SQL：获取具有相同值的连续行数最高的方法？

Question

我想获得具有连续值的行数的最高计数。例如，如果下面是一个表和我数个连续的“A的总数我会得到5.

1. 甲
2. 乙
3. 乙
4. 甲
5. 甲
6. A
7. A
8. A
9. 乙

我试图找到一种简洁的方式在SQL做到这一点。我想用PHP，但努力去做，并创建了一个凌乱的解决方案：

$streaksql = "SELECT * FROM `mytable`"; 
    $streaksql = $modx->query($streaksql); 

    $outcomecounter = 0; 
    $highestoutcome = 0; 

    while ($streak = $streaksql->fetch(PDO::FETCH_ASSOC)) { 

        $outcome = $row['outcome']; 

        if($outcome == 'A'){ 
         $outcomecounter = $outcomecounter +1; 

         if($outcomecounter > $highestoutcome){ 
          $highestoutcome = $outcomecounter; 
         } 

        } 
        else { 
         $outcomecounter = 0; 
        } 
       };//WHILE 

echo $highestoutcome; 

我想知道是否有人知道一个更合适的方法在SQL查询来做到这一点？

Answer 1

试试这个逻辑，

select top 1 col1 from myTable 
group by col1 
order by count(col2) desc 

Answer 2

尝试这个办法：从这篇文章

select COUNT(*)+1 
FROM your_table a1 
where value=(select value from your_table where your_table.id<a1.id order by id desc LIMIT 1) AND value= 'A' 

CMIIW :)

referrence Mysql Counting the consecutive number rows that match

Answer 3

另一个想法：

SELECT outcome, max(Amount) from (
    SELECT outcome, 
    IF(@a =outcome, @b := @b+1, @b := 1) as "Amount", 
    @a := outcome 
    FROM mytable 
    WHERE false = (@b:=0) 
    AND outcome = 'A'  
) as test 
GROUP by outcome;