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C#正则表达式替换无效字符以使其成为完美的浮点数

[英]C# Regex to replace invalid character to make it as perfect float number

例如,如果字符串是“ -234.24234.-23423.344”,则结果应为“ -234.2423423423344”

如果字符串为“ 898.4.44.4”,则结果应为“ 898.4444”

如果字符串为“ -898.4.-”,则结果应为“ -898.4”

结果应始终使场景成为双重类型

我能做的是:

string pattern = String.Format(@"[^\d\{0}\{1}]", 
NumberFormatInfo.CurrentInfo.NumberDecimalSeparator, 
NumberFormatInfo.CurrentInfo.NegativeSign);

string result = Regex.Replace(value, pattern, string.Empty); 
// this will not be able to deal with something like this "-.3-46821721.114.4"

有没有完美的方式来处理这些案件?

使用正则表达式本身来实现您的目标不是一个好主意,因为正则表达式缺少表达的ANDNOT逻辑。

尝试下面的代码,它将执行相同的操作。

var str = @"-.3-46821721.114.4";
var beforeHead = "";
var afterHead = "";

var validHead = new Regex(@"(\d\.)" /* use @"\." if you think "-.5" is also valid*/, RegexOptions.Compiled);
Regex.Replace(str, @"[^0-9\.-]", "");
var match = validHead.Match(str);

beforeHead = str.Substring(0, str.IndexOf(match.Value));
if (beforeHead[0] == '-')
{
    beforeHead = '-' + Regex.Replace(beforeHead, @"[^0-9]", "");
}
else
{
    beforeHead = Regex.Replace(beforeHead, @"[^0-9]", "");
}
afterHead = Regex.Replace(str.Substring(beforeHead.Length + 2 /* 1, if you use \. as head*/), @"[^0-9]", "");

var validFloatNumber = beforeHead + match.Value + afterHead;

字符串必须在操作前修剪。

这可能是个坏主意,但是您可以使用regex来做到这一点:

Regex.Replace(input, @"[^-.0-9]|(?<!^)-|(?<=\..*)\.", "")

正则表达式匹配:

[^-.0-9]    # anything which isn't ., -, or a digit.
|           # or
(?<!^)-     # a - which is not at the start of the string
|           # or
(?<=\..*)\. # a dot which is not the first dot in the string

这适用于您的示例,此外,这种情况下:“ 9-1.1”变为“ 91.1”。

如果您希望将“ asd-8”变为“ -8”而不是“ 8”,也可以将(?<!^)-更改为(?<!^[^-.0-9]*)-

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