PHP检查字符串是否包含一定百分比的字母数字字符

我有一个php字符串$comment有时$comment框会包含一些非字母数字字符，有没有办法找出$comment中字母数字的百分比

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谢谢

你可以试试这样的方法：（不是特别有效）

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但是Regex可能是一种更简单的方法，更简单

我的第一个想法是使用Regex剥离非字母字符，然后将字符串长度与原始字符串进行比较。数一数。除以字符总数。乘以100。我很好奇：在这个简单的过程中，哪两个步骤你自己做起来有困难？

$comment_alpha = preg_replace('/[^a-z\d]+/i', '', $comment);
$alpha_percent = 100 * strlen($comment_alpha) / strlen($comment);

<?php
$string = "TestItOut##@22383";



$all = array(
"0","1","2","3","4","5","6","7","8","9",          
"a","b","c","d","e","f","g","h","i","j","k","l","m","n","o","p","q","r","s","t","u","v","w","x","y","z",     
"A","B","C","D","E","F","G","H","I","J","K","L","M","N","O","P","Q","R","S","T","U","V","W","X","Y","Z"
);

$num_alphanum = 0;
foreach ( $all as $char ) {
    $num_alphanum += substr_count( $string, $char );
}

$percent = ( $num_alphanum / strlen( $string ) ) * 100;

echo $percent . "%";

?>