我想从字符串中删除所有特殊字符。允许输入A-Z(大写或小写)、数字(0-9)、下划线(_)或点符号(.)。

我有以下,它是有效的,但我怀疑(我知道!)它不是很有效:

    public static string RemoveSpecialCharacters(string str)
    {
        StringBuilder sb = new StringBuilder();
        for (int i = 0; i < str.Length; i++)
        {
            if ((str[i] >= '0' && str[i] <= '9')
                || (str[i] >= 'A' && str[i] <= 'z'
                    || (str[i] == '.' || str[i] == '_')))
                {
                    sb.Append(str[i]);
                }
        }

        return sb.ToString();
    }

最有效的方法是什么?正则表达式是什么样子的,它与普通字符串操作相比如何?

要清洗的字符串相当短,长度通常在10到30个字符之间。


当前回答

HashSet是O(1) 不确定它是否比现有的比较快

private static HashSet<char> ValidChars = new HashSet<char>() { 'a', 'b', 'c', 'A', 'B', 'C', '1', '2', '3', '_' };
public static string RemoveSpecialCharacters(string str)
{
    StringBuilder sb = new StringBuilder(str.Length / 2);
    foreach (char c in str)
    {
        if (ValidChars.Contains(c)) sb.Append(c);
    }
    return sb.ToString();
}

我测试了,这并不比公认的答案快。 如果你需要一组可配置的字符,我会把它留在这里,这将是一个很好的解决方案。

其他回答

I had to do something similar for work, but in my case I had to filter all that is not a letter, number or whitespace (but you could easily modify it to your needs). The filtering is done client-side in JavaScript, but for security reasons I am also doing the filtering server-side. Since I can expect most of the strings to be clean, I would like to avoid copying the string unless I really need to. This let my to the implementation below, which should perform better for both clean and dirty strings.

public static string EnsureOnlyLetterDigitOrWhiteSpace(string input)
{
    StringBuilder cleanedInput = null;
    for (var i = 0; i < input.Length; ++i)
    {
        var currentChar = input[i];
        var charIsValid = char.IsLetterOrDigit(currentChar) || char.IsWhiteSpace(currentChar);

        if (charIsValid)
        {
            if(cleanedInput != null)
                cleanedInput.Append(currentChar);
        }
        else
        {
            if (cleanedInput != null) continue;
            cleanedInput = new StringBuilder();
            if (i > 0)
                cleanedInput.Append(input.Substring(0, i));
        }
    }

    return cleanedInput == null ? input : cleanedInput.ToString();
}

我不确定这是最有效的方法,但对我来说很有效

 Public Function RemoverTildes(stIn As String) As String
    Dim stFormD As String = stIn.Normalize(NormalizationForm.FormD)
    Dim sb As New StringBuilder()

    For ich As Integer = 0 To stFormD.Length - 1
        Dim uc As UnicodeCategory = CharUnicodeInfo.GetUnicodeCategory(stFormD(ich))
        If uc <> UnicodeCategory.NonSpacingMark Then
            sb.Append(stFormD(ich))
        End If
    Next
    Return (sb.ToString().Normalize(NormalizationForm.FormC))
End Function
public string RemoveSpecial(string evalstr)
{
StringBuilder finalstr = new StringBuilder();
            foreach(char c in evalstr){
            int charassci = Convert.ToInt16(c);
            if (!(charassci >= 33 && charassci <= 47))// special char ???
             finalstr.append(c);
            }
return finalstr.ToString();
}

另一种试图通过减少分配来提高性能的方法,特别是在多次调用此函数的情况下。

它之所以有效,是因为可以保证结果不会比输入长,因此可以在不在内存中创建额外副本的情况下传递输入和输出。因此,您不能使用stackalloc来创建缓冲区数组,因为这需要从缓冲区中复制一个副本。

public static string RemoveSpecialCharacters(this string str)
{
    return RemoveSpecialCharacters(str.AsSpan()).ToString();
}

public static ReadOnlySpan<char> RemoveSpecialCharacters(this ReadOnlySpan<char> str)
{
    Span<char> buffer = new char[str.Length];
    int idx = 0;

    foreach (char c in str)
    {
        if (char.IsLetterOrDigit(c))
        {
            buffer[idx] = c;
            idx++;
        }
    }

    return buffer.Slice(0, idx);
}

HashSet是O(1) 不确定它是否比现有的比较快

private static HashSet<char> ValidChars = new HashSet<char>() { 'a', 'b', 'c', 'A', 'B', 'C', '1', '2', '3', '_' };
public static string RemoveSpecialCharacters(string str)
{
    StringBuilder sb = new StringBuilder(str.Length / 2);
    foreach (char c in str)
    {
        if (ValidChars.Contains(c)) sb.Append(c);
    }
    return sb.ToString();
}

我测试了,这并不比公认的答案快。 如果你需要一组可配置的字符,我会把它留在这里,这将是一个很好的解决方案。