double.Parse("3.5", CultureInfo.InvariantCulture)

double.Parse("3.5", CultureInfo.InvariantCulture)

如果没有指定要查找的小数分隔符，这是很困难的，但如果你这样做了，这就是我使用的:

    public static double Parse(string str, char decimalSep)
    {
        string s = GetInvariantParseString(str, decimalSep);
        return double.Parse(s, System.Globalization.CultureInfo.InvariantCulture);
    }

    public static bool TryParse(string str, char decimalSep, out double result)
    {
        // NumberStyles.Float | NumberStyles.AllowThousands got from Reflector
        return double.TryParse(GetInvariantParseString(str, decimalSep), NumberStyles.Float | NumberStyles.AllowThousands, System.Globalization.CultureInfo.InvariantCulture, out result);
    }

    private static string GetInvariantParseString(string str, char decimalSep)
    {
        str = str.Replace(" ", "");

        if (decimalSep != '.')
            str = SwapChar(str, decimalSep, '.');

        return str;
    }
    public static string SwapChar(string value, char from, char to)
    {
        if (value == null)
            throw new ArgumentNullException("value");

        StringBuilder builder = new StringBuilder();

        foreach (var item in value)
        {
            char c = item;
            if (c == from)
                c = to;
            else if (c == to)
                c = from;

            builder.Append(c);
        }
        return builder.ToString();
    }

    private static void ParseTestErr(string p, char p_2)
    {
        double res;
        bool b = TryParse(p, p_2, out res);
        if (b)
            throw new Exception();
    }

    private static void ParseTest(double p, string p_2, char p_3)
    {
        double d = Parse(p_2, p_3);
        if (d != p)
            throw new Exception();
    }

    static void Main(string[] args)
    {
        ParseTest(100100100.100, "100.100.100,100", ',');
        ParseTest(100100100.100, "100,100,100.100", '.');
        ParseTest(100100100100, "100.100.100.100", ',');
        ParseTest(100100100100, "100,100,100,100", '.');
        ParseTestErr("100,100,100,100", ',');
        ParseTestErr("100.100.100.100", '.');
        ParseTest(100100100100, "100 100 100 100.0", '.');
        ParseTest(100100100.100, "100 100 100.100", '.');
        ParseTest(100100100.100, "100 100 100,100", ',');
        ParseTest(100100100100, "100 100 100,100", '.');
        ParseTest(1234567.89, "1.234.567,89", ',');    
        ParseTest(1234567.89, "1 234 567,89", ',');    
        ParseTest(1234567.89, "1 234 567.89",     '.');
        ParseTest(1234567.89, "1,234,567.89",    '.');
        ParseTest(1234567.89, "1234567,89",     ',');
        ParseTest(1234567.89, "1234567.89",  '.');
        ParseTest(123456789, "123456789", '.');
        ParseTest(123456789, "123456789", ',');
        ParseTest(123456789, "123.456.789", ',');
        ParseTest(1234567890, "1.234.567.890", ',');
    }

这对任何文化都适用。它正确地无法解析具有多个小数分隔符的字符串，这与replace而不是swap的实现不同。

诀窍是使用不变区域性，来解析所有区域性中的dot。

double.Parse("3.5", System.Globalization.NumberStyles.AllowDecimalPoint, System.Globalization.NumberFormatInfo.InvariantInfo);

看，上面每个建议用常量字符串替换字符串的答案只可能是错误的。为什么?因为你不尊重Windows的区域设置!Windows保证用户可以自由地设置任何分隔符。他/她可以打开控制面板，进入区域面板，点击高级，随时改变角色。甚至在程序运行期间。想想这个。好的解决方案必须意识到这一点。

So, first you will have to ask yourself, where this number is coming from, that you want to parse. If it's coming from input in the .NET Framework no problem, because it will be in the same format. But maybe it was coming from outside, maybe from a external server, maybe from an old DB that only supports string properties. There, the db admin should have given a rule in which format the numbers are to be stored. If you know for example that it will be an US DB with US format you can use this piece of code:

CultureInfo usCulture = new CultureInfo("en-US");
NumberFormatInfo dbNumberFormat = usCulture.NumberFormat;
decimal number = decimal.Parse(db.numberString, dbNumberFormat);

这在世界上任何地方都适用。请不要使用“Convert.ToXxxx”。Convert类只被认为是任意方向转换的基类。此外:您也可以对DateTimes使用类似的机制。

如果数值来自用户输入，我认为100%正确的转换是不可能的。例如，如果值是123.456，它可以是一个分组，也可以是一个小数点。如果你真的需要100%，你必须描述你的格式，如果不正确就抛出异常。

但我改进了JanW的代码，所以我们提前了一点，达到100%。背后的思想是，如果最后一个分隔符是groupseparator，它将更像是一个整数类型，而不是double类型。

添加的代码在GetDouble的第一个if中。

void Main()
{
    List<string> inputs = new List<string>() {
        "1.234.567,89",
        "1 234 567,89",
        "1 234 567.89",
        "1,234,567.89",
        "1234567,89",
        "1234567.89",
        "123456789",
        "123.456.789",
        "123,456,789,"
    };

    foreach(string input in inputs) {
        Console.WriteLine(GetDouble(input,0d));
    }

}

public static double GetDouble(string value, double defaultValue) {
    double result;
    string output;

    // Check if last seperator==groupSeperator
    string groupSep = System.Globalization.CultureInfo.CurrentCulture.NumberFormat.NumberGroupSeparator;
    if (value.LastIndexOf(groupSep) + 4 == value.Count())
    {
        bool tryParse = double.TryParse(value, System.Globalization.NumberStyles.Any, System.Globalization.CultureInfo.CurrentCulture, out result);
        result = tryParse ? result : defaultValue;
    }
    else
    {
        // Unify string (no spaces, only . )
        output = value.Trim().Replace(" ", string.Empty).Replace(",", ".");

        // Split it on points
        string[] split = output.Split('.');

        if (split.Count() > 1)
        {
            // Take all parts except last
            output = string.Join(string.Empty, split.Take(split.Count()-1).ToArray());

            // Combine token parts with last part
            output = string.Format("{0}.{1}", output, split.Last());
        }
        // Parse double invariant
        result = double.Parse(output, System.Globalization.CultureInfo.InvariantCulture);
    }
    return result;
}