How to get Hourly data from mysql? - php

Question

I have a mysql database having my client details, i want to show hourly visits, page views in a array. I made below given mysql query.
<?php
$siteid=$_GET["site"];
$DB = mysqli_connect('localhost','username','password','databasename');
$Q = "SELECT time, COUNT(*), pw FROM clientstats WHERE siteid='$siteid' AND time > UNIX_TIMESTAMP(DATE_SUB(NOW(), INTERVAL 26970 SECOND)) GROUP BY HOUR( FROM_UNIXTIME( `time`) ) ORDER BY time ASC";
$R = mysqli_query($DB,$Q);
//start loop
//while or foreach
while($row = mysqli_fetch_assoc($R)){
date_default_timezone_set('Asia/Kolkata');
$timestamp = $row['time'];
$timee = date('H', $timestamp);
$times = $row['COUNT(*)']+$row['wh'];
echo ",['$timee', {$row['COUNT(*)']}, $times]\r\n";
}
?>
where 26970 is seconds passed from midnight to till now. This code works fine and outputs a array like this
,['02', 2, 3] ,['02', 1, 2] ,['04', 1, 2] ,['05', 1, 2] ,['06', 2, 3]
This array structure is ['HOUR', visits, pw] but the problem is that this does not show values for each hour continuously like 1,2,3,4,5,6.... etc. I want to show values for each hours. If values for that hours are nil then i want to output that hour value as ['hour', 0, 0], Besides there is doubling in above output for second hour. How to correct it also.
Please help me, Thanks in advance.

Answer 1

To get the result returned from MySQL, using a SELECT statement, there needs to be a row source that returns the hour values that are "missing" from the table we are getting the count from.
(Getting this result from a query is only one option; we could approach it differently, and do it in the client. For example, populating an array with the hours we want, then run the original query, and assign the count to the appropriate matching array element. This answer is a demonstration of how we can get MySQL to return the specified result.)
As one example of how we can do that, here's a query that returns 24 rows of integer values 0 thru 23.
SELECT 00 AS hr UNION ALL SELECT 01 UNION ALL SELECT 02 UNION ALL SELECT 03
UNION ALL SELECT 04 UNION ALL SELECT 05 UNION ALL SELECT 06 UNION ALL SELECT 07
UNION ALL SELECT 08 UNION ALL SELECT 09 UNION ALL SELECT 10 UNION ALL SELECT 11
UNION ALL SELECT 12 UNION ALL SELECT 13 UNION ALL SELECT 14 UNION ALL SELECT 15
UNION ALL SELECT 16 UNION ALL SELECT 17 UNION ALL SELECT 18 UNION ALL SELECT 19
UNION ALL SELECT 20 UNION ALL SELECT 21 UNION ALL SELECT 22 UNION ALL SELECT 23
We can wrap that query in parens, and reference that in place of a table in another query.
SELECT i.hr
FROM ( -- 0 thru 23
SELECT 00 AS hr UNION ALL SELECT 01 UNION ALL SELECT 02 UNION ALL SELECT 03
UNION ALL SELECT 04 UNION ALL SELECT 05 UNION ALL SELECT 06 UNION ALL SELECT 07
UNION ALL SELECT 08 UNION ALL SELECT 09 UNION ALL SELECT 10 UNION ALL SELECT 11
UNION ALL SELECT 12 UNION ALL SELECT 13 UNION ALL SELECT 14 UNION ALL SELECT 15
UNION ALL SELECT 16 UNION ALL SELECT 17 UNION ALL SELECT 18 UNION ALL SELECT 19
UNION ALL SELECT 20 UNION ALL SELECT 21 UNION ALL SELECT 22 UNION ALL SELECT 23
) i
ORDER BY i.hr
There are other ways to return the set of integers. This example is just an example of an inline view (derived table), that is contained within the query. This doesn't depend on any additional tables in the database.
If we have a table in the database that contains 24 rows with integer values 0 thru 23, we could reference that in place of the inline view.
We can also add a WHERE clause to the outer query, to exclude rows that hvae an hr value is later than the current hour.
And if we want to return a two digit hour (with a leading zero), we can use an expression in SELECT list. There are several expressions that will do the trick.
SELECT RIGHT(CONCAT('0',i.hr),2) AS hh
FROM ( -- 0 thru 23
SELECT 00 AS hr UNION ALL SELECT 01 UNION ALL SELECT 02 UNION ALL SELECT 03
UNION ALL SELECT 04 UNION ALL SELECT 05 UNION ALL SELECT 06 UNION ALL SELECT 07
UNION ALL SELECT 08 UNION ALL SELECT 09 UNION ALL SELECT 10 UNION ALL SELECT 11
UNION ALL SELECT 12 UNION ALL SELECT 13 UNION ALL SELECT 14 UNION ALL SELECT 15
UNION ALL SELECT 16 UNION ALL SELECT 17 UNION ALL SELECT 18 UNION ALL SELECT 19
UNION ALL SELECT 20 UNION ALL SELECT 21 UNION ALL SELECT 22 UNION ALL SELECT 23
) i
WHERE i.hr <= HOUR(NOW())
ORDER BY i.hr
We "match" the rows returned from that, with the counts returned from the original query. We can use an outer join operation so we return a row for an hour when there is no matching row from the count query.
To get a "missing" count returned as zero, we perform a conditional test, and replace a NULL value with a 0.
SELECT RIGHT(CONCAT('0',i.hr,2)) AS hh
, IFNULL(t.cnt,0) AS visits
, IFNULL(t.pw,0) AS pw
FROM ( -- 0 thru 23
SELECT 00 AS hr UNION ALL SELECT 01 UNION ALL SELECT 02 UNION ALL SELECT 03
UNION ALL SELECT 04 UNION ALL SELECT 05 UNION ALL SELECT 06 UNION ALL SELECT 07
UNION ALL SELECT 08 UNION ALL SELECT 09 UNION ALL SELECT 10 UNION ALL SELECT 11
UNION ALL SELECT 12 UNION ALL SELECT 13 UNION ALL SELECT 14 UNION ALL SELECT 15
UNION ALL SELECT 16 UNION ALL SELECT 17 UNION ALL SELECT 18 UNION ALL SELECT 19
UNION ALL SELECT 20 UNION ALL SELECT 21 UNION ALL SELECT 22 UNION ALL SELECT 23
) i
LEFT
JOIN ( -- counts by hour
SELECT HOUR(FROM_UNIXTIME(s.time)) AS hr
, COUNT(*) AS cnt
, MAX(s.pw) AS pw
FROM clientstats s
WHERE s.siteid = '$siteid'
AND s.time >= UNIX_TIMESTAMP(DATE(NOW()))
AND s.time < UNIX_TIMESTAMP(NOW())
GROUP BY HOUR(FROM_UNIXTIME(s.time))
) t
ON t.hr = i.hr
WHERE i.hr <= HOUR(NOW())
ORDER BY i.hr ASC
There are several different queries that will return the specified result. But the trick... to get zero counts returned, is to have a row source that returns the hour values we want returned, even when those hour values don't exist in the table we're getting a count from.