将多个正则表达式合并为一个
我正在尝试编写一个代码来将字符串连字符连字符拉丁字母连接起来。 有几个限制，我已经照顾，但我没有得到所需的输出。 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;
}
}
?>
我相信如果我将所有这些功能结合起来，就会产生所需的输出。 不过，我会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"
如果仔细观察每条规则，可以看到所有元音都在开头或前面的元音中包含元音。 一旦你意识到这一点，你可以尝试在开始时建立一个把[aeiou]放在因子上的模式：
$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';
这种模式仅用于匹配连字符的位置（除了规则2的特定情况外），这就是为什么它使用大量K开始在此位置的匹配结果，并且预测测试之后没有匹配字符。
$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 uo-lup-tas，
al-ma Ue-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-o-ni-am ge-nus om-ne a-ni-man-tum
请注意，我没有包括像拉丁字母中不存在的几个字母，如k，y和z，如果需要处理翻译的希腊字或其他字母，请随意添加它们。
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