JS到PHP的错误

问题描述：

I have some JS code which I want to convert to PHP code. I changed all the things that I know should, but it doesn't give the same result.

The PHP code I created from the JS code:

<?php
function hashCode($str){
    $hash = 0;

    if (strlen($str) == 0) return $hash;

    for ($i = 0; $i < strlen($str); $i++) {
          $char = $str[$i];
          $hash = (($hash<<5) - $hash) + $char;
          $hash = $hash & $hash;
    }
    return $hash;
}
?>

When entering a string it returns 0 and that is not because of the if function.
Original code:

hashCode = function(str){
    var hash = 0;
    if (str.length == 0) return hash;
    for (i = 0; i < str.length; i++) {
        char = str.charCodeAt(i);
        hash = ((hash<<5)-hash)+char;
        hash = hash & hash; // Convert to 32bit integer
    }
    return hash;
}

Can someone please help me?

Arend-Jan

我有一些JS代码，我想将其转换为PHP代码。我改变了所有我应该知道的东西，但它没有给出相同的结果。 p>

PHP代码 strong>我是用JS代码创建的： p>

 ＆lt;？php 
function hashCode（$ str）{
 $ hash = 0; 
 
 if if（strlen（$ str）== 0）return $ hash;  
 
 for（$ i = 0; $ i＆lt; strlen（$ str）; $ i ++）{
 $ char = $ str [$ i]; 
 $ hash =（（$ hash＆lt;＆lt; 5  ） -  $ hash）+ $ char; 
 $ hash = $ hash＆amp;  $ hash; 
} 
返回$ hash; 
} 
？＆gt; 
  code>  pre> 
 
 输入字符串时
it返回0，那不是因为 if函数。
原始代码： strong>  p> 
 
 
  hashCode = function（str）{
 var hash = 0; 
  if（str.length == 0）返回散列; 
 for（i = 0; i＆lt; str.length; i ++）{
 char = str.charCodeAt（i）; 
 hash =（（hash＆lt;＆lt;  ; 5）-hash）+ char; 
 hash = hash＆amp; 散列;  //转换为32位整数
} 
返回哈希; 
} 
  code>  pre> 
 
 有人可以帮助我吗？ p> 
 
 
 Arend-Jan  p> 
  div>

答

This is because, in the javascript, the char is the code of the caracter (charCodeAt). So this will do the same. Use ord($char)

function hashCode($str) {
    $hash = 0;
    if (strlen($str) == 0) {
        return $hash;
    }
    for ($i = 0; $i < strlen($str); $i++) {
        $char = $str[$i];
        $hash = (($hash << 5) - $hash) + ord($char);
        $hash = $hash & $hash;
    }
    return $hash;
}

If string is ABCD, the output will be this, in both case:

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