Question 109992 in Json Question posted in
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Json – How can I best limit the decimal places to 5 or shorten the longer ones?

I have a file with various characters and either words as well as numbers. These numbers can be integers like a 1 or 12 (as an example) or they have a comma and countless digits after the comma.
for example:

",{"n":"Box","p":[-4.0,4.0,0.0],"r":[270.0,0.0,0.0],"s":[1.0,1.000006,1.000006],"c":[0.448529363,0.4280135,0.412251264],"m":"wood_14"}"

The file is read with File.ReadAllText and then passed to NewtonsoftJson via JsonProperty accordingly
for example:

[JsonProperty("s")]
public double[] Scale
{
    get;
    set;
}

For example, with Scale, I want to limit the decimal places to a maximum of 5 digits.
Is this possible and if so, how?
I have been trying for days with different things, but nothing has worked yet.

My idea was to intervene in the string I build first before passing that to Json there. But unfortunately this does not work. I tried it with regular expression like Notepadd++ makes for example.

(edit)

Tags:

2

Answers


  1. 0

    double.ToString() will do it.

    const double d = 0.123456789;
string s = d.ToString("G5");
Console.WriteLine(s);

    That outputs 0.12346, showing that it rounds the 5 to 6.

    Documentation for Double.ToString shows much more detailed examples of how to format numbers.

    The number format (i.e. decimal separator) is culture-specific. If you want to use the current culture, then the above will work. If you always want a comma as the decimal separator, you’ll have to call the overload that accepts an IFormatProvider for the specific culture.

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  2. 0

    I don’t know if you want it to serialize or deserialize. So I made code for both cases. You have to leave one variant or another

    public class a
{
    private double[] val;
    
    [JsonProperty("s")]
    public double[] Scale
    {

        get {    return val;
                 // or
                 val?.Select(v => Math.Round(v, 5)).ToArray();  }
        set { 
                 val=value;
                 //or
                val = value?.Select(v => Math.Round(v, 5)).ToArray(); 
        }
    }
}

    if you want it in many properties you can make it as a function

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