Make an if-statement understand which "time-left" format is bigger - php

Question

I am trying to make an IF-statement that can calculate which "time-left" is bigger.
This is my code.
if('6h : 27m : 45s' > '6h : 27m : 15s') {
echo 'Timer to the left is bigger';
}
This code works fine, it actually prints Timer to the left is bigger.
However, when any time value (hour, minute or seconds) is only one digit instead of two. It doesn't work. Like this:
if('6h : 27m : 45s' > '6h : 27m : 5s') {
echo 'Timer to the left is bigger';
}
This time it does not print out Timer to the left is bigger. It is because the timer to the right is only 5s left, which is less than 15s, but it is because it is only one digit instead of two, and the IF-statement doesn't understand that.
How can I parse this to an understandable format for the if-statement?
NOTE the string is from an API, I can't edit it in anyway.

Answer 1

php function version_compare() is like strcmp(), but compares numbers as single unit. It has some side effects because of (virtually) converting chars, but I think it does not matter here:
So
if('6h : 27m : 45s' > '6h : 27m : 15s')
becomes
if( version_compare('6h : 27m : 45s','6h : 27m : 15s') > 0 )

Answer 2

Your post suggests that all three time units are always available. So just create two arrays of numbers how ever you wish. PHP7's spaceship operator will make numeric comparisons when handling two numeric values. It will compare the elements from left to right as you require.
No time parsing is necessary because the time units are ordered from greater to lesser.
Code: (Demo)
$left = '6h : 27m : 5s';
$right = '6h : 27m : 15s';
$lookup = [
-1 => 'less',
0 => 'equal',
1 => 'more'
];
$lArray = preg_split('/\D+/', $left, 0, PREG_SPLIT_NO_EMPTY);
$rArray = preg_split('/\D+/', $right, 0, PREG_SPLIT_NO_EMPTY);
$comparison = $lArray <=> $rArray;
echo $comparison . ' -> ' . $lookup[$comparison];
Output:
-1 -> less
You make write a conditional check for -1 if that is all that you require. The lookup was purely to clarify the demonstration.
In other words:
if ((preg_split('/\D+/', $left, 0, PREG_SPLIT_NO_EMPTY) <=> preg_split('/\D+/', $right, 0, PREG_SPLIT_NO_EMPTY)) == -1) {
// left is less than right
}
Or use a non-regex alternative that will cast each matched number as an integer then compare the two generated arrays containing three integers each. (Demo)
$format = '%dh : %dm %ds';
$comparison = sscanf($left, $format) <=> sscanf($right, $format);
I just noticed Wiimm's outside-the-box call of version_compare() to parse your time expressions. (+1 from me, but I think I need to explain more about it) While not its intended use, it is a clever and reliable non-regex technique because your time unit substrings (h, m, s) are not found in the list of development-stage substrings which hold special meaning uniformly positioned in both strings. This means that even if you had non-numeric substrings with special meaning, the comparison would still be evaluated correctly based solely on the numeric values.
Special substrings in order:
dev
alpha or a
beta or b
RC or rc
#
pl or p
This function is a three-way-comparing tool - like the spaceship operator. It also accepts a third parameter to command that it returns a boolean response. Using gt will return true if the first string is "greater than" the second string. While the third parameter is an abbreviation, I find it more intuitive than checking for -1.
Code: (Demo)
if (version_compare($left, $right, 'gt')) {
echo 'left is greater than right';
} else {
echo 'left is less than or equal to right';
}
Output:
left is less than or equal to right

Answer 3

It's not the most prettiest way of doing it, but since you get the data returned from an API you can't modify, you'll have to work with the data you got.
There are several ways of doing it. One way is to convert the time into seconds, and compare that. Another, is to convert them into objects - and work with that. The "gruntwork" is the same; you have to split up the data into hours, minutes and seconds, and use that - which is more usable than the format Xh : Ym : Zs.
Here we split up each time into pieces of three, by the delimiter :. This gets us the individual parts - but we still need to fetch the integer value, which is what we're interested in. We can grab that by using inval(), as the number comes first. Once we have the three pieces, we can either put them into a HH:mm:ss format, or you can multiply them up to everything being in seconds, and add them together (first is hours, second is minutes, third is seconds). Here we're not doing that, we glue them together in a readable format and use strtotime() to convert them into a timestamp - which we finally, after all this work, can compare!
$time_left = '6h : 27m : 45s';
$time_right = '6h : 27m : 5s';
$parts_left = explode(":", $time_left);
$parts_right = explode(":", $time_right);
$left = intval($parts_left[0]).":".intval($parts_left[1]).":".intval($parts_left[2]);
$right = intval($parts_right[0]).":".intval($parts_right[1]).":".intval($parts_right[2]);
if (strtotime($left) > strtotime($right)) {
echo 'Timer to the left is bigger';
}
Live demo at https://3v4l.org/uUQVh
According to your edited comment, the hours can be greater than 24 - in which case you need to convert the hours and minutes to seconds, and sum that up.
$time_left = '6h : 27m : 45s';
$time_right = '6h : 27m : 5s';
$parts_left = explode(":", $time_left);
$parts_right = explode(":", $time_right);
$left = intval($parts_left[0]) * 60*60 + intval($parts_left[1])*60 + intval($parts_left[2]);
$right = intval($parts_right[0]) * 60*60 + intval($parts_right[1])*60 + intval($parts_right[2]);
if ($left > $right) {
echo "Timer to the left is bigger";
}
Live demo at https://3v4l.org/DaINX