使用preg_replace（）替换特殊字符之间的内容

问题描述：

I have a paragraph as -

== one ===

==== two ==

= three ====

etc.

The number of = sign vary in every row of the paragraph.

I want to write a preg_replace() expression that will allow me to replace the texts between the = signs.

example:

== DONE ===

==== DONE ==

= DONE ====

I tried preg_replace("/\=+(.*)\=+/","DONE", $paragraph) but that doesn't work. Where am I going wrong?

我有一个段落为 - p>

  == one =  == 
 
 ====两个== 
 
 =三个==== 
  code>  pre> 
 
 等。 p> 
 
  
  =  code>符号的数量在段落的每一行都有所不同。 p> 
 
 
我想写一个  preg_replace（） 代码>  strong>表达式，允许我替换 =  code>符号之间的文本 strong>。   p> 
 
 
示例： p> 
 
 

  == DONE === 
 
 ==== DONE == 
 
 = DONE  ==== 
  code>  pre> 
 
 我试过 preg_replace（“/\=+(.*)\=+/”，“DONE”，$ paragraph）   code>但这不起作用。 我哪里错了？ p> 
  div>

答

You can use:

$str = preg_replace('/^=+\h*\K.+?(?=\h*=)/m', 'DONE', $str);

RegEx Demo

RegEx Breakup:

^          # Line start
=+         # Match 1 or more =
\h*        # Match or more horizontal spaces 
\K         # resets the starting point of the reported match
.+?        # match 1 or more of any character (non-greedy)
(?=\h*=)   # Lookahead to make sure 0 or more space followed by 1 = is there

答

You have to place the =s back.

Also, instead of .* use [^=]* (matches characters, which are not =) so that the =s don't get eaten up for the replacement.

Additionally, you don't have to escape =:

preg_replace("/(=+)([^=]*)(=+)/","$1 DONE $3", $paragraph);

See it in action

使用preg_replace（）替换特殊字符之间的内容

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