在数组中计数，其中值为0

I'm really stuck here. Im having an array looking like this below. And now I would like to count postStatus where postStatus = 0, for all of the arrays.

我真的被困在这里了。我有一个如下所示的阵列。现在我想计算postStatus，其中postStatus = 0，适用于所有数组。

So in this case there would be 2. But how do I do this?

所以在这种情况下会有2.但我该怎么做？

Array
(
[1] => Array
    (
        [postId] => 1
        [postHeader] => Post-besked #1
        [postContent] => Post content #1 
        [postDate] => 2011-12-27 17:33:11
        [postStatus] => 0
    )

[2] => Array
    (
        [postId] => 2
        [postHeader] => Post-besked #2 
        [postContent] => POst content #2
        [postDate] => 2011-12-27 17:33:36
        [postStatus] => 0
    )
)

3 个解决方案

#1

Just loop the outer array, check if there is a postStatus, increment a value to keep that count and you're done...

只需循环外部数组，检查是否有postStatus，增加一个值以保持计数，你就完成了......

$postStatus = 0;
foreach($myarray as $myarraycontent){
    if(isset($myarraycontent['postStatus']) && $myarraycontent['postStatus'] == 0){
        $postStatus++;
    }
}
echo $postStatus;

EDIT:

编辑：

I forgot to mention that isset() can be used but a better pratice is to use array_key_exists because if $myarraycontent['postStatus'] is NULL, it will return false. Thats the way isset() works...

我忘了提到可以使用isset（）但是更好的实践是使用array_key_exists，因为如果$ myarraycontent ['postStatus']为NULL，它将返回false。这就是isset（）的工作原理......

#2

$count = count(
  array_filter(
    $array, 
    function ($item) {
        return isset($item['postStatus']);
    }
  )
);

#3

How about this? Compact and concise :)

这个怎么样？小巧简洁:)

$postStatusCount = array_sum(array_map(
    function($e) { 
            return array_key_exists('postStatus', $e)  && $e['postStatus'] == 0 ? 1 :  0; 
    } , $arr)
);

#1