Php 使用嵌套数组创建唯一键和计数值

在PHP中，我想构造一个数组，每行使用不同的$raw['day']，并计算所有大于0的$raw['delay']：

开始

结果

array(
  [0] => ("day" => "2013-01-01", "misses" => 2),
  [1] => ("day" => "2013-01-02", "misses" => 1),
  [2] => ("day" => "2013-01-02", "misses" => 1),
)

我所尝试的：

我想首先我会得到所有独特的日子：

$all_days = array_column($raw, 'day');
$unique_days = array_unique($all_days);

现在我已经有了所有独特的日子，我可以再次循环$raw并增加未命中数。但这似乎过于冗长，我认为在PHP的数组函数中有一个更简单的解决方案

foreach ($raw as $row) {
  if ($row['delay'] > 0) {
    $unique_days[$row['day']]++
  }
}

---

您可以使用array_walk函数，但我想您唯一需要的是：

    $unique_days = array();
    foreach ($raw as $row) {
if (!isset($unique_days[$row['day']])) $unique_days[$row['day']] = 0; -- place 1
      if ($row['delay'] > 0) {
if (!isset($unique_days[$row['day']])) $unique_days[$row['day']] = 0; -- place 2
        $unique_days[$row['day']]++
      }
    }

它将只存储具有适当计数器的唯一天数

---

更新检查位置1和2-如果您需要最终阵列中的所有天数，则需要第一个，如果您只需要唯一的天数并且不需要警告，则需要第二个。这是一种可能的方法：

$sum=array();
foreach ($raw as $r) {
    if ($r["delay"]>0) $sum[$r["day"]]++;
}

然后

foreach ($a as $k => $v) {
 $sum[]=array("day"=>$k, "misses" => $v);
}
print_r($sum);

测验

---

你只是复制粘贴了有问题的代码，没有添加任何新的内容。@AmalMurali你错了，我添加了一行，并解释了其他电话不需要如此可爱，让人们在没有任何反馈的情况下投票。谢谢你们，对不起，不是我。。。向上投票：P我会看看这个谢谢你的发帖！

foreach ($a as $k => $v) {
 $sum[]=array("day"=>$k, "misses" => $v);
}
print_r($sum);

$raw = array(
  0 => array("day" => "2013-01-01", "delay" => 3),
  1 => array("day" => "2013-01-01", "delay" => 16),
  2 => array("day" => "2013-01-01", "delay" => 0),
  3 => array("day" => "2013-01-02", "delay" => 1),
  4 => array("day" => "2013-01-02", "delay" => 0),
  5 => array("day" => "2013-01-03", "delay" => 9),
);
$a=array();
foreach ($raw as $r) {
    if ($r["delay"]>0) {$a[$r["day"]]++;}
}

foreach ($a as $k => $v) {
print "$k = $v<br>";
}

foreach ($a as $k => $v) {
 $sum[]=array("day"=>$k, "misses" => $v);
}
print_r($sum);

Array (
 [0] => Array ( [day] => 2013-01-01 [misses] => 2 )
 [1] => Array ( [day] => 2013-01-02 [misses] => 1 )
 [2] => Array ( [day] => 2013-01-03 [misses] => 1 ) )