Count total number of occurrence of two consecutive values in a table - mysql

Question

My table structure
+----+--------+
| id | status |
+----+--------+
| 1 | 10 |
| 2 | 21 |
| 3 | 22 |
| 4 | 29 |
| 5 | 30 |
| 6 | 32 |
| 7 | 33 |
| 8 | 21 |
| 9 | 22 |
| 10 | 23 |
| 11 | 21 |
| 12 | 22 |
| 13 | 23 |
+----+--------+
I want to count total number of times when status 22 comes just after status 21.
In this case the query should return 3.
sql fiddle

Answer 1

Just use a Self Join with Conditional Aggregate
SELECT Sum(CASE WHEN a.status = 22 AND b.status = 21 THEN 1 END) As Stat_Count
FROM testTable a
LEFT OUTER JOIN testTable b
ON a.id = b.id + 1
SQLFIDDLE DEMO

Answer 2

If you can have gaps in your id's you can use a subquery to check whether the previous status of a 22 row is 21
select count(*)
from testtable a
where a.status = 22 and (select status from testtable b
where b.id < a.id order by id desc limit 1) = 21
http://sqlfiddle.com/#!2/9d567/2
Another way gets all id's of previous rows of rows with a status of 22 in derived table and then joins the ids to count how many have a status of 21
select count(*) from (
select max(b.id) max_b_id
from testtable a join testtable b on b.id < a.id
where a.status = 22
group by a.id
) t1 join testtable a on a.id = t1.max_b_id
where a.status = 21

Answer 3

I have tried to solve it in php
$q="";
$q= mysqli_query("select *from testTable");
while($r=mysqli_fetch_assoc($q)){
$rows[]=$r;
}
$success=0;
for ($i=0;$i<count($rows);$i++){
if($rows[$i]['status']==21 and $rows[$i+1]['status']==22 ){
$success+=1;
}
}
echo $success;