的正则表达式^ |在C＃中 [英] Regex for ^ | in C#

问题描述

我工作的HL7消息，我需要一个正则表达式。这不起作用：

  HL7消息= MSH | ^〜\&放大器; | DATACAPTOR | 123 | 123 | 20100816171948 | ORU ^ R01 | 081617194802900 | p | 2.3 | 8859/1 
  

我的正则表达式是：

  MSH | ^〜\&放大器; | DATACAPTOR | \d {3} | \d {3} |（\d { 4} \d {2} \d {2} \d {2} \d {2} \d {2}）| ORU\\ ^ R01 | \d {20} | P | 2.3 | 8859/1 
  

任何人可以提出对特殊字符的正则表达式？
我使用这个代码：

  strRegex =\\vMSH | ^〜\\&放; | DATACAPTOR | \\d {3} | \\d {3} | 
（\\d {4} \\d {2} \\d {2 } \\d {2} \\d {2} \\d {2}）| ORU\\ ^ R01 | \\d {20} | P | 2.3 | 8859 / 1; 
正则表达式RX =新的正则表达式（strRegex，RegexOptions.Compiled | RegexOptions.IgnoreCase）;   

解决方案

$ C>， ^ 和 \ 是的正则表达式，所以你必须用 \ 。记住 \ 也是一个普通字符串中的转义字符，所以你不得不逃脱，也：

  VAR strRegex =\\vMSH\\ | \\ ^〜\\\\&放大器; \\ | DATACAPTOR\\\ \\ | ... 
  

但它通常一个更容易使用逐字字符串（ @...）：

  VAR strRegex = @\vMSH\ | \\ \\ ^〜\\&放大器; \ | DATACAPTOR\ | ... 
  

最后，请注意（\d {4} \d {2} \d {2} \d {2} \d {2} \d {2}）可以简化为（\d {14}）。

不过，对于这样的结构，它可能更容易只使用 < 。code>拆分 法

  VAR段=MSH | ^ 〜\&放大器; | DATACAPTOR ......; 
 VAR栏= segment.Split（'|'）; 
变种时间戳=字段[5]; 
  

^{警告：HL7消息可能会使用不同的控制字符&MDASH;启动在MSH段的第四个字符作为一个字段分隔符（在这种情况下， | ^〜\&安培; 是控制字符）。最好是先解析控制字符，如果你不控制你的输入，这些控制字符可能会改变。}

I am working on HL7 messages and I need a regex. This doesn't work:

HL7 message=MSH|^~\&|DATACAPTOR|123|123|20100816171948|ORU^R01|081617194802900|P|2.3|8859/1

My regex is:

MSH|^~\&|DATACAPTOR|\d{3}|\d{3}|(\d{4}\d{2}\d{2}\d{2}\d{2}\d{2})|ORU\\^R01|\d{20}|P|2.3|8859/1

Can anybody suggest a regex for special characters? I am using this code:

strRegex = "\\vMSH|^~\\&|DATACAPTOR|\\d{3}|\\d{3}|
(\\d{4}\\d{2}\\d{2}\\d{2}\\d{2}\\d{2})|ORU\\^R01|\\d{20}|P|2.3|8859/1";
Regex rx = new Regex(strRegex, RegexOptions.Compiled | RegexOptions.IgnoreCase );

解决方案

|, ^, and \ are all special characters in regular expressions, so you'd have to escape them with \. Remember \ is also an escape character within a regular string literal so you'd have to escape that, too:

var strRegex = "\\vMSH\\|\\^~\\\\&\\|DATACAPTOR\\|…

But it's generally a lot easier to use a verbatim string literal (@"…"):

var strRegex = @"\vMSH\|\^~\\&\|DATACAPTOR\|…

Finally, note that (\d{4}\d{2}\d{2}\d{2}\d{2}\d{2}) can be simplified to (\d{14}).

However, for a structure like this, it's probably easier to just use the Split method.

var segment = "MSH|^~\&|DATACAPTOR…";
var fields = segment.Split('|');
var timestamp = fields[5];

^{Warning: HL7 messages may use different control characters—starting the 4th character in the MSH segment as a field separator (in this case |^~\& are the control characters). It's best to parse the control characters first if you don't control your input and these control characters may change.}

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