1

可能重复：
以编程方式确定是用“a”还是“an”来描述对象？

我需要输出这样的句子：

Are you An American?

或者：

Are you A German?

我有国家名称，但我需要为该国家名称找到正确的文章 A 或 An。

是否有任何 php 库/实用程序函数可以实现该语法规则，除了此处所述的所有例外情况？我想到了 Zend_Locale，但没有找到足够的东西。

我也用谷歌搜索过，但是“A vs An PHP”的结果并没有真正的帮助。

4

4 回答 4

8

请参阅此答案以获取一些有用的解决方案。答案包含 Perl 模块的摘录Lingua::EN::Inflect，似乎在确定使用哪个不定冠词方面做得很好：

A("cat")        # -> "a cat"
AN("cat")       # -> "a cat"
A("euphemism")      # -> "a euphemism"
A("Euler number")   # -> "an Euler number"
A("hour")       # -> "an hour"
A("houri")      # -> "a houri"

规则被定义为正则表达式，因此移植到 PHP 应该不会太难。

编辑：我最终将其转换为 PHP（也可在github上找到）。

用法：print IndefiniteArticle::A("umbrella"); // an umbrella

<?php

class IndefiniteArticle
{

    public static function AN($input, $count=1) {
        return self::A($input, $count);
    }

    public static function A($input, $count=1) {
        $matches = array();
        $matchCount = preg_match("/\A(\s*)(?:an?\s+)?(.+?)(\s*)\Z/i", $input, $matches);
        list($all, $pre, $word, $post) = $matches;
        if(!$word)
            return $input;
        $result = self::_indef_article($word, $count);  
                return $pre.$result.$post;
    }

    # THIS PATTERN MATCHES STRINGS OF CAPITALS STARTING WITH A "VOWEL-SOUND"
    # CONSONANT FOLLOWED BY ANOTHER CONSONANT, AND WHICH ARE NOT LIKELY
    # TO BE REAL WORDS (OH, ALL RIGHT THEN, IT'S JUST MAGIC!)

    private static $A_abbrev = "(?! FJO | [HLMNS]Y.  | RY[EO] | SQU
          | ( F[LR]? | [HL] | MN? | N | RH? | S[CHKLMNPTVW]? | X(YL)?) [AEIOU])
            [FHLMNRSX][A-Z]
        ";

    # THIS PATTERN CODES THE BEGINNINGS OF ALL ENGLISH WORDS BEGINING WITH A
    # 'y' FOLLOWED BY A CONSONANT. ANY OTHER Y-CONSONANT PREFIX THEREFORE
    # IMPLIES AN ABBREVIATION.

    private static $A_y_cons = 'y(b[lor]|cl[ea]|fere|gg|p[ios]|rou|tt)';

    # EXCEPTIONS TO EXCEPTIONS

    private static $A_explicit_an = "euler|hour(?!i)|heir|honest|hono";

    private static $A_ordinal_an = "[aefhilmnorsx]-?th";

    private static $A_ordinal_a = "[bcdgjkpqtuvwyz]-?th";

    private static function _indef_article($word, $count) {
        if($count != 1) // TODO: Check against $PL_count_one instead
            return "$count $word";

            # HANDLE USER-DEFINED VARIANTS
        // TODO


        # HANDLE ORDINAL FORMS
        if(preg_match("/^(".self::$A_ordinal_a.")/i", $word))       return "a $word";
        if(preg_match("/^(".self::$A_ordinal_an.")/i", $word))      return "an $word";

        # HANDLE SPECIAL CASES

        if(preg_match("/^(".self::$A_explicit_an.")/i", $word))         return "an $word";
        if(preg_match("/^[aefhilmnorsx]$/i", $word))        return "an $word";
        if(preg_match("/^[bcdgjkpqtuvwyz]$/i", $word))      return "a $word";

            # HANDLE ABBREVIATIONS

        if(preg_match("/^(".self::$A_abbrev.")/x", $word))          return "an $word";
        if(preg_match("/^[aefhilmnorsx][.-]/i", $word))         return "an $word";
        if(preg_match("/^[a-z][.-]/i", $word))          return "a $word";

        # HANDLE CONSONANTS

        if(preg_match("/^[^aeiouy]/i", $word))                  return "a $word";

            # HANDLE SPECIAL VOWEL-FORMS

        if(preg_match("/^e[uw]/i", $word))                      return "a $word";
        if(preg_match("/^onc?e\b/i", $word))                    return "a $word";
        if(preg_match("/^uni([^nmd]|mo)/i", $word))     return "a $word";
        if(preg_match("/^ut[th]/i", $word))                     return "an $word";
        if(preg_match("/^u[bcfhjkqrst][aeiou]/i", $word))   return "a $word";

            # HANDLE SPECIAL CAPITALS

        if(preg_match("/^U[NK][AIEO]?/", $word))                return "a $word";

        # HANDLE VOWELS

        if(preg_match("/^[aeiou]/i", $word))            return "an $word";

        # HANDLE y... (BEFORE CERTAIN CONSONANTS IMPLIES (UNNATURALIZED) "i.." SOUND)

        if(preg_match("/^(".self::$A_y_cons.")/i", $word))      return "an $word";

        # OTHERWISE, GUESS "a"
        return "a $word";
    }
}
于 2012-10-09T11:09:06.550 回答
0

自己动手很简单：

function an($str) {

    $vowels = "aeiou";

    if (strpos($vowels, strtolower(substr($str, 0, 1))) !== false) {

        return "An " . $str;

    }
    else {

        return "A " . $str;

    }

}

本质上，如果它以元音开头，则使用“An”，否则使用“A”。

可能有一些边缘情况不是这样，但我想不出任何想法。如果你找到了一个，在你做一般测试之前明确地测试它：

function an($str) {

    if ($str == "foobarbaz") {

        return "An";

    }

    $vowels = "aeiou";

    if (strpos($vowels, strtolower(substr($str, 0, 1))) !== false) {

        return "An " . $str;

    }
    else {

        return "A " . $str;

    }

}
于 2012-10-09T10:47:13.273 回答
-2

试试这个：-

$vowel_arr=array('a','e','i','o','u');
$string="America";
$len=strtolower(substr($string,0,1));

if(in_array($len,$vowel_arr))
  $pre = "an";
else
  $pre = "a";


 echo  $pre." ".$string;
于 2012-10-09T10:45:36.963 回答
-2

我建议创建一个非常简单的函数。

此函数将国家名称作为参数并查看第一个字符。如果字符来自这个数组

$an_char_array = array('a', 'e', 'i', 'o');

那么你会return "an {$country}";否则return "a {$country}";。

您应该首先考虑任何异常并处理它们，然后执行一般部分。

函数可能如下所示：

function prependAricle($word) {
    // handle exceptions here, e.g.:
    $an_exceptions = array('umbrella'/*, ...*/);
    $a_exceptions = array('pity'/*, ...*/);
    if(in_array(strtolower($word), $an_exceptions))
        return "an {$word}";
    if(in_array(strtolower($word), $a_exceptions))
        return "a {$word}";

    // handle general rule
    $an_char_array = array('a', 'e', 'i', 'o', 'A', 'E', 'I', 'O');
    if(in_array(substr($word, 0, 1), $an_char_array))
        return "an {$word}";
    else
        return "a {$word}";
}
于 2012-10-09T10:49:19.587 回答