count_chars

(PHP 4, PHP 5, PHP 7, PHP 8)

count_chars — 返回字串中使用的字元資訊

說明

count_chars(字串 $string, 整數 $mode = 0): 陣列|字串

計算 string 中每個位元組值 (0..255) 出現的次數，並以各種方式返回。

參數

string: 被檢查的字串。
mode: 請參閱返回值。

返回值

根據 mode，count_chars() 會返回以下其中一種

0 - 一個陣列，以位元組值作為鍵，每個位元組出現的頻率作為值。
1 - 與 0 相同，但只列出頻率大於零的位元組值。
2 - 與 0 相同，但只列出頻率等於零的位元組值。
3 - 傳回一個包含所有唯一字元的字串。
4 - 傳回一個包含所有未使用的字元的字串。

更新日誌

版本	說明
8.0.0	在此版本之前，函式失敗時會傳回 `false`。

範例

範例 #1 count_chars() 範例

<?php
$data = "Two Ts and one F.";

foreach (count_chars($data, 1) as $i => $val) {
 echo "字串中有 $val 個 \"" , chr($i) , "\"。\n";
}
?>

上述範例將輸出：

There were 4 instance(s) of " " in the string.
There were 1 instance(s) of "." in the string.
There were 1 instance(s) of "F" in the string.
There were 2 instance(s) of "T" in the string.
There were 1 instance(s) of "a" in the string.
There were 1 instance(s) of "d" in the string.
There were 1 instance(s) of "e" in the string.
There were 2 instance(s) of "n" in the string.
There were 2 instance(s) of "o" in the string.
There were 1 instance(s) of "s" in the string.
There were 1 instance(s) of "w" in the string.

另請參閱

strpos() - 尋找子字串在字串中第一次出現的位置
substr_count() - 計算子字串出現的次數

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＋新增筆記

使用者貢獻的筆記 8 則筆記

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marcus33cz ¶

12 年前

如果您在多位元組字串上使用 count_chars 遇到問題，您可以更改頁面編碼。或者，您也可以使用這個 mb_count_chars 版本的函式。基本上它是原始函式的模式「1」。

<?php
/**
 * 計算多位元組字串中字元的出現次數
 * @param string $input UTF-8 資料
 * @return array 字元的關聯陣列。
 */
function mb_count_chars($input) {
 $l = mb_strlen($input, 'UTF-8');
 $unique = array();
 for($i = 0; $i < $l; $i++) {
 $char = mb_substr($input, $i, 1, 'UTF-8');
 if(!array_key_exists($char, $unique))
 $unique[$char] = 0;
 $unique[$char]++;
 }
 return $unique;
}

$input = "Let's try some Greek letters: αααααΕεΙιΜμΨψ, Russian: ЙЙЫЫЩН, Czech: ěščřžýáíé";
print_r( mb_count_chars($input) ); 
//回傳：Array ( [L] => 1 [e] => 7 [t] => 4 ['] => 1 [s] => 5 [ ] => 9 [r] => 3 [y] => 1 [o] => 1 [m] => 1 [G] => 1 [k] => 1 [l] => 1 [:] => 3 [α] => 5 [Ε] => 1 [ε] => 1 [Ι] => 1 [ι] => 1 [Μ] => 1 [μ] => 1 [Ψ] => 1 [ψ] => 1 [,] => 2 [R] => 1 [u] => 1 [i] => 1 [a] => 1 [n] => 1 [Й] => 2 [Ы] => 2 [Щ] => 1 [Н] => 1 [C] => 1 [z] => 1 [c] => 1 [h] => 1 [ě] => 1 [š] => 1 [č] => 1 [ř] => 1 [ž] => 1 [ý] => 1 [á] => 1 [í] => 1 [é] => 1 )
?>

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Eric Pecoraro ¶

19 年前

<?php

// 要求字串中包含 (n) 個不重複的字元
// 修改以下函式，增加指定字串中所需的不重複字元數的彈性。

$pass = '123456' ; // true
$pass = '111222' ; // false

req_unique($pass,3);

function req_unique($string,$unique=3) {
 if ( count(count_chars($string,1)) < $unique) {
 echo 'false';
 }else{
 echo 'true';
 }
}

