Question

我正在尝试编写一个脚本，该脚本将采用文本字符串并允许我替换随机单词。例如：

$str = "The quick brown fox jumps over the lazy dog";

我会用 as 代替这样的几个词：

<块引用>

敏捷的______狐狸跳过____狗

我可能可以通过首先将字符串拆分为数组来做到这一点

$arr = str_word_count($str, 1);

然后替换 $arr[2] 和 $arr[7]。

如果字符串中有非单词（例如标点符号），我认为我会遇到的问题：

$str = "The quick brown fox, named Jack, jumps over the lazy dog; and Bingo was his...";

我该如何解决这个问题？想法？

Answer 1

你可以这样做：

   $test1 = "test1";
    $test2 = "test2";
    $test3 = "Bingo2";
    // set new words


    $str = "The quick brown fox, named Jack, jumps over the lazy dog; and Bingo was his...";
    $re = explode(" ", $str);
    // split them on space in array $re
    echo $str  . "<br>";
    $num = 0;

    foreach ($re as $key => $value) {
        echo $value . "<br>";
        $word = "";

        switch (true) {
            case (strpos($value, 'Jack') !== false):
                // cheak if each value in array has in it wanted word to replace 
                // and if it does
                $new = explode("Jack", $value);
                // split at that word just to save punctuation
                $word = $test1 . $new[1];
                //replace the word and add back punctuation
                break;
            case (strpos($value, 'dog') !== false):
                $new1 = explode("dog", $value);
                $word = $test2 . $new1[1];
                break;
            case (strpos($value, 'Bingo') !== false):
                $new2 = explode("Bingo", $value);
                $word = $test3 . $new2[1];
                break;
            default:
                $word = $value;
                // if no word are found to replace just leave it
        }

        $re[$num++] = $word;
        //push new words in order back into array
    };


    echo  implode(" ", $re);
        // join back with space

结果：

The quick brown fox, named test1, jumps over the lazy test2; and Bingo2 was his...

它可以使用或不使用标点符号。

但请记住，如果您有 Jack 和 Jacky，例如，您将需要添加额外的逻辑，例如检查标点部分是否没有任何带有 Regex to match only letters 的字母，如果确实跳过它，则意味着它不是完全匹配。或者舒缓类似的。

编辑（基于评论）：

$wordstoraplce = ["Jacky","Jack", "dog", "Bingo","dontreplace"];
$replacewith = "_";
$word = "";
$str = "The quick brown fox, named Jack, jumps over the lazy dog; and Bingo was his...";
echo $str . "<br>";
foreach ($wordstoraplce as $key1 => $value1) {
    $re = explode(" ", $str);
    foreach ($re as $key => $value) {
        if((strpos($value, $value1) !== false)){
            $countn=strlen($value1);
            $new = explode($value1, $value);
            if (!ctype_alpha ($new[1])){
                $word = " " . str_repeat($replacewith,$countn) . $new[1]. " ";
            }else{
                $word = $value;
            }
        }else{
            $word = $value;
        };
        //echo  $word;  
        $re[$key] = $word;      
    };
    $str =  implode(" ", $re);
};
echo $str;

结果：

The quick brown fox, named Jack, jumps over the lazy dog; and Bingo was his...
The quick brown fox, named ____, jumps over the lazy ___; and _____ was his...

Answer 2

我认为更好的方法是使用正则表达式，因为您不仅允许使用逗号，还允许使用不是单词字符的所有内容。此外，正则表达式比循环中的正常拆分或子字符串快得多。我的解决方案是：

<?php
function randomlyRemovedWords($str)
{
    $sentenceParts = [];

    $wordCount = preg_match_all("/([\w']+)([^\w']*)/", $str, $sentenceParts, PREG_SET_ORDER);

    for ($i = 0;$i < $wordCount / 4;$i++)
    { //nearly every fourth word will be changed
        $index = rand(0, $wordCount - 1);

        $sentenceParts[$index][1] = preg_replace("/./", "_", $sentenceParts[$index][1]);
    }

    $str = "";
    foreach ($sentenceParts as $part)
    {
        $str .= $part[1] . $part[2];
    }

    return $str;
}

echo randomlyRemovedWords("The quick brown fox, doesn't jumps over, the lazy dog.");
echo "\n<br>\n";
echo randomlyRemovedWords("The quick brown fox, jumps over, the lazy dog.");

结果

The quick brown ___, _______ jumps over, the ____ dog.
<br>
The quick brown fox, jumps ____, ___ ____ dog.

这样您就可以确保忽略所有非单词字符并随机删除单词。

随机替换字符串中的单词

2 个答案: