比较两个字符串，突出重复的单词在PHP

Question

见图片：

我真的想知道什么是比较上重复的单词两个字符串（长文本文件）的最佳方法，那么我需要在第二个字符串中突出显示它们。就像copyscape一样。它用于我们内部的内部数据库。

我是否缺少一个简单的PHP函数？任何人都可以指向正确的方向吗？

我所知道的是制作两个数组并将它们与foreach循环进行比较。但它没有任何意义，我的脚本越来越没有突出显示40行

Answer 1

我认为https://github.com/gorhill/PHP-FineDiff可能会完成这项工作。如果需要，它甚至可以将字符级别的各种粒度的文本进行比较。

实际上，你可以找到共同点重复的短语，如果他们在finediff.php加入

static $commons; 

public static function renderCommonsFromOpcodes($from, $opcodes) 
{ 
    FineDiff::renderFromOpcodes($from, $opcodes, array('FineDiff', 'renderCommonsFromOpcode')); 
} 

private static function renderCommonsFromOpcode($opcode, $from, $from_offset, $from_len) 
{ 
    if ($opcode === 'c') { 
     self::$commons[] = substr($from, $from_offset, $from_len); 
    } 
} 

到FineDiff ::类出现以相同的顺序。

用法：

你可以玩的是

include 'finediff.php'; 

$from_text = "PHP FPM is a popular general-purpose scripting language that is especially suited to web development."; 
$to_text = "Fast, flexible and pragmatic, PHP FPM powers everything from your blog to the most popular websites in the world"; 

$opcodes = FineDiff::getDiffOpcodes($from_text, $to_text, FineDiff::wordDelimiters); 

FineDiff::renderCommonsFromOpcodes($from_text, $opcodes); 

print_r(FineDiff::$commons); 

/* 
Array 
(
    [0] => PHP FPM 
    [1] => popular 
) 
*/ 

Answer 2

一种方法使用array_intersect其中两个阵列从要比较，然后用一个字符串替换功能来突出常用词的两个字符串产生。

$str1='PHP is a popular general-purpose scripting language that is especially suited to web development.'; 
$str2='Fast, flexible and pragmatic, PHP powers everything from your blog to the most popular websites in the world.'; 

$a1=explode(' ',$str1); 
$a2=explode(' ',$str2); 

function longenough($word){ 
    return strlen($word) > 3; 
} 

$a1=array_filter($a1,'longenough'); 
$a2=array_filter($a2,'longenough'); 

$common=array_intersect($a1, $a2); 

foreach($common as $word){ 
    $str2=preg_replace("@($word)@i",'<span style="color:red">$1</span>', $str2); 
} 

echo $str2;