How to reference named capture groups in replacement parameters in preg_replace?

一般的捕获组可以用 \0 \1 \N 来表名,如果是用 P 这样命名的捕获结果, 在替换参数中用什么来表示 ?

<code>  <?php $text="<span style='some' > AAAA ";
  $text.="<img    style="max-width:90%" alt="How to reference named capture groups in replacement parameters in preg_replace?" >";
  $text.="<p style="some">";
  $text.="<img    style="max-width:90%" alt="How to reference named capture groups in replacement parameters in preg_replace?" >";
  $text.="</p>
<p style="some"> some </p>";
  
  
  $out=preg_replace("/span|p)[^>]+>(.+))>/","[\P=
  TAG]\\1[\/\k<tag>]",$text);
 
print_r( $out);</tag></code>

好象直接用k 这样不行

回复内容：

一般的捕获组可以用 \0 \1 \N 来表名,如果是用 P 这样命名的捕获结果, 在替换参数中用什么来表示 ?

<code>  <?php $text="<span style='some' > AAAA ";
  $text.="<img    style="max-width:90%" alt="How to reference named capture groups in replacement parameters in preg_replace?" >";
  $text.="<p style="some">";
  $text.="<img    style="max-width:90%" alt="How to reference named capture groups in replacement parameters in preg_replace?" >";
  $text.="</p>
<p style="some"> some </p>";
  
  
  $out=preg_replace("/span|p)[^>]+>(.+))>/","[\P=
  TAG]\\1[\/\k<tag>]",$text);
 
print_r( $out);</tag></code>

好象直接用k 这样不行

http://php.net/manual/en/function.preg-replace.php

replacement may contain references of the form \n or (since PHP
4.0.4) $n, with the latter form being the preferred one. Every such reference will be replaced by the text captured by the n'th
parenthesized pattern. n can be from 0 to 99, and \0 or $0 refers to
the text matched by the whole pattern. Opening parentheses are counted
from left to right (starting from 1) to obtain the number of the
capturing subpattern. To use backslash in replacement, it must be
doubled ("\\" PHP string).

When working with a replacement pattern where a backreference is
immediately followed by another number (i.e.: placing a literal number
immediately after a matched pattern), you cannot use the familiar \1
notation for your backreference. \11, for example, would confuse
preg_replace() since it does not know whether you want the \1
backreference followed by a literal 1, or the \11 backreference
followed by nothing. In this case the solution is to use ${1}1. This
creates an isolated $1 backreference, leaving the 1 as a literal.

When using the deprecated e modifier, this function escapes some
characters (namely ', ", and NULL) in the strings that replace the

1. This is done to ensure that no syntax errors arise

2. backreference usage with either single or double quotes (e.g.

'strlen('$1')+strlen("$2")'). Make sure you are aware of PHP's
string syntax to know exactly how the interpreted string will look.