用除了之外的一个替换多个特殊字符

I need to replace:

我需要更换:

All consecutive special characters by its single occurrence.

单次出现的所有连续特殊字符。

I need to exclude ..

我需要排除..

e.g.

$about = "   .....I......####    ";

should become

$about = " .....I......# ";

I wrote a regular expression:

我写了一个正则表达式:

$about = preg_replace("/[^\w^.]{2,}/","$1",$about);

but it is not working. How can I achieve this?

但它不起作用。我怎样才能做到这一点?

1 个解决方案

#1

Your pattern [^\w^.]{2,} matches any 2 or more occurrences of a char other than word, ^ and ..

您的模式[^ \ w ^。] {2,}匹配除了word,^和..之外的任何2个或多个char的匹配项。

To match consecutive identical chars other than word and a dot, you may use

要匹配除字和点之外的连续相同字符,您可以使用

([^\w.])\1+

and replace with $1.

并以$ 1替换。

See the regex demo.

请参阅正则表达式演示。

Details:

([^\w.]) - matches and captures into Group 1 a char other than word char and a dot

([^ \ w。]) - 匹配并捕获除了字char和点之外的字符组1

\1+ - matches 1 or more occurrences of the value captured into Group 1.

\ 1+ - 匹配1次或多次捕获到组1中的值。

The replacement pattern only inserts a single occurrence of the matched char.

替换模式仅插入匹配的char的单个匹配项。

PHP demo:

$re = '/([^\w.])\1+/';
$str = '   .....I......####    ';
$subst = '$1';
$result = preg_replace($re, $subst, $str);
echo "The result of the substitution is '".$result . "'";

#1