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Home/ Questions/Q 3497618
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Editorial Team
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Editorial Team
Asked: May 18, 20262026-05-18T12:23:33+00:00 2026-05-18T12:23:33+00:00

I have a sql table : date (Y-m-d) / time (00:00:00) / power (INT)

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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

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  1. Editorial Team
    Editorial Team
    2026-05-18T12:23:33+00:00Added an answer on May 18, 2026 at 12:23 pm

    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

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