$array1 = array(0,1,3,0);
$array2 = array(2,0,3,2);
$array3 = array(0,4,5,1);
echo min($array1) . "\n";
echo min($array2) . "\n";
echo min($array3) . "\n";
this return me:
这个返回我:
0
0
0
but i want receive:
但是我希望得到:
1
2
1
How to skip 0 in function MIN?
如何在函数MIN中跳过0 ?
2 个解决方案
#1
13
Try array_filter:
试试array_filter:
If no callback is supplied, all entries of input equal to FALSE will be removed.
如果没有提供回调，则将删除所有等于FALSE的输入项。
so it will remove 0 for you.
它将为你移除0。
echo min(array_filter($array1)) . "\n";
#2
7
In addition to array_filter, you can be more explicit with
除了array_filter之外，您还可以使用更显式的
$min = min(array_diff($array1, array(0)); // result = 1
This can let you disregard more values very easily, e.g.
这很容易让你忽视更多的价值。
$min = min(array_diff($array1, array(0, 1)); // result = 3
#1
13
Try array_filter:
试试array_filter:
If no callback is supplied, all entries of input equal to FALSE will be removed.
如果没有提供回调，则将删除所有等于FALSE的输入项。
so it will remove 0 for you.
它将为你移除0。
echo min(array_filter($array1)) . "\n";
#2
7
In addition to array_filter, you can be more explicit with
除了array_filter之外，您还可以使用更显式的
$min = min(array_diff($array1, array(0)); // result = 1
This can let you disregard more values very easily, e.g.
这很容易让你忽视更多的价值。
$min = min(array_diff($array1, array(0, 1)); // result = 3