字符串與普通Exprerssion查找指數開始與一些指數

Question

我要搜索的字符串\ n \ n或\ n \ n或\ n \ S * \ n搜索從一些指標開始喜歡

mystring.LastIndexof(regularexpression, mystartindex) 

指數

，同樣

mystring.Indexof(regularexpression, mystartindex) 

Answer 1

使用this Regex.Match overload

public Match Match(
    string input, 
    int startat 
) 

如

Regex re("\n\s*\n"); 
var index = re.Match(mystring, mystartindex).Index; 

注意您可以使用Match.NextMatch拿到下一場比賽：

Match m = Regex.Match(mystring, re); 
while (m.Success) { 
    Console.WriteLine("'{0}' found at index {1}.", m.Value, m.Index); 
    m = m.NextMatch(); 
} 

更新從自己的解決方案改編：

public static class RegexMatchExtensions 
{ 
    public static int LastIndexOf(this MatchCollection matches, int index) 
    { 
     var match = matches.Cast<Match>().LastOrDefault(m => m.Index <= index); 
     return (match == null)? -1 : match.Index; 
    } 

    public static int IndexOf(this MatchCollection matches, int index) 
    { 
     var match = matches.Cast<Match>().FirstOrDefault(m => m.Index > index); 
     return (match == null)? -1 : match.Index; 
    } 
} 

用法：

/*private static readonly*/ Regex re = new Regex(@"\n\s*\n", 
       RegexOptions.Compiled 
      | RegexOptions.Multiline 
      | RegexOptions.CultureInvariant); 

var doubleLineIndexes = re.Matches(wholeDocumentText); 

var first = doubleLineIndexes.IndexOf(73); 
var last = doubleLineIndexes.LastIndexOf(73); 

Answer 2

Regex re = new Regex（@「\ n \ s * \ n」）; MatchCollection DoubleLineIndexes = re.Matches（wholeDocumentText）;

INT LastIndexOf（MatchCollection MATHES，INT指數） {

 for (int i = 0; i < mathes.Count; i++) 
     { 
      if (mathes[i].Index > index) return mathes[i-1].Index; 
     } 

     return -1; 

    } 

    int IndexOf(MatchCollection mathes, int index) 
    { 

     for (int i = 0; i < mathes.Count; i++) 
     { 
      if (mathes[i].Index > index) return mathes[i].Index; 
     } 

     return -1; 

    }