[英]How to format numbers using Regular Expression in PHP
I would need to change the format to 123.456.789.11
, so far I have only managed to do it as shown in the example https://regex101.com/r/sY1nH4/1 , but I would need it to always have 3 digits at the beginning, thank you for your help我需要将格式更改为
123.456.789.11
,到目前为止,我只能按照示例https://regex101.com/r/sY1nH4/1中所示进行操作,但我需要它始终具有 3 位数字一开始,谢谢你的帮助
$repl = preg_replace('/(?!^)(?=(?:\d{3})+$)/m', '.', $input);
You should assert that it is not the end of the string instead to prevent adding a dot at the end:您应该断言它不是字符串的末尾,以防止在末尾添加点:
\d{3}(?!$)\K
\d{3}
Match 3 digits \d{3}
匹配3个数字(?!$)
Negative lookahead, assert not the end of the string to the right (?!$)
否定前瞻,断言不是字符串的结尾到右边\K
Forget what is matched so far \K
忘记到目前为止匹配的内容$re = '/\d{3}(?!$)\K/m';
$str = '111222333444
11222333444';
$result = preg_replace($re, ".", $str);
echo $result;
Output Output
111.222.333.444
112.223.334.44
I would use this approach, using a capture group:我会使用这种方法,使用捕获组:
$input = "12345678911";
$output = preg_replace("/(\d{3})(?=\d)/", "$1.", $input);
echo $output; // 123.456.789.11
The above replaces every 3 numbers, starting from the left, with the same numbers followed by a dot, provided that at least one other digit follows.上面的代码从左边开始每 3 个数字替换为相同的数字后跟一个点,前提是至少要跟一个其他数字。
Here is a possible solution using \B
(opposite of \b
):这是使用
\B
(与\b
相反)的可能解决方案:
\d{3}\B\K
Replace it with .
将其替换为
.
By using \B
after 3 digits we ensure that we don't match the last position.通过在 3 位数字后使用
\B
,我们确保不匹配最后的 position。
No need for a regular expression actually.实际上不需要正则表达式。 split into equal parts, join with comma:
分成相等的部分,用逗号连接:
$formatted = implode(',', str_split($str, 3));
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