我需要将'in'和'ou', [a]到[t]相加数组。
Array
(
[1] => Array
(
[in] => Array
(
[a] => 3
[b] => 0
[c] => 0
[d] => 0
[e] => 0
[f] => 0
[o] => 0
[t] => 3
)
[ou] => Array
(
[a] => 0
[b] => 0
[c] => 1
[d] => 0
[e] => 0
[f] => 0
[o] => 0
[t] => 1
)
)
[2] => Array
(
[in] => Array
(
[a] => 0
[b] => 0
[c] => 0
[d] => 0
[e] => 0
[f] => 0
[o] => 0
[t] => 0
)
[ou] => Array
(
[a] => 0
[b] => 0
[c] => 0
[d] => 1
[e] => 2
[f] => 0
[o] => 0
[t] => 3
)
)
)
下面是我如何计算合计'in'+'ou'。然而,当涉及到"in"a,b,c,d,e,f,t和"ou"a,b,c,d,e,f,t的个人总数时,我似乎陷入了窠臼。
//get day total
foreach($arr as $array){
foreach($array as $inou){
foreach(array_keys($inou) as $value){
if(isset($total[$value])){
$total[$value] += $inou[$value];
}else{
$total[$value] = $inou[$value];
}
}
}
}
输出应该类似于
in(
[a] => 3
[b] => 0
[c] => 0
...
[t] => 3
)
ou(
[a] => 0
[b] => 0
[c] => 1
[d] => 1
[e] => 2
[f] => 0
[t] => 4
)
这应该可以让你开始:
$sumIN = 0;
$sumOU = 0;
foreach($arr as $innerArr)
{
$sumIN += array_sum($innerArr['in']);
$sumOU += array_sum($innerArr['ou']);
}