SQL: Way to get highest count of consecutive rows with same value?

Question

I am trying to get the highest count of the number of rows with a consecutive value. For example, if below were a table and I counted the total number of consecutive 'A's I would get 5.

1. A
2. B
3. B
4. A
5. A
6. A
7. A
8. A
9. B

I am trying to find a neat way to do this in SQL. I am trying to do it with PHP but struggling, and have created a messy solution:

   $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;

I am wondering if anyone know a neater way to do this in an SQL query?

Answer 1

Try this logic,

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

Answer 2

try this :

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 from this post Mysql Counting the consecutive number rows that match

Answer 3

Another Idea:

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;