Php 循环内搜索数组

假设我有以下数组

$diff

a, a, a, a, b, b, b, a, a, b, b, b, a, a, a, b

A represents a value inside $diff.
B represents an Array inside $diff.

现在，如果一个

在其序列中出现两次以上，并且
不是一个数组，我必须对其进行计数（
而是一个值）。否则，不要理会它

对于上述输入，代码的功能如下

[a] = not an array; 0
[a,a] = not an array; 0
[a,a,a] = not an array; 3
[a,a,a,a] = not an array; 4
[b] = array;
[b,b] = array;
[b,b,b] = array;
[a] = not an array; 0
[a,a] = not an array; 0
[b] = array;
[b,b] = array;
[b,b,b] = array;
[a] = not an array; 0
[a,a] = not an array; 0
[a,a,a] = not an array; 3
[b] = array;

这是我的尝试，但没有成功！，由于值被替换，因此值被更改

<?php

foreach($diff as $key => $val)  {

    if (!is_array($diff[$key])) { // THIS MEANS THAT THE CURRENT ELEMENT IS NOT AN ARRAY. 
       if(is_array($diff[$key-1]) ) {   //START OF SEQ. IF THE PREVIOUS ELEMENT IS AN ARRAY AND CURRENT ELEMENT IS NOT AN ARRAY.

        $SEQ_START=$key;
        $n=1;

            for($i=0; $i<=count($diff); $i+=1) { // I AM CHECKING HERE IF THE NEXT 3 ELEMENTS are NOT ARRAY, HENCE I CAN INCREMENT IT

            if(!is_array($diff[$SEQ_START+$i])) $n+=1;
            else $n=0;
            }
        }
    }
}

?>

您只需要一个计数器来计算连续的非数组值。使用每个数组值增加它，并使用每个非数组值重置它：

$seqLength = 0;
foreach ($arr as $index => $value) {
    if (is_array($value)) {
        $seqLength++;
        echo 'array';
    } else {
        $seqLength = 0;
        echo 'not an array';
    }
    if ($seqLength > 2) {
        echo '; '.$seqLength;
    } else {
        echo '; 0';
    }
}

也许：

根据下面@Grexis的评论更新

$diff = array(array(),'a', 'a', 'a', 'a', array(), array(), array(), 'a', 'a', array(), array(), array(), 'a', 'a', 'a', array());

// Counter to hold current sequence total
$count = 0;
// Array to hold results
$counted = array();

// Loop array
foreach ($diff as $key => $val) {
  if (is_array($val)) { // If it is an array
    if ($count > 2) { // If the counter is more than 2
      $counted[(isset($seq)) ? $seq + 1 : 0] = $count; // add it to the array
    }
    // Reset the counter
    $count = 0;
    $seq = $key;
  } else {
    // Increment the counter
    $count++;
  }
} 
// If there is a >2 counter at the end, add one more result
if ($count > 2) {
  $counted[(isset($seq)) ? $seq + 1 : 0] = $count;
}

print_r($counted);
// Outputs:
// Array
// (
//     [0] => 4
//     [12] => 3
// )

// or if you want the total count
$total = array_sum($counted);
echo $total; // 7

我理解计数的逻辑，但您想要的最终结果是什么？连续段的数量（>2）或连续段中的元素数量（>2）？（e.x.如果是前者，答案将是2，如果是后者，答案将是7）在上面的示例中，我需要答案是7。这是不必要的，但是如果在foreach中包含索引，则可以将序列开始的索引存储为以$counted为单位的索引。类似于
$counted[$index-$count]=$count和
$counted[计数（$diff）-$count]=$count第21行不会说谎，你的实现肯定更整洁。。。炫耀：p
0
0
3
4
5

0
0
3
4
5

$diff = array(array(),'a', 'a', 'a', 'a', array(), array(), array(), 'a', 'a', array(), array(), array(), 'a', 'a', 'a', array());

// Counter to hold current sequence total
$count = 0;
// Array to hold results
$counted = array();

// Loop array
foreach ($diff as $key => $val) {
  if (is_array($val)) { // If it is an array
    if ($count > 2) { // If the counter is more than 2
      $counted[(isset($seq)) ? $seq + 1 : 0] = $count; // add it to the array
    }
    // Reset the counter
    $count = 0;
    $seq = $key;
  } else {
    // Increment the counter
    $count++;
  }
} 
// If there is a >2 counter at the end, add one more result
if ($count > 2) {
  $counted[(isset($seq)) ? $seq + 1 : 0] = $count;
}

print_r($counted);
// Outputs:
// Array
// (
//     [0] => 4
//     [12] => 3
// )

// or if you want the total count
$total = array_sum($counted);
echo $total; // 7