[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AND
和NOT
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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