我想把“3.5”这样的字符串解析为double。然而,
double.Parse("3.5")
收益率为35和
double.Parse("3.5", System.Globalization.NumberStyles.AllowDecimalPoint)
抛出一个FormatException。
现在我的计算机的区域设置为德语，其中逗号用作小数分隔符。它可能需要做一些事情，double.Parse()期望“3,5”作为输入，但我不确定。
如何解析包含十进制数字的字符串，该十进制数字可能是或可能不是我当前地区指定的格式?
当前回答
var doublePattern = @"(?<integer>[0-9]+)(?:\,|\.)(?<fraction>[0-9]+)";
var sourceDoubleString = "03444,44426";
var match = Regex.Match(sourceDoubleString, doublePattern);
var doubleResult = match.Success ? double.Parse(match.Groups["integer"].Value) + (match.Groups["fraction"].Value == null ? 0 : double.Parse(match.Groups["fraction"].Value) / Math.Pow(10, match.Groups["fraction"].Value.Length)): 0;
Console.WriteLine("Double of string '{0}' is {1}", sourceDoubleString, doubleResult);
其他回答
下面的方法效率较低，但我使用这种逻辑。这只在小数点后有两位数字时有效。
double val;
if (temp.Text.Split('.').Length > 1)
{
val = double.Parse(temp.Text.Split('.')[0]);
if (temp.Text.Split('.')[1].Length == 1)
val += (0.1 * double.Parse(temp.Text.Split('.')[1]));
else
val += (0.01 * double.Parse(temp.Text.Split('.')[1]));
}
else
val = double.Parse(RR(temp.Text));
我对这个话题的看法，试图提供一个通用的，双重转换方法:
private static double ParseDouble(object value)
{
double result;
string doubleAsString = value.ToString();
IEnumerable<char> doubleAsCharList = doubleAsString.ToList();
if (doubleAsCharList.Where(ch => ch == '.' || ch == ',').Count() <= 1)
{
double.TryParse(doubleAsString.Replace(',', '.'),
System.Globalization.NumberStyles.Any,
CultureInfo.InvariantCulture,
out result);
}
else
{
if (doubleAsCharList.Where(ch => ch == '.').Count() <= 1
&& doubleAsCharList.Where(ch => ch == ',').Count() > 1)
{
double.TryParse(doubleAsString.Replace(",", string.Empty),
System.Globalization.NumberStyles.Any,
CultureInfo.InvariantCulture,
out result);
}
else if (doubleAsCharList.Where(ch => ch == ',').Count() <= 1
&& doubleAsCharList.Where(ch => ch == '.').Count() > 1)
{
double.TryParse(doubleAsString.Replace(".", string.Empty).Replace(',', '.'),
System.Globalization.NumberStyles.Any,
CultureInfo.InvariantCulture,
out result);
}
else
{
throw new ParsingException($"Error parsing {doubleAsString} as double, try removing thousand separators (if any)");
}
}
return result;
}
与预期工作:
1.1 1, 1 1000000000 1.000.000.000 1000000000.99 1.000.000.000, 99 5000111.3 5.000.111, 3 0.99, 000111,88 0,99.000. 111.88
没有实现默认转换，因此试图解析1.3、14、1、3.14或类似情况会失败。
string testString1 = "2,457";
string testString2 = "2.457";
double testNum = 0.5;
char decimalSepparator;
decimalSepparator = testNum.ToString()[1];
Console.WriteLine(double.Parse(testString1.Replace('.', decimalSepparator).Replace(',', decimalSepparator)));
Console.WriteLine(double.Parse(testString2.Replace('.', decimalSepparator).Replace(',', decimalSepparator)));
我不能写评论，所以我在这里写:
double.Parse("3.5"， CultureInfo.InvariantCulture)不是一个好主意，因为在加拿大我们写3,5而不是3.5，这个函数的结果是35。
我在自己的电脑上进行了测试:
double.Parse("3.5", CultureInfo.InvariantCulture) --> 3.5 OK
double.Parse("3,5", CultureInfo.InvariantCulture) --> 35 not OK
这是Pierre-Alain Vigeant提到的正确方法
public static double GetDouble(string value, double defaultValue)
{
double result;
// Try parsing in the current culture
if (!double.TryParse(value, System.Globalization.NumberStyles.Any, CultureInfo.CurrentCulture, out result) &&
// Then try in US english
!double.TryParse(value, System.Globalization.NumberStyles.Any, CultureInfo.GetCultureInfo("en-US"), out result) &&
// Then in neutral language
!double.TryParse(value, System.Globalization.NumberStyles.Any, CultureInfo.InvariantCulture, out result))
{
result = defaultValue;
}
return result;
}
下面的代码在任何场景下都可以完成这项工作。这是一点解析。
List<string> inputs = new List<string>()
{
"1.234.567,89",
"1 234 567,89",
"1 234 567.89",
"1,234,567.89",
"123456789",
"1234567,89",
"1234567.89",
};
string output;
foreach (string input in inputs)
{
// Unify string (no spaces, only .)
output = input.Trim().Replace(" ", "").Replace(",", ".");
// Split it on points
string[] split = output.Split('.');
if (split.Count() > 1)
{
// Take all parts except last
output = string.Join("", split.Take(split.Count()-1).ToArray());
// Combine token parts with last part
output = string.Format("{0}.{1}", output, split.Last());
}
// Parse double invariant
double d = double.Parse(output, CultureInfo.InvariantCulture);
Console.WriteLine(d);
}
