SELECT WEEK(MAX(complaintRaisedDate)) FROM events LIMIT 1))
          week_ranges
LEFT JOIN (  SELECT count(id) AS n, WEEK(complaintRaisedDate) AS weeknum
               FROM events
              WHERE categoryId=1
           GROUP BY weeknum) weekly_tallies
       ON week_ranges.i = weekly_tallies.weeknum
 ORDER BY `week` ASC;

SQL小提琴

于 2013-10-19T21:17:51.550 回答
1

试试：http ://sqlfiddle.com/#!2/5dfbf/36

CREATE  TABLE weeks (
         id INT
       );
INSERT INTO weeks (id) VALUES (0), (1), (2), (3), (4), (5), (6), (7), (8), (9), (10), (11), (12), (13), (14), (15), (16), (17), (18), (19), (20), (21), (22), (23), (24), (25), (26), (27), (28), (29), (30), (31), (32), (33), (34), (35), (36), (37), (38), (39), (40), (41), (42), (43), (44), (45), (46), (47), (48), (49), (50), (51), (52), (53), (54);



    SELECT count(events.id), ifnull(CONCAT('Week ',WEEK(complaintRaisedDate)),0) week
FROM events RIGHT OUTER JOIN weeks ON WEEK(events.complaintRaisedDate) = weeks.id
GROUP BY weeks.id 
HAVING weeks.id>=(SELECT MIN(WEEK(events.complaintRaisedDate)) FROM events)
AND weeks.id<=(SELECT MAX(WEEK(events.complaintRaisedDate)) FROM events);
于 2013-10-19T19:38:27.407 回答