php 在PHP中计算字符串中元音的简单方法？
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Simple way to count the vowels in a string in PHP?
提问by Jake
If I define a simple string variable, how would I count and output the number of vowels in the string in the simplest possible way?
如果我定义一个简单的字符串变量，我将如何以最简单的方式计算和输出字符串中的元音数量？
I have searched and found a number of similar ways to do so, but most seem more complex than necessary. They are all functional and maybe the complexity IS necessary, but I am looking for the simplest solution possible.
我已经搜索并找到了许多类似的方法，但大多数似乎比必要的更复杂。它们都是功能性的，也许复杂性是必要的，但我正在寻找最简单的解决方案。
My string variable would be something like:
我的字符串变量类似于：
$someString = "This is some text with some more text and even more text."
I just want to display the total instances of a,e,i,o,u. Thank you in advance.
我只想显示 a,e,i,o,u 的总实例。先感谢您。
回答by Nathan
Here's an easy solution:
这是一个简单的解决方案：
<?php
$string = "This is some text with some more text and even more text.";
echo "There are <strong>".preg_match_all('/[aeiou]/i',$string,$matches)." vowels</strong> in the string <strong>".$string."</strong>";
?>
回答by Gricey
Why not just
为什么不只是
$Vowels = substr_count($someString, 'a')+substr_count($someString, 'e')+substr_count($someString, 'i')+substr_count($someString, 'o')+substr_count($someString, 'u');
I would, however, encase it in a function otherwise you would have to change the names of the variables every time you want to reuse it:
但是，我会将它封装在一个函数中，否则每次要重用它时都必须更改变量的名称：
function CountVowels($String) {
return substr_count($String, 'a')+substr_count($String, 'e')+substr_count($String, 'i')+substr_count($String, 'o')+substr_count($String, 'u');
}
echo CountVowels('This is some text with some more text and even more text.');
//Echos 17
--
——
2018 Update:
2018 更新：
This could be much neater as
这可能会更整洁，因为
function count_vowels(string $s): int {
return array_sum(array_map(function ($vowel) use ($s) {
return substr_count($s, $vowel);
}, ['a', 'e', 'i', 'o', 'u']));
}
count_vowels('This is some text with some more text and even more text.');
However, as others pointed out, this may not be the fastest on long strings since it has to iterate through the string 5 times.
然而，正如其他人指出的那样，这可能不是长字符串最快的，因为它必须遍历字符串 5 次。
回答by evan
untested but should work:
未经测试，但应该工作：
$matches = preg_match_all('/[aeiou]/i', $someString, $ignore);
回答by alex
Here is another way to do it...
这是另一种方法来做到这一点......
$vowels = array('a', 'e', 'i', 'o', 'u');
$vowelsCounts = array_sum(
array_intersect_key(
array_count_values(
str_split(
strtolower($str)
)
),
array_flip($vowels)
)
);
键盘。
But, this works too, just make sure you don't operate this on the original string, as it will change it.
但是，这也有效，只要确保不要在原始字符串上操作它，因为它会改变它。
str_replace($vowels, FALSE, $str, $vowelsCounts);
键盘。
回答by Johnny Jiang
I know it's quite late, but I've tested all answers above and performance wasn't good when handling big string.
我知道现在已经很晚了，但是我已经测试了上面的所有答案，并且在处理大字符串时性能不佳。
I believe this code is better on memory and execution time
我相信这段代码在内存和执行时间上更好
$counter = 0;
$a = array("a","e","i","o","u");
foreach (count_chars($str, 1) as $k => $v) {
//echo "There were $v instance(s) of \"" .$k."/". chr($k) . "\" in the string.\n";
if(in_array(strtolower(chr($k)), $a))
$counter += $v;
}
回答by FtDRbwLXw6
preg_match('#[aeiou]#i', $someString, $matches);
echo count($matches) - 1, "\n";
Something like this ought to work. I can't think of a simpler method.
像这样的事情应该有效。我想不出更简单的方法。
回答by Vivek parikh
Count the vowel in string using function in php.
使用php中的函数计算字符串中的元音。
echo substr_count("￡string","a")+substr_count("￡string","e")+substr_count("￡string","i")+substr_count("￡string","o")+substr_count("￡string","u");
回答by Tamer Shlash
$someString = "This is some text with some more text and even more text.";
$total = 0;
$vowels = Array('a','e','i','o','u');
for ($i=0;$i<strlen($someString);$i++)
{
for ($j = 0;$j<5;$j++)
if ($someString[$i] == $vowels[$j])
{
$total++;
break;
}
}
echo $total;
回答by galchen
$someString = "This is some text with some more text and even more text.";
echo preg_match_all('/[aouie]/i', $someString, $out);
回答by Hammerite
Not suited for strings in multibyte character sets.
不适合多字节字符集中的字符串。
function countSpecificChars ($string, $charsOfInterest) {
$count = 0;
$len = strlen($string);
for ($i = 0; $i < $len; $i++) {
if (in_array($string[$i], $charsOfInterest)) {
$count++;
}
}
return $count;
}
function countVowels ($string) {
return countSpecificChars($string, array('a', 'e', 'i', 'o', 'u'));
}
echo countVowels('This is some text with some more text and even more text.');
// echoes '17'
