将正则表达式条件放在流写C＃的位置 [英] Where to put regex conditions on streamwriting C#

问题描述

我想在我的代码中添加新的正则表达式，从列表中打印（使用streamreader）。

在哪里添加正则表达式，从文件和正则表达式中删除这些字符（ - ：_ |）以添加标题为文件。



正则表达式替换char：line.Replace（@ - ：_ |，string.Empty）;



正则表达式添加标题：

  using （StreamWriter file =  new  StreamWriter（fs））
 {
 file.WriteLine（ 数字，用户，功能，消息）; 
 
  for （ int  index =  0 ; index <  pr.Length; index ++）
 {
 
 file.WriteLine（  {0}，\{1} \，{2}，{3}，
 pr [s] .Number，pr [s] .User，
 pr [s] .Function，pr [s] .Message; 
} 
} 

我尝试了什么：

 使用（StreamWriter writer =  new  StreamWriter（path2））
 {
  foreach （ string  l  in 不活动）
 {
 
  //   line.Replace（@ - ：_ | ，string.Empty）;  
 writer.WriteLine（l）; 
} 
} 

解决方案

引用：

我想删除不替换那些字符

这就是用空字符串替换它们的方法 - 它会删除字符。

如果你想要删除所有这些字符，无论rgey出现在哪里，速度比单个字符串，你有错误的正则表达式。

你的正则表达式会改变这个

你好 - ：_ |那里！

对此：

你好！

但是会保持不变：

你好 - ：有！_ | 

如果要删除所有实例任何角色，你需要这个正则表达式：

 [ - ：_ |] + 

获取 Expresso 的副本[ ^ ] - 它是免费的，它检查并生成正则表达式。



但该代码看起来非常像一个成员开始变成帮助吸血鬼/巨魔的代码，因为他拒绝学习任何事情并为自己思考......我在这里闻到了一个袜子傀儡......

你应该尝试替换

 @ - ：_ |

with

 @[ - ：_ |]

只是一些有趣的链接来帮助构建和调试RegEx。

以下是RegEx文档的链接：

perlre - perldoc.perl.org [ ^ ]

以下是帮助构建RegEx并调试它们的工具的链接：

.NET正则表达式测试程序 - 正则表达式风暴 [ ^ ]

Expresso正则表达式工具 [ ^ ]

RegExr：Learn，Build，&测试RegEx [ ^ ]

此节目RegEx是一个很好的图表，它非常有助于理解RegEx的作用：

Debuggex：在线可视正则表达式测试器。 JavaScript，Python和PCRE。 [ ^ ]

你不能这样使用string.Replace（使用全局字符串）;你必须使用string.Replace来替换每个字符。

示例：

 l = l.Replace（   - ， ）。替换（ ：， ）。替换（  _， ）。替换（  |， ）; 

注意：

- string.Replace不会进行变量的就地编辑，你必须再次将方法的结果强制转换为变量（ l = ... ）。

- 您可以将调用链接到替换方法。

- 在代码块中你在评论行中表示，你似乎使用的行与 l 中定义的变量不匹配你的foreach循环。

还有其他选项可以做同样的事情，例如使用正则表达式，但我认为在其他一些帖子中已经有一些关于这个主题的指示。

希望这会有所帮助。麻烦。

I want to add new regex to my code, which prints from list (with streamreader).
Where to add regex which removes these characters( - : _ |) from file and regex to add header to the file.

regex of replacing char: line.Replace(@"- : _ |", string.Empty);

regex for adding header:

using (StreamWriter file = new StreamWriter(fs))                    
{
    file.WriteLine( "Number,User,Function,Message"); 

    for (int index = 0; index < pr.Length; index++)
    {
      
        file.WriteLine("{0},\"{1}\",{2},{3}",
           pr[s].Number, pr[s].User,
           pr[s].Function, pr[s].Message; 
    }
}

What I have tried:

using (StreamWriter writer = new StreamWriter(path2))
            {
                foreach (string l in inactive)
                {

           //line.Replace(@"- : _ |", string.Empty); 
                    writer.WriteLine(l);
}
}

解决方案

Quote:

i want to remove not replace those characters

That's what replacing them with an empty string does - it removes the characters.
If you want to remove all those characters regardless of where rgey appear, ratehr than as a single string, you have teh wrong regex.
Your regex will change this

Hello -:_|There!

To this:

Hello There!

But will leave this unchanged:

Hello -:There!_|

If you want to remove all instances of any of the characters, you need this regex:

[-:_|]+

Get a copy of Expresso[^] - it's free, and it examines and generates Regular expressions.

But that code is looking very much like the code used by a member who is starting to become rather a Help Vampire / Troll because he refuses to learn anything and think for himself ... I smell a sock puppet here...

You should try to replace

@"- : _ |"

with

@"[- : _ |]"

Just a few interesting links to help building and debugging RegEx.
Here is a link to RegEx documentation:
perlre - perldoc.perl.org[^]
Here is links to tools to help build RegEx and debug them:
.NET Regex Tester - Regex Storm[^]
Expresso Regular Expression Tool[^]
RegExr: Learn, Build, & Test RegEx[^]
This one show you the RegEx as a nice graph which is really helpful to understand what is doing a RegEx:
Debuggex: Online visual regex tester. JavaScript, Python, and PCRE.[^]

You cannot use string.Replace this way (using a global string); you have to use a string.Replace for each character to replace.
Example:

l = l.Replace("-", "").Replace(":", "").Replace("_", "").Replace("|", "");

Notes:
- string.Replace does not proceed to an in-place editing of the variable, you have to cast the result of the method to the variable again (l = ...).
- You can chain the calls to the Replace method.
- In the code block you showed, in the commented line, you seem to use a line which does not match the l variable defined in your foreach loop.
There are other options to do the same, using regular expressions for example, but I think some directions already have been given to you on this subject in some other posts.
Hope this helps. Kindly.

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