如果存在连续的相似字符串，则只打印一个字符串

如果我理解正确，你可以记住最后一个数字：

$lastB = null;
foreach ($foo as $r) //I'm guessing that $r is the dummy variable, based on your code
{
        $b = substr($r['pon_port'],4,2); //You wrote this
        //This part prevents repetitions from happening
        if ($lastB === null || $b != $lastB) var_dump($lastB = $b);
}

（在这种情况下，我把赋值偷偷地放进了争论中，用了var_dump($lastB = $b)而不是两个单独的语句。我觉得它更整洁，但如果你愿意，你可以把它们分开。）

如果你想做一些事情而不是打印变量，作为最终结果，你可以使用以下代码：

$lastB = null;
foreach ($foo as $r) //I'm guessing that $r is the dummy variable, based on your code
{
        $b = substr($r['pon_port'],4,2); //You wrote this
        //This part prevents repetitions from happening
        if ($lastB === null || $b != $lastB)
        {
            //do something with $b...
            $lastB = $b;
        }
}

我自己在序列"2 2 7 8 8 7"上尝试第一个代码，我得到了：

string(1) "2" string(1) "7" string(1) "8" string(1) "7"

我（可能是愚蠢的）基于上下文的假设是它们总是个位数。为了可读性，您可以将其单行：

$nums = array(0,0,1,2,2,3,4,0,1,2,2,1,2,1,2,3,3,4,4,4);
print_r($nums);
$nums = implode('',$nums);
$nums = preg_replace('~('d)((?='1).)+~','$1',$nums);
$nums = str_split($nums);
print_r($nums)

输出：

Array
(
    [0] => 0
    [1] => 0
    [2] => 1
    [3] => 2
    [4] => 2
    [5] => 3
    [6] => 4
    [7] => 0
    [8] => 1
    [9] => 2
    [10] => 2
    [11] => 1
    [12] => 2
    [13] => 1
    [14] => 2
    [15] => 3
    [16] => 3
    [17] => 4
    [18] => 4
    [19] => 4
)
Array
(
    [0] => 0
    [1] => 1
    [2] => 2
    [3] => 3
    [4] => 4
    [5] => 0
    [6] => 1
    [7] => 2
    [8] => 1
    [9] => 2
    [10] => 1
    [11] => 2
    [12] => 3
    [13] => 4
)