How to store data for a field with a storage time

How to store data in a time field? there is a field that requires storage time, and all the values of this field need to be added, and then converted to the correct time. For example

5.45 minutes and 45 seconds for storage
2,120 minutes 01 seconds storage is 120.01
89.19 for minutes and 19 seconds

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The sum is 5.45 + 120.01 + 89.19 = 214.65, but it is different from the original value.

Is there any good way to solve this problem? I don't think it can be stored like this. for example, it is very reliable to convert all the time input by users into seconds and store them in the database, simply add the number of seconds. No errors.

But how can we convert the value entered by the user into seconds?

Reply to discussion (solution)

$ Input = 1234.5678; strtok (string) $ input ,'. '); $ s = strtok (string) $ input ,'. ') * 60 + strtok ('. '); echo $ s. 'second ';

$ Input = 1234.5678; $ s = strtok (string) $ input, '.') * 60 + strtok ('.'); echo $ s. 'second ';

$ Input = 1234.5678; $ s = strtok (string) $ input, '.') * 60 + strtok ('.'); echo $ s. 'second ';

Well, I see, what if there is an hour? for example, it's 3H45. 26, that is, 3 hours, 45 minutes, and 26 seconds.
But sometimes there are only minutes without hours. in this case, you cannot convert the hour into minutes by yourself. do you need to enter 205.26 for three hours and 45 minutes ~

I just don't know how to do it, and I have no idea.

The author studies the strtotime () function to see if it can be implemented.

5.45 + 120.01 + 89.19 = 214.65
60? 10? System Plus ,? But no ?.

? All first? Seconds ?, And then add.

$t = array('3H45.26','1H5.10','30.59');echo sumT($t); // 5H21.35function sumT($t){$sum = 0;if($t){foreach($t as $v){$sum += getSecond($v);}}return tostr($sum);}function getSecond($str){if(strstr($str,'H')==''){$str = '0H'.$str;}$str = str_replace('H','.',$str);list($h, $m, $s) = explode('.', $str);return $h*3600+$m*60+$s;}function tostr($t){$h = (int)($t/3600);$m = (int)($t%3600/60);$s = $t%3600%60;$h = $h>0? $h.'H' : '';return $h.$m.'.'.$s;}

Ask the user to write 034526 in 3 hours, 45 minutes, 26 seconds. isn't that too much? Easier than yours
Therefore

echo strtotime('034526') - strtotime(date('Y-m-d')), 

13526