可能重复:
Programmatically determine whether to describe an object with “a” or “an”?
我需要输出这样一句话:
Are you An American?
或者:
Are you A German?
我有国家/地区名称,但我需要为该国家/地区名称找到正确的文章A或An。
是否有任何php库/实用程序函数实现了语法规则,其中包含here所述的所有异常?我想过Zend_Locale,但没有找到足够的东西。
我也用谷歌搜索,但“A vs An PHP”结果并没有真正的帮助。
答案 0 :(得分:5)
有关可行的解决方案,请参阅this answer。答案中包含Lingua::EN::Inflect
Perl模块的摘录,该模块似乎在确定使用哪个不定文章方面表现相当不错:
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不应该太难。
编辑:我最终将其转换为PHP(也可在github上获得)。
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 $word";
}
}
答案 1 :(得分:0)
滚动自己非常简单:
function an($str) {
$vowels = "aeiou";
if (strpos($vowels, strtolower(substr($str, 0, 1))) !== false) {
return "An " . $str;
}
else {
return "A " . $str;
}
}
基本上,如果它以元音开头,则使用'An',否则使用'A'。
可能有一些边缘情况并非如此,但我无法想到任何一个问题。如果找到一个,请在进行常规测试之前对其进行明确测试:
function an($str) {
if ($str == "foobarbaz") {
return "An";
}
$vowels = "aeiou";
if (strpos($vowels, strtolower(substr($str, 0, 1))) !== false) {
return "An " . $str;
}
else {
return "A " . $str;
}
}
答案 2 :(得分:-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;
答案 3 :(得分:-2)
我建议创建一个非常简单的函数。
此功能将国家/地区名称作为参数,并查看第一个字符。 如果字符来自此数组
$an_char_array = array('a', 'e', 'i', 'o');
然后你会return "an {$country}";
否则return "a {$country}";
。
您应该考虑任何异常并首先处理它们,然后执行一般部分。
功能可能如下所示:
function prependAricle($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}";
}