?>

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seb at synchrocide dot net ¶

20 年前

在經過多次嘗試錯誤，試圖建立一個函式來找出字串中不重複字元的數量後，我找到了 count_chars() 函式 - 我 20 多行的無用程式碼被它取代了

<?
函式 unichar($string) {
$two = strtolower(str_replace(' ', '', $string));
$res = count(count_chars($two, 1));
return $res;
}

/* 範例 :: */

echo unichar("bob"); // 2
echo unichar("Invisibility"); //8
echo unichar("The quick brown fox slyly jumped over the lazy dog"); //26

?>

我不知道這可以用在哪裡，但它蠻有趣的

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匿名 ¶

8 年前

count_chars 支援多位元組字元。

<?php

function mb_count_chars ($string, $mode = 0) {
 
 $result = array_fill(0, 256, 0);
 
 for ($i = 0, $size = mb_strlen($string); $i < $size; $i++) {
 $char = mb_substr($string, $i, 1);
 if (strlen($char) > 1) {
 continue;
 }
 
 $code = ord($char);
 if ($code >= 0 && $code <= 255) {
 $result[$code]++;
 }
 }
 
 switch ($mode) {
 case 1: // same as 0 but only byte-values with a frequency greater than zero are listed.
 foreach ($result as $key => $value) {
 if ($value == 0) {
 unset($result[$key]);
 }
 }
 break;
 case 2: // same as 0 but only byte-values with a frequency equal to zero are listed.
 foreach ($result as $key => $value) {
 if ($value > 0) {
 unset($result[$key]);
 }
 }
 break;
 case 3: // a string containing all unique characters is returned.
 $buildString = '';
 foreach ($result as $key => $value) {
 if ($value > 0) {
 $buildString .= chr($key);
 }
 }
 return $buildString;
 case 4: // a string containing all not used characters is returned.
 $buildString = '';
 foreach ($result as $key => $value) {
 if ($value == 0) {
 $buildString .= chr($key);
 }
 }
 return $buildString;
 }
 
 // change key names...
 foreach ($result as $key => $value) {
 $result[chr($key)] = $value;
 unset($result[$key]);
 }
 
 return $result;
 
}
?>

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-1

Andrey G ¶

4 年前

檢查兩個字串是否為異位構詞 (Anagram)

<?php

function isAnagram($string1, $string2)
{
 return count_chars($string1, 1) === count_chars($string2, 1);
}

isAnagram('act', 'cat'); // true

?>

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-1

pzb at novell dot com ¶

17 年前

這個函式很適合用於輸入驗證。我經常需要檢查字串中的所有字元是否都是 7 位元 ASCII 字元（且不為 null）。這是我目前找到最快的函式

<?php
function is7bit($string) {
 // 空字串視為 7 位元乾淨
 if (!strlen($string)) {
 return true;
 }
 // count_chars 會以遞增的八位元組順序返回字元
 $str = count_chars($str, 3);
 // 檢查是否有 null 字元
 if (!ord($str[0])) {
 return false;
 }
 // 檢查是否有 8 位元字元
 if (ord($str[strlen($str)-1]) & 128) {
 return false;
 }
 return true;
}
?>

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-3

phpC2007 ¶

17 年前

這是一個用來計算字串中不同字元數量個數的函式。它可以作為一個支援 utf8 的簡化版 count_chars（但僅限於單一模式）...

<?php
函式 utf8_count_strings($stringChar)
{
 $num = -1;
 $lenStringChar = strlen($stringChar);
 
 for ($lastPosition = 0; 
 $lastPosition !== false;
 $lastPosition = strpos($textSnippet, $stringChar, $lastPosition + $lenStringChar))
 {
 $num++;
 }
 
 return $num;
}
?>

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-5

mlong at mlong dot org ¶

22 年前

// 這兩個函式的用途


<?php 
 $string="aaabbc"; 
 
 // 你只想計算字母 a 的數量
 $acount=substr_count($string,"a"); 
 
 // 你想計算字母 a 和字母 b 的數量
 $counts=count_chars($string,0); 
 $acount=$counts[ord("a")]; 
 $bcount=$counts[ord("b")]; 
?>

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