24 hours of values - php

Question

I have a sql table : date (Y-m-d) / time (00:00:00) / power (INT)
When I select a date from an inline datepicker, I am trying to post 3 HighCharts graph (one-24 hours, two-31 days of month, three-12 months of year) and I need to get the values out of the table for the chart to be created.
For the day, I need the 24 values for each hour '100,200,300,200,300 etc..'
Here is the PHP for the "day" but it is not working...
<?php
$choice = (isset($_POST['choice']))
? date("Y-m-d",strtotime($_POST['choice']))
: date("Y-m-d");
$con = mysql_connect("localhost","root","xxxxxx");
if (!$con) {
die('Could not connect: ' . mysql_error());
}
mysql_select_db("inverters", $con);
$sql = "SELECT HOUR(time), COUNT(power)
FROM feed
WHERE time = DATE_SUB('".$choice."', INTERVAL 24 HOUR)
GROUP BY HOUR(time)
ORDER BY HOUR(time)";
$res = mysql_query($sql) or die('sql='.$sql."\n".mysql_error());
$row = mysql_fetch_assoc($res);
echo $row['choice'].'<br />';
?>
This has been confirmed by another individual that the code does not work, would anyone have a helpful solution to fix the error ?
Alan

Answer 1

At the moment, your SELECT gives you only the results which happened exactly 24 hours before the current moment. What you need is a range. Example for 1 hour (indentation added for clarity):
WHERE `time` BETWEEN
DATE_SUB('".$choice."', INTERVAL 24 HOUR)
AND DATE_SUB('".$choice."', INTERVAL 23 HOUR)
This way, you'll get results with time in the 1-hour range of "now - 24 hours" and "now - 23 hours". The BETWEEN operator is equivalent to this:
WHERE `time` >= DATE_SUB('".$choice."', INTERVAL 24 HOUR)
AND `time` <= DATE_SUB('".$choice."', INTERVAL 23 HOUR)

Answer 2

You need a loop to walk over the rows:
$sql = "
SELECT HOUR(time) as h, power
FROM feed
WHERE date = '".$choice."'
ORDER BY HOUR(time)";
$res = mysql_query($sql) or die('sql='.$sql."\n".mysql_error());
while($row = mysql_fetch_assoc($res)) {
echo $row['h'].":".$row['power']'<br />';
}
This will give you the power per day for a given month:
$sql = "
SELECT DAY(date) as d, SUM(power) as powerday
FROM feed
WHERE MONTH(date) = '".$month."'
GROUP BY DAY(date)
ORDER BY DAY(date)";
$res = mysql_query($sql) or die('sql='.$sql."\n".mysql_error());
while($row = mysql_fetch_assoc($res)) {
echo $row['d'].":".$row['powerday']'<br />';
}

Answer 3

Thank you everyone for all your help !
The problem was in the first string, I only had to change the date format in addition to your wonderful examles !
$choice = (isset($_POST['choice'])) ? date("m",strtotime($_POST['choice'])) : date("m");
Thank You Very Much !
Alan