Extracting a value from Json object in C#

Question

I have a Json document and I am trying to get the value for AnalogInput for each channel from 1 to 4. I have tried this code:

 JObject originalObject = JObject.Parse(testJsonObject);
 var analogInputTrueValues = originalObject.Descendants().OfType<JProperty>().Where(p => p.Name == "DigitalInput").Select(x => x.Value).ToArray();

where testJsonObject is the Json file that gets loaded by another method.

Debugging the code, the value for analogInputTrueValues is:

{Newtonsoft.Json.Linq.JToken[4]}
    [0]: {13}
    [1]: {13}
    [2]: {14}
    [3]: {14}

,which is correct. but I am interested to have an array or a list like {"13","13","14","14"}. This is where I can not move forward since I can not extract those exact values and have them in a list or an array. Even when I do:

digitalInputTrueValues.GetValue(0)
{13}
    base: {13}
    HasValues: false
    Type: String
    Value: "13"

I can't extract the Value, which is what I am interested in. How can I get around this kind of problem and extract my desired values? The object that I am working with is as follows:

{
        "module": {
            "serial": "3",
            "label": "A",
            "lat": "B",
            "long": "C",
            "channels": [
{"channel": "1", "label": "Channel 1", "AnalogInput": "13", "AnalogInputRaw": "13", "AnalogInputScale": "Raw", "DigitalInput": "Off"},
{"channel": "2", "label": "Channel 2", "AnalogInput": "13", "AnalogInputRaw": "13", "AnalogInputScale": "Raw", "DigitalInput": "On"},
{"channel": "3", "label": "Channel 3", "AnalogInput": "14", "AnalogInputRaw": "14", "AnalogInputScale": "Raw", "DigitalInput": "On"},
{"channel": "4", "label": "Channel 4", "AnalogInput": "14", "AnalogInputRaw": "14", "AnalogInputScale": "Raw", "DigitalInput": "On"}
            ],
            "variables": [
             {"1": "0"},
             {"2": "0"},
             {"3": "1"},
             {"4": "0"}
            ]
        }
}

Brian Rogers · Accepted Answer · 2014-10-30 00:55:46Z

2

You just need to include ToString() in your Select expression after x.Value:

JObject originalObject = JObject.Parse(json);
var analogInputTrueValues = originalObject.Descendants()
                                          .OfType<JProperty>()
                                          .Where(p => p.Name == "AnalogInput")
                                          .Select(x => x.Value.ToString())
                                          .ToArray();

Working example: https://dotnetfiddle.net/tU5Mc8

Alternative method using strongly-typed classes: https://dotnetfiddle.net/US4Bs0

edited Oct 30, 2014 at 0:55

answered Oct 30, 2014 at 0:30

Brian Rogers

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3 Comments

Payam Over a year ago

Thanks, it worked. Is the way that I am extracting the AnalogInput value a good way? Is there a better way to do it?

Brian Rogers Over a year ago

Using LINQ-to-JSON, the way you are doing, is correct and good. An alternative method would be to create strongly-typed classes corresponding to your JSON and then use JsonConvert.DeserializeObject() to deserialize the JSON into your classes. It comes down to personal preference.

Brian Rogers Over a year ago

I added some fiddles to my answer so you can compare the difference between the two methods. Using LINQ-to-JSON is much less code, but some people find the strongly typed classes easier to work with.

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