Question

我想替换一个字符串中的一个随机字。

所以让我们说字符串是

$str = 'I like blue, blue is my favorite colour because blue is very nice and blue is pretty';

让我们说我想用红色取代蓝色这个词，但在随机位置只取代2次。

因此在完成一个函数后，输出可能就像

I like red, blue is my favorite colour because red is very nice and blue is pretty

另一个可能是

I like blue, red is my favorite colour because blue is very nice and red is pretty

所以我想多次替换同一个单词，但每次都在不同的位置。

我想过使用preg_match，但是没有选择peing替换的单词的位置也是随机的。

有没有人知道如何实现这个目标？

Answer 1

就像我厌恶使用正则表达式来表达它非常简单的东西一样，为了保证完全 n 取代我觉得它可以帮到这里，因为它允许使用array_rand()，它完全符合您的要求 - 从不确定长度列表中选择 n 随机项目（改进）。< / p>

<?php

    function replace_n_occurences ($str, $search, $replace, $n) {

        // Get all occurences of $search and their offsets within the string
        $count = preg_match_all('/\b'.preg_quote($search, '/').'\b/', $str, $matches, PREG_OFFSET_CAPTURE);

        // Get string length information so we can account for replacement strings that are of a different length to the search string
        $searchLen = strlen($search);
        $diff = strlen($replace) - $searchLen;
        $offset = 0;

        // Loop $n random matches and replace them, if $n < 1 || $n > $count, replace all matches
        $toReplace = ($n < 1 || $n > $count) ? array_keys($matches[0]) : (array) array_rand($matches[0], $n);
        foreach ($toReplace as $match) {
            $str = substr($str, 0, $matches[0][$match][1] + $offset).$replace.substr($str, $matches[0][$match][1] + $searchLen + $offset);
            $offset += $diff;
        }

        return $str;

    }

    $str = 'I like blue, blue is my favorite colour because blue is very nice and blue is pretty';

    $search = 'blue';
    $replace = 'red';
    $replaceCount = 2;

    echo replace_n_occurences($str, $search, $replace, $replaceCount);

See it working

Answer 2

echo preg_replace_callback('/blue/', function($match) { return rand(0,100) > 50 ? $match[0] : 'red'; }, $str);

Answer 3

嗯，你可以使用这个算法：

计算要替换字符串的随机次数
将字符串分解为数组
仅当1到100之间的随机值为％3（对于istance）时，该数组的
替换字符串出现
减少第1点计算的数字。
重复直到数字达到0。

Answer 4

<?php
$amount_to_replace = 2;
$word_to_replace = 'blue';
$new_word = 'red';

$str = 'I like blue, blue is my favorite colour because blue is very nice and blue is pretty';

$words = explode(' ', $str); //convert string to array of words
$blue_keys = array_keys($words, $word_to_replace); //get index of all $word_to_replace

if(count($blue_keys) <= $amount_to_replace) { //if there are less to replace, we don't need to randomly choose.  just replace them all
    $keys_to_replace = array_keys($blue_keys);
}
else {
    $keys_to_replace = array();
    while(count($keys_to_replace) < $amount_to_replace) { //while we have more to choose
        $replacement_key = rand(0, count($blue_keys) -1);
        if(in_array($replacement_key, $keys_to_replace)) continue; //we have already chosen to replace this word, don't add it again
        else {
            $keys_to_replace[] = $replacement_key;
        }
    }
}

foreach($keys_to_replace as $replacement_key) {
    $words[$blue_keys[$replacement_key]] = $new_word;
}

$new_str = implode(' ', $words); //convert array of words back into string
echo $new_str."\n";
?>

N.B。我刚刚意识到这不会取代第一个蓝色，因为它以“蓝色”的形式输入到单词数组中，因此在array_keys调用中不匹配。

PHP替换字符串的随机单词

4 个答案: