Reputation: 1
I have a HTML form which takes the data from the user as open_date and close_date. When the form is submitted, the data will be loaded into a MYSQL database.
What i want is: I have a column in my database which stores difference between these two timestamps in hours.
Could you please help me in getting the solution?
Upvotes: 0
Views: 411
Reputation: 35572
function dateDiff($time1, $time2, $precision = 6) {
// If not numeric then convert texts to unix timestamps
if (!is_int($time1)) {
$time1 = strtotime($time1);
}
if (!is_int($time2)) {
$time2 = strtotime($time2);
}
// If time1 is bigger than time2
// Then swap time1 and time2
if ($time1 > $time2) {
$ttime = $time1;
$time1 = $time2;
$time2 = $ttime;
}
// Set up intervals and diffs arrays
$intervals = array('year','month','day','hour','minute','second');
$diffs = array();
// Loop thru all intervals
foreach ($intervals as $interval) {
// Create temp time from time1 and interval
$ttime = strtotime('+1 ' . $interval, $time1);
// Set initial values
$add = 1;
$looped = 0;
// Loop until temp time is smaller than time2
while ($time2 >= $ttime) {
// Create new temp time from time1 and interval
$add++;
$ttime = strtotime("+" . $add . " " . $interval, $time1);
$looped++;
}
$time1 = strtotime("+" . $looped . " " . $interval, $time1);
$diffs[$interval] = $looped;
}
$count = 0;
$times = array();
// Loop thru all diffs
foreach ($diffs as $interval => $value) {
// Break if we have needed precission
if ($count >= $precision) {
break;
}
// Add value and interval
// if value is bigger than 0
if ($value > 0) {
// Add s if value is not 1
if ($value != 1) {
$interval .= "s";
}
// Add value and interval to times array
$times[] = $value . " " . $interval;
$count++;
}
}
// Return string with times
return implode(", ", $times);
}
echo dateDiff("2010-01-26", "2004-01-26") . "\n";
echo dateDiff("2006-04-12 12:30:00", "1987-04-12 12:30:01") . "\n";
echo dateDiff("now", "now +2 months") . "\n";
echo dateDiff("now", "now -6 year -2 months -10 days") . "\n";
echo dateDiff("2009-01-26", "2004-01-26 15:38:11") . "\n";
the link tell you all about date difference
Upvotes: 2
Reputation: 10074
You can:
$timeDiff = strtotime('2012-07-08 13:43:22') - strtotime('2012-07-07 12:13:12');
echo round($timeDiff / 60 / 60);
Upvotes: 0