C# serialize decimal to xml

user_50227 0
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I got a decimal property, like

[XmlElementAttribute(DataType = "decimal")] decimal Price

The problem is that I wanna force it to serialize always with precision of 2, but if the price is 10.50 it will be serialized to XML like <Price>10.5</Price>.

Theres any way to force it (without creating a new property or changing the get of this property? I'm looking for some way to do this only sending a pattern to the XmlSerializer (or the XmlElementAttribute) or any smart way to do this ?

Thanks

serialization xml c#
3 Answers

I was having the opposite problem. My decimals were serializing with 4 decimal places, even though they were all 4 zeroes. I discovered that if I call decimal.Round(value, 2) then it serializes to 2 decimal places. It would appear that the Decimal type remembers what you last rounded it too when it is serialized.

I was suspicious of the suggestion, but it worked that simply. Even though the value didn't need rounding, calling Round changed how many decimal places showed up in serialization.

user_2550

I had a few ints that I assigned to decimal properties (of the object-to-serialize). They needed to be written with 2 decimals by the XML Serializer, but came out without decimals.

Assuming

var myInt = 3;

No success:

Setting a property that needs serialization via

obj.Property = decimal.Round((decimal)myInt, 2);

results in no decimals being written at serialization.

Success:

With

obj.Property = decimal.Parse(d.ToString("n2", CultureInfo.InvariantCulture), NumberStyles.Any, CultureInfo.InvariantCulture);

I get two decimals.

Now I now use these extension methods at setting the properties:

/// <summary>
/// For XML serialization.
/// </summary>
public static class DecimalSerializationHelper
{
    public static decimal WithTwoDecimals(this decimal d)
        => decimal.Parse(d.ToString("n2", CultureInfo.InvariantCulture), NumberStyles.Any, CultureInfo.InvariantCulture);

    public static decimal WithTwoDecimals(this int x) => ((decimal)x).WithTwoDecimals();
}

so that it reads like

obj.Property = myInt.WithTwoDecimals();
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