只保留数值从字符串？ [英] Keep only numeric value from a string?

问题描述

我有一些字符串这样

string phoneNumber = "[phone]";

我想带出的parethenses和仪表板，在其他的词，仅仅保留数值。

I would like to strip out the parethenses and the dash, in other word just keep the numeric values.

所以输出可能看起来像这样

So the output could look like this

9143951430

我如何获得所需的输出？

How do I get the desired output ?

推荐答案

您请执行下列操作之一：

You do any of the following:

• 使用正则表达式。您可以使用正则表达式为

• Use regular expressions. You can use a regular expression with either

• 一个否定的字符类，定义是你不想要的东西（人物不是小数位数其他那些字符）：

• A negative character class that defines the characters that are what you don't want (those characters other than decimal digits):

private static readonly Regex rxNonDigits = new Regex( @"[^\d]+");

在这种情况下，你可以做采取这些方法的执行：

In which case, you can do take either of these approaches:

// simply replace the offending substrings with an empty string
private string CleanStringOfNonDigits_V1( string s )
{
  if ( string.IsNullOrEmpty(s) ) return s ;
  string cleaned = rxNonDigits.Replace(s, "") ;
  return cleaned ;
}

// split the string into an array of good substrings
// using the bad substrings as the delimiter.
// Then aggregate that into a StringBuilder to get the result.
private string CleanStringOfNonDigits_V2( string s )
{
  if (string.IsNullOrEmpty(s)) return s;
  string cleaned = rxNonDigits
                   .Split(s)
                   .Aggregate(new StringBuilder(), (sb,ss) => sb.Append(ss) )
                   .ToString()
                   ;
  return cleaned ;
}

一个积极的字符集，定义你想要什么（十进制数字）：

a positive character set that defines what you do want (decimal digits):

private static Regex rxDigits = new Regex( @"[\d]+") ;

在这种情况下，你可以做这样的事情：

In which case you can do something like this:

private string CleanStringOfNonDigits_V3( string s )
{
  if ( string.IsNullOrEmpty(s) ) return s ;
  StringBuilder sb = new StringBuilder() ;
  for ( Match m = rxDigits.Match(s) ; m.Success ; m = m.NextMatch() )
  {
    sb.Append(m.Value) ;
  }
  string cleaned = sb.ToString() ;
  return cleaned ;
}

您再不需要使用正则表达式，无论是。

You're not required to use a regular expression, either.

• 您可以直接使用LINQ，因为一个字符串是的IEnumerable<焦炭> ：

private string CleanStringOfNonDigits_V4( string s )
{
  if ( string.IsNullOrEmpty(s) ) return s;
  string cleaned = new string( s.Where( char.IsDigit ).ToArray() ) ;
  return cleaned;
}

如果你只对付西方的字母，唯一的十进制你会看到数字是ASCII，跳过 char.IsDigit 将有可能给你买一个小的性能：

If you're only dealing with western alphabets where the only decimal digits you'll see are ASCII, skipping char.IsDigit will likely buy you a little performance:

private string CleanStringOfNonDigits_V5( string s )
{
  if (string.IsNullOrEmpty(s)) return s;
  string cleaned = new string(s.Where( c => c-'0' < 10 ).ToArray() ) ;
  return cleaned;
}

最后，你可以简单地遍历字符串，夹紧你不想要的数字，像这样的：

Finally, you can simply iterate over the string, chucking the digits you don't want, like this:

private string CleanStringOfNonDigits_V6( string s )
{
  if (string.IsNullOrEmpty(s)) return s;
  StringBuilder sb = new StringBuilder(s.Length) ;
  for (int i = 0; i < s.Length; ++i)
  {
    char c = s[i];
    if ( c < '0' ) continue ;
    if ( c > '9' ) continue ;
    sb.Append(s[i]);
  }
  string cleaned = sb.ToString();
  return cleaned;
}

或者这样的：

private string CleanStringOfNonDigits_V7(string s)
{
  if (string.IsNullOrEmpty(s)) return s;
  StringBuilder sb = new StringBuilder(s);
  int j = 0 ;
  int i = 0 ;
  while ( i < sb.Length )
  {
    bool isDigit = char.IsDigit( sb[i] ) ;
    if ( isDigit )
    {
      sb[j++] = sb[i++];
    }
    else
    {
      ++i ;
    }
  }
  sb.Length = j;
  string cleaned = sb.ToString();
  return cleaned;
}

从一个角度看清晰度和代码清洁，版本1是你想要的。它是很难被击败一个衬垫。

From a standpoint of clarity and cleanness of code, the version 1 is what you want. It's hard to beat a one liner.

如果性能问题，我怀疑是版本7，最后的版本，是赢家。它会创建一个临时—一个的StringBuilder（）并执行就地StringBuilder中的就地缓冲区内的改造。

If performance matters, my suspicion is that the version 7, the last version, is the winner. It creates one temporary — a StringBuilder() and does the transformation in-place within the StringBuilder's in-place buffer.

其他的选项所有做更多的工作。

The other options all do more work.

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