用正则表达式拆分字符串并保留分隔符 [英] Splitting a string with Regex and keep the separator
问题描述
每当存在一个分隔符字符串时，我需要分割一个字符串（让我们看看
sey #Key（val），其中val是一个变量）并将其重新加入到正确的
$ b $中b进行一些处理后的订单。
有没有办法使用Regex.Split函数在#Key（val）出现时拆分字符串
但是这样可以保持#Key（val）
的出现，我可以在对每个令牌执行某些操作后重构最终字符串（我需要基本上转换每个令牌） />
将字符串转换为字符数组但我需要以不同的方式执行此操作
是字符串是#Key（val）
谢谢。
Andrea
I need to split a string whenever a separator string is present (lets
sey #Key(val) where val is a variable) and rejoin it in the proper
order after doing some processing.
Is there a way to use the Regex.Split function to split the string
whenever the #Key(val) occurrs but that keeps the #Key(val)
occurrences to that I can reconstruct the final string after doing
certain operations on each token (I need to basically convert each
string into an array of characters but I need to do this differently
is the string is a #Key(val)
Thanks.
Andrea
推荐答案
* na ***@community.nosp 上午写道，2007年4月6日23:16：
* na***@community.nospam wrote, On 4-6-2007 23:16:
我需要在分隔符字符串时拆分字符串存在（允许
sey #Key（val），其中val是变量）和rejo在做了一些处理之后用适当的
命令。
有没有办法使用Regex.Split函数来分割字符串
每当#Key（val）出现时
但是保持#Key（val）
出现的次数我可以在执行某些操作后重建最终字符串
在每个令牌上（我需要基本上将每个
字符串转换为字符数组但我需要以不同方式执行此操作
是字符串是#Key（val）
谢谢。
Andrea
I need to split a string whenever a separator string is present (lets
sey #Key(val) where val is a variable) and rejoin it in the proper
order after doing some processing.
Is there a way to use the Regex.Split function to split the string
whenever the #Key(val) occurrs but that keeps the #Key(val)
occurrences to that I can reconstruct the final string after doing
certain operations on each token (I need to basically convert each
string into an array of characters but I need to do this differently
is the string is a #Key(val)
Thanks.
Andrea
此代码以不同方式分割每一行：
正则表达式rx =新正则表达式（@"（？=＃\ w + \（\ w + \））|（？< =＃\ w + \（\ w + \\ \\））"，
RegexOptions.None）;
string [] arr = rx.Split（输入）;
它寻找模式开头或结尾的每一点
你正在寻找。我认为它不是大输入的最快，但是我没有测试过
。
你可能会更好地尝试使用MatchEvaluator还有一个很好的
写的替换电话，但是为了帮助你，我需要更多的信息来确定你将要做什么样的字符串操作。
Jesse
This code splits each line differently:
Regex rx = new Regex(@"(?=#\w+\(\w+\))|(?<=#\w+\(\w+\))",
RegexOptions.None);
string[] arr = rx.Split(input);
It looks for every point at the beginning or the end of the pattern
you''re looking for. it isn''t the fastest on large inputs I guess, but I
haven''t tested.
You might be better off experimenting with a MatchEvaluator and a well
written replace call, but to help you with that I''d need a little more
info on what kind of string manipulation you''d be doing.
Jesse
* na ** *@community.nosp 上午写道，在4-6-2007 23:16：
* na***@community.nospam wrote, On 4-6-2007 23:16:
每当有一个分隔符字符串时我需要拆分一个字符串（允许
sey #Key（val），其中val是一个变量）并在进行一些处理后以适当的
顺序重新加入。
有没有办法使用Regex.Split函数在#Key（val）出现时拆分字符串
但保持#Key（val）
出现之后，我可以在e上执行某些操作后重建最终字符串ach令牌（我需要基本上将每个
字符串转换为字符数组但我需要以不同方式执行此操作
字符串是#Key（val）
谢谢。
Andrea
I need to split a string whenever a separator string is present (lets
sey #Key(val) where val is a variable) and rejoin it in the proper
order after doing some processing.
Is there a way to use the Regex.Split function to split the string
whenever the #Key(val) occurrs but that keeps the #Key(val)
occurrences to that I can reconstruct the final string after doing
certain operations on each token (I need to basically convert each
string into an array of characters but I need to do this differently
is the string is a #Key(val)
Thanks.
Andrea
这应该更好：
Regex rx2 = new
正则表达式（@"（？< keyval>＃\ w + \（\ w + \））|（？< other>（ （？！＃\ w + \（\ w + \））。）*）"，
RegexOptions.None）;
string result = rx2。替换（输入，新的
MatchEvaluator（ManipulateString））;
私有字符串ManipulateString（匹配目标）
{
if（target.Groups [" keyval"]。成功）
{
返回ManupulateKeyVal（target.Groups [" keyval" ] .Value）;
}
else if（target.Groups [" other"]。Success）
{
返回ManupulateOther（target.Groups [" other"]。Value）;
}
}
这将传递找到的碎片以便操作函数和
完成后将结果传递给新的字符串。
亲切的问候，
Jesse
This should work even better:
Regex rx2 = new
Regex(@"(?<keyval>#\w+\(\w+\))|(?<other>((?!#\w+\( \w+\)).)*)",
RegexOptions.None);
string result = rx2.Replace("input", new
MatchEvaluator(ManipulateString));
private string ManipulateString(Match target)
{
if (target.Groups["keyval"].Success)
{
return ManupulateKeyVal(target.Groups["keyval"].Value);
}
else if (target.Groups["other"].Success)
{
return ManupulateOther(target.Groups["other"].Value);
}
}
This will pass the found pieces in order to the manipulate function and
pass the result into a new string when done.
Kind regards,
Jesse
嗨Andrea，
我不确定我是否完全理解你的问题。如果Jesse的回复有帮助，请您告诉我们
吗？谢谢。
问候，
Walter Wang（wa****@online.microsoft.com，删除''在线。''）
Microsoft在线社区支持
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Hi Andrea,
I''m not sure if I fully understand your question. Would you please let us
know if Jesse''s reply helps? Thanks.
Regards,
Walter Wang (wa****@online.microsoft.com, remove ''online.'')
Microsoft Online Community Support
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