Question

I am trying to save the time in the database for when a user add an entry. Every time I run the time() function it prints (or returns) 1277155717 which represents 1969.

I was wondering if there is a way to save the time to the database in a way that it represents the actual date today at this moment.

I am using the function

/* Works out the time since the entry post, takes a an argument in unix time (seconds) */
function time_since($original) {
    // array of time period chunks
    $chunks = array(
        array(60 * 60 * 24 * 365 , 'year'),
        array(60 * 60 * 24 * 30 , 'month'),
        array(60 * 60 * 24 * 7, 'week'),
        array(60 * 60 * 24 , 'day'),
        array(60 * 60 , 'hour'),
        array(60 , 'minute'),
    );

    $today = time(); /* Current unix time  */
    $since = $today - $original;

    // $j saves performing the count function each time around the loop
    for ($i = 0, $j = count($chunks); $i < $j; $i++) {

        $seconds = $chunks[$i][0];
        $name = $chunks[$i][1];

        // finding the biggest chunk (if the chunk fits, break)
        if (($count = floor($since / $seconds)) != 0) {
            // DEBUG print "<!-- It's $name -->\n";
            break;
        }
    }

    $print = ($count == 1) ? '1 '.$name : "$count {$name}s";

    if ($i + 1 < $j) {
        // now getting the second item
        $seconds2 = $chunks[$i + 1][0];
        $name2 = $chunks[$i + 1][1];

        // add second item if it's greater than 0
        if (($count2 = floor(($since - ($seconds * $count)) / $seconds2)) != 0) {
            $print .= ($count2 == 1) ? ', 1 '.$name2 : ", $count2 {$name2}s";
        }
    }
    return $print;
}

In order to display the number of minutes, years, months, etc since the comment was posted and it is returning (40 years, 6 months ago) when I pass the value of the function time();

Answer 1

Why won’t you just use sql’s timestamp type, i.e.

INSERT INTO posts (content, created) VALUES ("Sample post", NOW());