Question

我正在尝试编写一个代码来将字符串连接成拉丁文。我已经处理了一些限制，但是我没有获得所需的输出。我的代码如下：

<?php

$string = "impulerittantaenanimis caelestibusirae";

$precedingC = precedingConsonant($string);
$xrule = xRule($precedingC);
$consonantc = consonantCT($xrule);
$consonantp = consonantPT($consonantc);
$cbv = CbetweenVowels($consonantp);
$tv = twoVowels($cbv);

echo $tv;

function twoVowels($string)
{
    return preg_replace('/([aeiou])([aeiou])/', '$1-$2', $string);
}
function CbetweenVowels($string)
{
    return preg_replace('/([aeiou])([^aeiou])([aeiou])/', '$1-$2$3', $string);
}
function consonantPT($string)
{
    return preg_replace('/([^aeiou]p)(t[aeiou])/', '$1-$2', $string);
}
function consonantCT($string)
{
    return preg_replace('/([^aeiou]c)(t[aeiou])/', '$1-$2', $string);
}
function precedingConsonant($string)
{
    $arr1 = str_split($string);
    $length = count($arr1);
    for($j=0;$j<$length;$j++)
    {
        if(isVowel($arr1[$j]) && !isVowel($arr1[$j+1]) && !isVowel($arr1[$j+2]) && isVowel($arr1[$j+3]))
        {
            $pc++;  
        }
    }

    function strAppend2($string)
    {
        $arr1 = str_split($string);
        $length = count($arr1);


        for($i=0;$i<$length;$i++)
        {
            $check = $arr1[$i+1].$arr1[$i+2];
            $check2 = $arr1[$i+1].$arr1[$i+2].$arr1[$i+3];
            if($check=='br' || $check=='cr' || $check=='dr' || $check=='fr' || $check=='gr' || $check=='pr' || $check=='tr' || $check=='bl' || $check=='cl' || $check=='fl' || $check=='gl' || $check=='pl' || $check=='ch' || $check=='ph' || $check=='th' || $check=='qu' || $check2=='phl' || $check2=='phr')
            {
                if(isVowel($arr1[$i]) && !isVowel($arr1[$i+1]) && !isVowel($arr1[$i+2]) && isVowel($arr1[$i+3]))
                {
                    $updatedString = substr_replace($string, "-", $i+1, 0);
                    return $updatedString;
                }
            }
            else
            {
                if(isVowel($arr1[$i]) && !isVowel($arr1[$i+1]) && !isVowel($arr1[$i+2]) && isVowel($arr1[$i+3]))
                {
                    $updatedString = substr_replace($string, "-", $i+2, 0);
                    return $updatedString;
                }
            }
        }
    }
    $st1 = $string;
    for($k=0;$k<$pc;$k++)
    {
        $st1 = strAppend2($st1);
    }

    return $st1;
}
function xRule($string)
{
    return preg_replace('/([aeiou]x)([aeiou])/', '$1-$2', $string);
}
function isVowel($ch)
{
    if($ch=='a' || $ch=='e' || $ch=='i' || $ch=='o' || $ch=='u')
    {
        return true;
    }
    else
    {
        return false;
    }
}
function isConsonant($ch)
{
    if($ch=='a' || $ch=='e' || $ch=='i' || $ch=='o' || $ch=='u')
    {
        return false;
    }
    else
    {
        return true;
    }
}

?>

我相信如果我结合所有这些功能，它将产生所需的输出。但是，我将在下面指定我的约束：

Rule 1 : When two or more consonants are between vowels, the first consonant is joined to the preceding vowel; for example - rec-tor, trac-tor, ac-tor, delec-tus, dic-tator, defec-tus, vic-tima, Oc-tober, fac-tum, pac-tus, 

Rule 2 : 'x' is joined to the preceding vowel; as, rex-i. 

However we give a special exception to the following consonants - br, cr, dr, fr, gr, pr, tr; bl, cl, fl, gl, pl, phl, phr, ch, ph, th, qu. These consonants are taken care by adding them to the later vowel for example - con- sola-trix
n- sola-trix. 

Rule 3 : When 'ct' follows a consonant, that consonant and 'c' are both joined to the first vowel for example - sanc-tus and junc-tum

Similarly for 'pt' we apply the same rule for example - scalp-tum, serp-tum, Redemp-tor. 

Rule 4 : A single consonant between two vowels is joined to the following vowel for example - ma-ter, pa-ter AND Z is joined to the following vowel. 

Rule 5 : When two vowels come together they are divided, if they be not a diphthong; as au-re-us. Diaphthongs are - "ae","oe","au"

Answer 1

如果仔细查看每条规则，您可以看到所有规则都包含元音开头或前一个元音。一旦你意识到这一点，你可以尝试构建一个单独的模式，将[aeiou]放在开头的因子中：

$pattern = '~
    (?<=[aeiou]) # each rule involves a vowel at the beginning (also called a
                 # "preceding vowel")
    (?:
        # Rule 2: capture particular cases
        ( (?:[bcdfgpt]r | [bcfgp] l | ph [lr] | [cpt] h | qu ) [aeiou] x )
      |
        [bcdfghlmnp-tx]
        (?:
            # Rule 3: When "ct" follows a consonant, that consonant and "c" are both
            # joined to the first vowel
            [cp] \K (?=t)
          |
            # Rule 1: When two or more consonants are between vowels, the first
            # consonant is joined to the preceding vowel
            \K (?= [bcdfghlmnp-tx]+ [aeiou] )
        )   
      |
        # Rule 4: a single consonant between two vowels is joined to the following
        # vowel
        (?:
            \K (?= [bcdfghlmnp-t] [aeiou] )
          | 
            # Rule 2: "x" is joined to the preceding vowel
            x \K (?= [a-z] | (*SKIP)(*F) ) 
        )
      |
        # Rule 5: When two vowels come together they are divided, if they not be a
        # diphthong ("ae", "oe", "au")
        \K (?= [aeiou] (?<! a[eu] | oe ) )
    )
~xi';

此模式仅用于匹配连字符的位置（规则2的特定情况除外），这就是为什么它使用大量\K来启动此位置的匹配结果并向前看在不匹配字符的情况下测试以下内容。

$string = <<<EOD
Aeneadum genetrix, hominum diuomque uoluptas,
alma Uenus, caeli subter labentia signa
quae mare nauigerum, quae terras frugiferentis
concelebras, per te quoniam genus omne animantum
EOD;

$result = preg_replace($pattern, '-$1', $string);

Ae-ne-a-dum ge-ne-trix，ho-mi-num di-u-om-qu-e u-o-lup-tas，
  al-ma U-e-nus，cae-li sub-ter la-ben-ti-a sig-na
  qu-ae ma-re nau-i-ge-rum，qu-ae ter-ras fru-gi-fe-ren-tis
  con-ce-leb-ras，per te qu-o-ni-am ge-nus om-ne a-ni-man-tum

请注意，我没有包含拉丁字母表中不存在的几个字母，如k，y和z，如果您需要处理翻译的希腊语或其他字词，请随意添加它们。

将多个正则表达式合并为一个

1 个答案: