獲取范圍從天

Question

我有這個表結構：

編輯更復雜的例子：添加隱藏范圍

category|   day      |   a   |
--------|------------|-------|
1       | 2012-01-01 |   4   |
1       | 2012-01-02 |   4   |
1       | 2012-01-03 |   4   |
1       | 2012-01-04 |   4   |
1       | 2012-01-05 |   5   |
1       | 2012-01-06 |   5   |
1       | 2012-01-07 |   5   |
1       | 2012-01-08 |   4   |
1       | 2012-01-09 |   4   |
1       | 2012-01-10 |   4   |
1       | 2012-01-11 |   5   |
1       | 2012-01-12 |   5   |
1       | 2012-01-16 |   5   |
1       | 2012-01-17 |   5   |
1       | 2012-01-18 |   5   |
1       | 2012-01-19 |   5   |
...

'category-day'作為唯一鍵。 我會根據列“a”和給定的限制范圍為每個類別提取一系列日期，如下所示：

1,2012-01-01|2012-01-04,4
1,2012-01-05|2012-01-07,5
1,2012-01-08|2012-01-10,4
1,2012-01-11|2012-01-12,5
1,2012-01-13|2012-01-15,0
1,2012-01-16|2012-01-19,5

或類似的。

我搜索最好的方法來做到這一點。 最好只用mysql而且還帶一點點php。

注1：不是全天都插入：兩天之內非連續性不可能是其他日子。 在這種情況下，我將輸出錯過的范圍，列“a”= 0。

注意2：我用一個簡單的查詢和一些PHP行，但我不喜歡它，因為我的簡單算法需要一個周期，每個范圍內的每個類別都會找到。 如果范圍太大而且類別太多，那就不太好了。

最終編輯：好的！ 閱讀完所有評論和答案后，我認為不存在有效，高效且同時可讀的解決方案。 所以Mosty Mostacho的答案是100％有效的解決方案，但它有100％有效的建議。 謝謝你們。

Answer 1

新編輯：

正如我在評論中告訴你的那樣，我強烈建議你使用快速查詢，然后在PHP中處理缺少的日期，因為它會更快，更易讀：

select
  concat(@category := category, ',', min(day)) col1,
  concat(max(day), ',', @a := a) col2
from t, (select @category := '', @a := '', @counter := 0) init
where @counter := @counter + (category != @category or a != @a)
group by @counter, category, a

但是，如果您仍想使用查詢版本，請嘗試以下操作：

select
  @counter := @counter + (category != @category or a != @a) counter,
  concat(@category := category, ',', min(day)) col1,
  concat(max(day), ',', @a := a) col2
from (
  select distinct s.day, s.category, coalesce(t1.a, 0) a
  from (
    select (select min(day) from t) + interval val - 1 day day, c.category
    from seq s, (select distinct category from t) c
    having day <= (select max(day) from t)
  ) s
  left join t t1 on s.day = t1.day and s.category = t1.category
  where s.day between (
    select min(day) from t t2
    where s.category = t2.category) and (
    select max(day) from t t2
    where s.category = t2.category)
  order by s.category, s.day
) t, (select @category := '', @a := '', @counter := 0) init
group by counter, category, a
order by category, min(day)

需要注意的是MySQL不會允許你動態創建的數據，除非你硬編碼UNIONS ，為例子 。 這是一個昂貴的過程，這就是為什么我強烈建議你創建一個只有integer字段的表，其中值為1到X ，其中X是， 至少是分隔min(day)和max(day)的最大日期數量從你的桌子。 如果您不確定該日期，只需添加100,000數字，您就可以生成超過200年的測距期。 在上一個查詢中，此表是seq ，它具有的列是val 。

這導致：

+--------------+--------------+
|     COL1     |     COL2     |
+--------------+--------------+
| 1,2012-01-01 | 2012-01-04,4 |
| 1,2012-01-05 | 2012-01-07,5 |
| 1,2012-01-08 | 2012-01-10,4 |
| 1,2012-01-11 | 2012-01-12,5 |
| 1,2012-01-13 | 2012-01-15,0 |
| 1,2012-01-16 | 2012-01-19,5 |
+--------------+--------------+

好吧，我在說謊。 結果實際上是返回一個counter列。 只是忽略它，因為刪除它（使用派生表）會更低性能！

Answer 2

這里有一個單行的暴行:)（注意：更改“datt”表名。）

select dd.category,
dd.day as start_day,
(select dp.day from 
    (
        select 1 as n,d1.category,d1.day,d1.a from datt d1 where not exists (
            select * from datt where day = d1.day - INTERVAL 1 DAY and a=d1.a
        )
        union
        select 2 as n,d1.category,d1.day,d1.a from datt d1 where not exists (
            select * from datt where day = d1.day + INTERVAL 1 DAY and a=d1.a
        )
    ) dp where dp.day >= dd.day - INTERVAL (n-2) DAY order by day asc limit 0,1) 
as end_day,
dd.a from (
    select 1 as n,d1.category,d1.day,d1.a from datt d1 where not exists (
        select * from datt where day = d1.day - INTERVAL 1 DAY and a=d1.a
    )
    union
    select 2 as n,d1.category,d1.day,d1.a from datt d1 where not exists (
        select * from datt where day = d1.day + INTERVAL 1 DAY and a=d1.a
    )
) dd
where n=1

它的輸出是：

|| 1 || 2012-01-01 || 2012-01-01 || 4 ||
|| 1 || 2012-01-03 || 2012-01-04 || 4 ||
|| 1 || 2012-01-05 || 2012-01-07 || 5 ||
|| 1 || 2012-01-08 || 2012-01-10 || 4 ||
|| 1 || 2012-01-11 || 2012-01-12 || 5 ||

注意：這是01-12天表中不存在的2012-01-02的結果。

Answer 3

不需要PHP或臨時表或任何東西。

免責聲明：我這樣做只是為了好玩 。 這種特技可能太瘋狂，無法在生產環境中使用。 因此，我不會將此作為“真正的”解決方案發布。 此外，我不願意解釋它是如何工作的:)我並沒有重新思考/重構它。 可能有更優雅的方式，名稱/別名可以提供更多信息。 所以請不要火焰或任何東西。

這是我的解決方案。 看起來比它復雜。 我認為比其他答案更容易理解，沒有冒犯:)

設置測試數據：

drop table if exists test;
create table test(category int, day date, a int);
insert into test values
(1       , '2012-01-01' ,   4   ),
(1       , '2012-01-02' ,   4   ),
(1       , '2012-01-03' ,   4   ),
(1       , '2012-01-04' ,   4   ),
(1       , '2012-01-05' ,   5   ),
(1       , '2012-01-06' ,   5   ),
(1       , '2012-01-07' ,   5   ),
(1       , '2012-01-08' ,   4   ),
(1       , '2012-01-09' ,   4   ),
(1       , '2012-01-10' ,   4   ),
(1       , '2012-01-11' ,   5   ),
(1       , '2012-01-12' ,   5   ),
(1       , '2012-01-16' ,   5   ),
(1       , '2012-01-17' ,   5   ),
(1       , '2012-01-18' ,   5   ),
(1       , '2012-01-19' ,   5   );

