MySQL select rows from exactly 7 days ago

Tim Blackburn picture Tim Blackburn · Oct 13, 2011 · Viewed 56.4k times · Source

I'm completely stumped on this one, being trying for hours but with no success, hoping someone can help. Trying to build a cron script to run daily that returns the rows that are exactly 7 days older than the current date.

The thing is, my query is not returning anything. No error messges, nothing (I know there are entries in the DB from the last 7 days - we get about 7000 new entries a day, so they are there!) I've tried a SELECT * and echo out the edit date with success, so everything is working, apart from my SQL script.

The column I'm referencing (edit_date) is type 'datetime' formated with Y-m-d h-m-s. This column always has a datetime value assigned on both create and edit.

function get_ad_sql($table){
    $sql = "SELECT 
                * 
            FROM 
                ".$table." 
            WHERE 
                edit_date = DATE_SUB(NOW(), INTERVAL 7 DAY)
            ";  
    return $sql;
}

And calling the function and 'trying' to echo the primary_key:

$sqlAng = get_ad_sql('angebote');
$result = mysql_query($sqlAng);
while($row = mysql_fetch_array($result)){
    echo $row['primary_key'];
}

I've tried every variation of DATE_SUB(NOW(), INTERVAL 7 DAY), including CURDATE(), DATE_FORMAT(edit_date, '%m/%d/%Y') that I could find on here and online, but couldn't get anything to work. Hope someone can help me!

Answer

Ghazanfar Mir picture Ghazanfar Mir · Oct 13, 2011

It is very rare to get same datetime entries which gives date and time upto seconds. Therefore, for getting appropriate results we need to ignore the time part, and deal with date part, thus, using CURDATE() function.

You could do that ignoring the time part and compare with the date using following:

function get_ad_sql($table){
    $sql = "SELECT 
                * 
            FROM 
                ".$table." 
            WHERE 
                DATE(edit_date) = DATE_SUB(CURDATE(), INTERVAL 7 DAY)
            ";  
    return $sql;
}