加快SQL循环

Question

I have a nested loop in my PHP code that is taking a long time and wondered of there was a way to speed this up:

$sql = "SELECT * FROM table_1";
$result = mysqli_query($sql);
while($row = mysqli_fetch_array($result)){

    $sql2 = "
    SELECT count(*) 
FROM user_modules  
WHERE 
begin_ymdhi >= ".$date_from." AND 
begin_ymdhi <= ".$date_to." AND 
(
    completed_ymdhi IS NULL OR  
    completed_ymdhi = '' 
) AND 
user_id = ".$row['user_id']." AND 
task_id = ".$row['task_id']." AND 
    module_discon = 'N' 
    ";
}

The outer query gets 1000 rows, the inner has to count across 10,000 rows - the run-time is around 115 seconds ! Is there any way to improve this method, either using a different technique or combined SQL query?

Answer 1

Don't use nested queries, combine them into a single query with a join:

SELECT t1.*, COUNT(u.user_id) ct
FROM table_1 t1
LEFT JOIN user_modules AS u ON u.user_id = t1.user_id AND u.task_id = t1.task_id
    AND u.begin_ymdhi BETWEEN '$date_from' AND '$date_to'
    AND u.module_discon = 'N'
GROUP BY t1.user_id, t1.task_id

Answer 2

SELECT user_modules.user_id, user_modules.task_id, count(*) 
FROM user_modules LEFT JOIN table_1 USING (user_id, task_id) 
WHERE  
begin_ymdhi >= ".$date_from." AND 
begin_ymdhi <= ".$date_to." AND 
    module_discon = 'N' AND
(
    completed_ymdhi IS NULL OR  
    completed_ymdhi = '' 
) 
GROUP BY user_modules.user_id, user_modules.task_id

Append EXPLAIN before that entire SELECT statement (ie EXPLAIN SELECT count(*)... ) and MySQL will give you a run-down on what the select is doing.

Make sure the begin_ymdhi field is indexed properly. SHOW INDEX FROM table_2 to see.

Answer 3

Are the task_id's unique? if so, the most straight forward would be something like:

 $sql2 = "
    SELECT count(task_id) AS TaskCount
FROM user_modules  
WHERE 
begin_ymdhi >= ".$date_from." AND 
begin_ymdhi <= ".$date_to." AND 
(
    completed_ymdhi IS NULL OR  
    completed_ymdhi = '' 
) AND module_discon = 'N'  
group by user_id
";

$result = mysqli_query($sql2);