它來了：

SELECT category, MIN(`day`) AS firstDayInRange, max(`day`) AS lastDayInRange, a
, COUNT(*) as howMuchDaysInThisRange /*<-- as a little extra*/
FROM
(
SELECT 
IF(@prev != qr.a, @is_a_changing:=@is_a_changing+1, @is_a_changing) AS is_a_changing, @prev:=qr.a, qr.* /*See if column a has changed. If yes, increment, so we can GROUP BY it later*/
FROM
(
SELECT 
test.category, q.`day`, COALESCE(test.a, 0) AS a /*When there is no a, replace NULL with 0*/
FROM
test
RIGHT JOIN
(
SELECT
DATE_SUB(CURDATE(), INTERVAL number_days DAY) AS `day` /*<-- Create dates from now back 999 days. This query is surprisingly fast. And adding more numbers to create more dates, i.e. 10000 dates is also no problem. Therefor a temporary dates table might not be necessary?*/
FROM
(
SELECT (a + 10*b + 100*c) AS number_days FROM
  (SELECT 0 AS a UNION ALL SELECT 1 UNION ALL SELECT 2 UNION ALL SELECT 3 UNION ALL SELECT 4 UNION ALL SELECT 5 UNION ALL SELECT 6 UNION ALL SELECT 7 UNION ALL SELECT 8 UNION ALL SELECT 9) aa
, (SELECT 0 AS b UNION ALL SELECT 1 UNION ALL SELECT 2 UNION ALL SELECT 3 UNION ALL SELECT 4 UNION ALL SELECT 5 UNION ALL SELECT 6 UNION ALL SELECT 7 UNION ALL SELECT 8 UNION ALL SELECT 9) bb
, (SELECT 0 AS c UNION ALL SELECT 1 UNION ALL SELECT 2 UNION ALL SELECT 3 UNION ALL SELECT 4 UNION ALL SELECT 5 UNION ALL SELECT 6 UNION ALL SELECT 7 UNION ALL SELECT 8 UNION ALL SELECT 9) cc
)sq /*<-- This generates numbers 0 to 999*/
)q USING(`day`) 
, (SELECT @is_a_changing:=0, @prev:=0) r
/*This WHERE clause is just to beautify. It may not be necessary*/
WHERE q.`day` >= (SELECT MIN(test.`day`) FROM test) AND q.`day` <= (SELECT MAX(test.`day`) FROM test) 
)qr
)asdf
GROUP BY is_a_changing
ORDER BY 2

結果如下所示：

category    firstDayInRange     lastDayInRange      a   howMuchDaysInThisRange
--------------------------------------------------------------------------
1           2012-01-01          2012-01-04          4   4
1           2012-01-05          2012-01-07          5   3
1           2012-01-08          2012-01-10          4   3
1           2012-01-11          2012-01-12          5   2
            2012-01-13          2012-01-15          0   3
1           2012-01-16          2012-01-19          5   4

Answer 4

首先，這是@Mosty解決方案的擴展。

為了使Mosty的解決方案能夠包括類別/日期組合而不是表中沒有，我采取了以下方法 -

首先獲取一個不同的類別列表，然后將其加入整個日期范圍 -

SELECT category, `start` + INTERVAL id DAY AS `day`
FROM dummy,(SELECT DISTINCT category FROM t) cats, (SELECT MIN(day) `start`, MAX(day) `end` FROM t) tmp
WHERE id <= DATEDIFF(`end`, `start`)
ORDER BY category, `day`

上面的查詢使用帶有單個字段id的表dummy來構建完整的日期范圍。 id字段包含0,1,2,3，.... - 它需要有足夠的值來覆蓋所需日期范圍內的每一天。 然后可以將其連接回原始表，以創建所有日期的所有類別的完整列表以及 - 的適當值 -

SELECT cj.category, cj.`day`, IFNULL(t.a, 0) AS a
FROM (
    SELECT category, `start` + INTERVAL id DAY AS `day`
    FROM dummy,(SELECT DISTINCT category FROM t) cats, (SELECT MIN(day) `start`, MAX(day) `end` FROM t) tmp
    WHERE id <= DATEDIFF(`end`, `start`)
    ORDER BY category, `day`
) AS cj
LEFT JOIN t
    ON cj.category = t.category
    AND cj.`day` = t.`day`

然后可以將其應用於Mosty的查詢來代替表格t -

SELECT
    CONCAT(@category := category, ',', MIN(`day`)) col1,
    CONCAT(MAX(`day`), ',', @a := a) col2
FROM (
    SELECT cj.category, cj.day, IFNULL(t.a, 0) AS a
    FROM (
        SELECT category, `start` + INTERVAL id DAY AS `day`
        FROM dummy,(SELECT DISTINCT category FROM t) cats, (SELECT MIN(day) `start`, MAX(day) `end` FROM t) tmp
        WHERE id <= DATEDIFF(`end`, `start`)
        ORDER BY category, `day`
    ) AS cj
    LEFT JOIN t
        ON cj.category = t.category
        AND cj.`day` = t.day) AS t, (select @category := '', @a := '', @counter := 0) init
WHERE @counter := @counter + (category != @category OR a != @a)
GROUP BY @counter, category, a

