PHP：每次将数字增加5

Question

I'm trying to reproduce the following as a one-liner.

if($l < 10) $next = 5; return;
if($l < 20) $next = 10; return;
if($l < 30) $next = 15; return;
if($l < 40) $next = 20; return;
if($l < 50) $next = 25; return;
if($l < 60) $next = 30; return;
if($l < 70) $next = 35; return;
if($l < 80) $next = 40; return;
if($l < 90) $next = 45; return;
if($l < 100) $next = 50; return;

(not syntactically correct but you get the idea)

So that if the number is less than 10, $next is 5, and if the number is less than 20 then it's 10.

$next = ((round($l, -1)-5)); is as close as I can get to it but that gives

5
15
25
35
45
55
65
75
85

not the desired 5, 10, 15, 20 .. etc

What is the correct way to write this?

Answer 1

add 10 to your number, then divide the result by 10 and round it down to the nearest (floor) integer, you will then have the number by which to multiply 5, which will yield your result... so... let's say your number is 47.

47 + 10 = 57

57 / 10 = 5.7

floor 5.7 = 5

5x5 = 25

return floor(($i + 10)/10) * 5

Answer 2

(($l + 10) / 10) * 5可以解决问题

Answer 3

除以十（并四舍五入为整数），然后乘以5 ... 100/10 = 10 => 10 * 5 = 50

Answer 4

This might do:

$i=10;
while($l < $i){
    $next = $i / 2;
    $i+=10;
}

Answer 5

如果我正确理解了您的问题，那并不比难。

$next = floor($l/10)*5+5;