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
乘以这个数，然后除以之前乘以的数。
例如,
perc = double.Parse("3.555)*1000;
result = perc/1000
其他回答
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。
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);
}
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。
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)));
我正在开发一个。net Maui应用程序，可以在Windows、Mac、Android和iPhone上运行。我有3个双精度值，我解析和存储使用'。(如。“32.5”)在所有情况下:纬度、经度和海拔。我碰巧将Android和iPhone设置为西班牙语，并注意到Android会解析'。字符串很好。但是，iPhone拒绝正确解析它，除非我用'，'代替'.'。否则，结果总是一个巨大的数字。
我没有处理复杂的本地化问题，而是提出了一个简单的解决方案，即利用双数字的特定限制。
case "Lat":
waypoint.Lat = ParseDouble(xmlVal, 90);
break;
case "Lon":
waypoint.Lon = ParseDouble(xmlVal, 180);
break;
case "Alt":
waypoint.Alt = ParseDouble(xmlVal, 32000);
public static double ParseDouble(string val, double limit)
{
double result;
if (double.TryParse(val, out result))
{
if (Math.Abs(result) <= limit)
return result;
else if (double.TryParse(val.Replace('.', ','), out result))
{
if (Math.Abs(result) <= limit)
return result;
}
}
return 0;
}