Answer 5

完全在mysql方面將有性能adv：一旦創建了程序，它將在0.35 - 0.37秒內運行

create procedure fetch_range()
begin
declare min date;
declare max date;

create  table testdate(
    date1 date
);

select min(day) into min
from category;

select max(day) into max
from category;

while min <= max do

insert into testdate values(min);
set min = adddate(min,1);
end while;

select concat(category,',',min(day)),concat(max(day),',',a) 
from(
SELECT if(isNull(category),@category,category) category,if(isNull(day),date1,day) day,@a,if(isNull(a) || isNull(@a),if(isNull(a) && isNull(@a),@grp,@grp:=@grp+1),if(@a!=a,@grp:=@grp+1,@grp)) as sor_col,if(isNull(a),0,a) as a,@a:=a,@category:= category
FROM  `category` 
RIGHT JOIN testdate ON date1 = category.day) as table1
group by sor_col;

drop table testdate;

end 

O / P：

1,2012-01-01|2012-01-04,4
1,2012-01-05|2012-01-07,5
1,2012-01-08|2012-01-10,4
1,2012-01-11|2012-01-12,5
1,2012-01-13|2012-01-15,0
1,2012-01-16|2012-01-19,5

---

這是mysql解決方案，它將提供所需的結果，僅排除錯過的范圍。

PHP：可以通過php添加缺少的范圍。

$sql = "set @a=0,@grp=0,@datediff=0,@category=0,@day='';";
mysql_query($sql);

$sql= "select category,min(day)min,max(day) max,a
from(
select category,day,a,concat(if(@a!=a,@grp:=@grp+1,@grp),if(datediff(@day,day) < -1,@datediff:=@datediff+1,@datediff)) as grp_datediff,datediff(@day,day)diff, @day:= day,@a:=a
FROM  category
order by day)as t
group by grp_datediff";

$result = mysql_query($sql);

$diff = 0;
$indx =0;
while($row = mysql_fetch_object($result)){
    if(isset($data[$indx - 1]['max'])){
    $date1 = new DateTime($data[$indx - 1]['max']);
    $date2 =  new DateTime($row->min);
    $diff = $date1->diff($date2);
    }
    if ($diff->days > 1) {

        $date = new DateTime($data[$indx-1]['max']);
        $interval = new DateInterval("P1D");
        $min = $date->add($interval);

        $date = new DateTime($data[$indx-1]['max']);
        $interval = new DateInterval("P".$diff->days."D");
        $max = $date->add($interval);

        $data[$indx]['category'] = $data[$indx-1]['category'];
        $data[$indx]['min'] = $min->format('Y-m-d');
        $data[$indx]['max'] = $max->format('Y-m-d');
        $data[$indx++]['a'] = 0;

         $data[$indx]['category'] = $row->category;
    $data[$indx]['min'] = $row->min;
    $data[$indx]['max'] = $row->max;
    $data[$indx]['a'] = $row->a;
    }else{


    $data[$indx]['category'] = $row->category;
    $data[$indx]['min'] = $row->min;
    $data[$indx]['max'] = $row->max;
    $data[$indx]['a'] = $row->a;
    }

$indx++;
}

Answer 6

要使其按照您的意願工作，您應該有兩個表：

1. 期間
2. 好幾天

通過FOREIGN KEY每個時期可以有多天與之相關。 使用當前的表結構，您可以做的最好的事情是檢測PHP端的連續周期。

Answer 7

你是這個意思嗎？

SELECT
    category,
    MIN(t1.day),
    MAX(t2.day),
    a
FROM
    `table` AS t1
INNER JOIN `table` AS t2 USING (category, a)

Answer 8

如果我理解你的問題，我會使用一些東西：

SELECT MAX(day), MIN(day) FROM `YourTable` WHERE `category`= $cat AND `A`= $increment;

......而且......

$dateRange = $cat.","."$min"."|"."$max".",".$increment;