我正在做一些事情,我意识到我想要在一个字符串中找到多少个/s,然后我突然想到,有几种方法可以做到这一点,但不能决定哪种是最好的(或最简单的)。

目前我想说的是:

string source = "/once/upon/a/time/";
int count = source.Length - source.Replace("/", "").Length;

但我一点都不喜欢,有人愿意吗?

我并不想为此挖掘出正则表达式,对吧?

我知道我的字符串将包含我要搜索的项,所以你可以假设…

当然对于长度为> 1的字符串,

string haystack = "/once/upon/a/time";
string needle = "/";
int needleCount = ( haystack.Length - haystack.Replace(needle,"").Length ) / needle.Length;

当前回答

string s = "65 fght 6565 4665 hjk";
int count = 0;
foreach (Match m in Regex.Matches(s, "65"))
  count++;

其他回答

string Name = "Very good nice one is very good but is very good nice one this is called the term";
bool valid=true;
int count = 0;
int k=0;
int m = 0;
while (valid)
{
    k = Name.Substring(m,Name.Length-m).IndexOf("good");
    if (k != -1)
    {
        count++;
        m = m + k + 4;
    }
    else
        valid = false;
}
Console.WriteLine(count + " Times accures");

字符串出现的泛型函数:

public int getNumberOfOccurencies(String inputString, String checkString)
{
    if (checkString.Length > inputString.Length || checkString.Equals("")) { return 0; }
    int lengthDifference = inputString.Length - checkString.Length;
    int occurencies = 0;
    for (int i = 0; i < lengthDifference; i++) {
        if (inputString.Substring(i, checkString.Length).Equals(checkString)) { occurencies++; i += checkString.Length - 1; } }
    return occurencies;
}

对于任何想要使用String扩展方法的人,

以下是我使用的基于张贴的最好的答案:

public static class StringExtension
{    
    /// <summary> Returns the number of occurences of a string within a string, optional comparison allows case and culture control. </summary>
    public static int Occurrences(this System.String input, string value, StringComparison stringComparisonType = StringComparison.Ordinal)
    {
        if (String.IsNullOrEmpty(value)) return 0;

        int count    = 0;
        int position = 0;

        while ((position = input.IndexOf(value, position, stringComparisonType)) != -1)
        {
            position += value.Length;
            count    += 1;
        }

        return count;
    }

    /// <summary> Returns the number of occurences of a single character within a string. </summary>
    public static int Occurrences(this System.String input, char value)
    {
        int count = 0;
        foreach (char c in input) if (c == value) count += 1;
        return count;
    }
}
int count = new Regex(Regex.Escape(needle)).Matches(haystack).Count;

如果你使用的是。net 3.5,你可以用LINQ在一行代码中完成:

int count = source.Count(f => f == '/');

如果你不想使用LINQ,你可以用:

int count = source.Split('/').Length - 1;

您可能会惊讶地发现,您原来的技术似乎比这两种方法都快30% !我刚刚用“/once/upon/a/time/”做了一个快速的基准测试,结果如下:

你的原稿= 12s 源。计数= 19秒 源。分裂= 17秒 Foreach(来自bobwienholt的答案)= 10s

(迭代次数为50,000,000次,因此在现实世界中您不太可能注意到太多差异。)