如何在没有 ord() 的情况下在 PHP 中将字符串转换为 ASCII 值？
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How to convert a string to ASCII value in PHP without ord()?
提问by Rani
I am looking to convert a string say 'Hello' world to its ASCII value in php.
But I don't want to use ord(). Are there any other solutions for printing ascii value without using ord()?
我正在寻找在 php 中将字符串 say 'Hello' world 转换为它的 ASCII 值。但我不想使用ord(). 是否有任何其他解决方案可以在不使用的情况下打印 ascii 值ord()？
回答by user3942918
Unpacks from a binary string into an array according to the given format.
根据给定的格式将二进制字符串解包为数组。
Use the formatC*to return everything as what you'd get from ord().
使用该格式C*将所有内容作为您从ord().
print_r(unpack("C*", "Hello world"));
Array
(
[1] => 72
[2] => 101
[3] => 108
[4] => 108
[5] => 111
[6] => 32
[7] => 119
[8] => 111
[9] => 114
[10] => 108
[11] => 100
)
回答by Purple Lady
You could try the iconvnative function:
您可以尝试本iconv机功能：
string iconv ( string $in_charset , string $out_charset , string $str )
So it would be:
所以它会是：
<?php
$string = "This is the Euro symbol ''.";
echo iconv("UTF-8", "ASCII", $text), PHP_EOL;
?>
Taken from: http://php.net/manual/en/function.iconv.php
回答by Gavin Hymanson
You could iterate over each character in the string, find its offset into a dictionary string using say strpos, then add on a base number eg 65if your dictionary started with "ABC...
You'd need to handle unfound characters, so maybe better to use a dictionary of "#ABC... then add a base of 64, obviously you'd need to test for "#" as a special character then.
You could even test against multiple distionary strings for limited character sets "#A..Z", "#a..z", "#0..9"
You get the idea, but without knowing why you want to limit yourself I can't tell whther this is useful to you.
您可以迭代字符串中的每个字符，使用 say strpos找到它在字典字符串中的偏移量，然后添加一个基数，例如65如果您的字典以“ ABC
开头...
您需要处理未找到的字符，所以也许最好使用“ #ABC...然后添加一个基数64的字典，显然你需要测试“#”作为一个特殊字符。
您甚至可以针对有限字符集“ #A..Z”、“ #a..z”、“ #0..9”的
多个字典字符串进行测试，
您明白了，但不知道为什么要限制自己我能够'
