有没有办法简单的正则表达式？ [英] Is there a way to simply a regular expression?

问题描述

我有如下XML：

 <   break     name   =  article_1-1 >  
 <   h1  >  
 < span class =code-keyword><   page     num   =  1    /  > 某些标题<   / h1  >  
  <  <跨度class =code-leadattribute> bl  >  
 人名撰稿人<   / bl  >  
 <   h3  >  
 意见 
 <   / h3  >  
 <   p  > 第一段<   / p  >  
 <   p  > 第二段<   / p  >  
 <   p  > 第三段<   / p  >  
 <   bq  >  
 某些值 
 <   / bq  >  
 <   p  >  
 第四段用斜体值 
 < span class =code-keyword><   / p  >  
 <   fig  >  
 <   img     src   =  images / img_1-1.jpg    width   =  1553    height   =  1050    alt   =     /  >  
 <   fc  >  
 图片标题 
 <   / fc  >  
 <   cr  > 某些人的摄影<   / cr  >  
 <   / fig  >  
 < ;   h3  >  
  CITY，STATE  
 <   / h3  >  
 <   / break  >  

我希望如下：

 <   break     name   =  article_1-1 >  
 <   h1  >  <   page     num   =  1    /  > 某些标题<   / h1  >  
 <   bl  >  人名撰稿er <   / bl  >  
 <   h3  >   OPINION  <   / h3   >  
 <   p  > 第一段<   / p  >  
 <   p  > 第二段<   / p  >  
 <   p  > 第三段<   / p  >  
 <   bq  >  某些值 <   / bq  >  
 <   p  >  第四段用斜体值 <   / p  >  
 <   fig  >  <   img     src   =   images / img_1-1.jpg    width   =  1553    height   =  1050 < span class =code-attribute>   alt   =     /  >  <   fc  >  图片标题 <   / fc  >  <   cr  > ; 一些人的摄影<   / cr  >  <   / fig  >  
 <   h3  >   CITY，STATE  <   / h3  >  
 <   / break  >  

我将在稍后阶段删除缩进但我主要关注的是在同一行中打开和关闭XML标签。



我想要一个正则表达式。我尝试了一些东西，但我认为有更好的方法。



请帮忙。



问候



我尝试过的事情：

  string  pattern =  @ （？:(？： （小于\w>）|（小于\w>）|（小于\w ..> |（小于p为H.）|（\ />）））（\ S +）|（（小于\ /（（标题）|（头）|（中断）|（主体））\w + GT;？！）（\s +）（小于\ /（？ （标题）|（头）|（中断）|（主体））\w + GT））|（（小于\ / FC>）（\s +）（小于CR>）））< /跨度>; 
 
  string  substitution2 =  @  $ 1 $ 2 $ 3 $ 8 $ 14 $ 20 $ 22; 

解决方案

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I have an XML as follows:

<break name="article_1-1">
<h1>
  <page num="1" />Some heading</h1>
<bl>
  Human name Contributing Writer</bl>
<h3>
  OPINION
</h3>
<p>First Paragraph</p>
<p>Second Paragraph</p>
<p>Third Paragraph</p>
<bq>
  Some value
</bq>
<p>
  Fourth Paragraph with italic values
</p>
<fig>
  <img src="images/img_1-1.jpg" width="1553" height="1050" alt="" />
  <fc>
	Image caption
  </fc>
  <cr>PHOTOGRAPHS BY SOME HUMAN</cr>
</fig>
<h3>
  CITY, STATE
</h3>
</break>

I want to make it like:

<break name="article_1-1">
<h1><page num="1" />Some heading</h1>
<bl>Human name Contributing Writer</bl>
<h3>OPINION</h3>
<p>First Paragraph</p>
<p>Second Paragraph</p>
<p>Third Paragraph</p>
<bq>Some value</bq>
<p>Fourth Paragraph with italic values</p>
<fig><img src="images/img_1-1.jpg" width="1553" height="1050" alt="" /><fc>Image caption</fc><cr>PHOTOGRAPHS BY SOME HUMAN</cr></fig>
<h3>CITY, STATE</h3>
</break>

I am removing the indentation at a later stage but my main focus is on bringing the opening and closing XML tags in the same line.

I want a regex for this. I have tried something but I think there is a better way.

Please help.

Regards

What I have tried:

string pattern = @"(?:(?:(<\w.>)|(<\w>)|(<\w..>|(<p>)|(\/>)))(\s+)|((<\/(?!(title)|(head)|(break)|(body))\w+>)(\s+)(<\/(?!(title)|(head)|(break)|(body))\w+>))|((<\/fc>)(\s+)(<cr>)))";

string substitution2 = @"$1$2$3$8$14$20$22";

解决方案

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