C# get value from deserialized json object
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I'm currently Deserializing a json string using the Newtonsoft.Json nuget packet using the following code:
var data = (JObject)JsonConvert.DeserializeObject(json);
Now I'm receiving an object in the following format:
{{ "meta": { "rap": 2098, "count": 5 }, "data": [ { "name": "Gold Tetramino of Mastery", "rap": 735, "uaid": "16601901", "link": "https://www.roblox.com/Gold-Tetramino-of-Mastery-item?id=5786047", "img": "https://t4.rbxcdn.com/081337d7ea86e6a406512aaa83bbcdeb", "serial": "---", "count": 1 }, { "name": "Silver Tetramino of Accomplishment", "rap": 385, "uaid": "16601900", "link": "https://www.roblox.com/Silver-Tetramino-of-Accomplishment-item?id=5786026", "img": "https://t1.rbxcdn.com/60da69cd76f8dad979326f63f4a5b657", "serial": "---", "count": 1 }, { "name": "Subzero Ski Specs", "rap": 370, "uaid": "155175547", "link": "https://www.roblox.com/Subzero-Ski-Specs-item?id=19644587", "img": "https://t4.rbxcdn.com/8ead2b0418ef418c7650d34103d39b6d", "serial": "---", "count": 1 }, { "name": "Rusty Tetramino of Competence", "rap": 319, "uaid": "16601899", "link": "https://www.roblox.com/Rusty-Tetramino-of-Competence-item?id=5785985", "img": "https://t2.rbxcdn.com/968ad11ee2f4ee0861ae511c419148c8", "serial": "---", "count": 1 }, { "name": "Bluesteel Egg of Genius", "rap": 289, "uaid": "16601902", "link": "https://www.roblox.com/Bluesteel-Egg-of-Genius-item?id=1533893", "img": "https://t7.rbxcdn.com/48bf59fe531dd1ff155e455367e52e73", "serial": "---", "count": 1 } ]}}
Now I'm trying to get the following value from it:
"rap": 2098,
I just need 2098 and I've been trying the following code:
string rap = data["rap"].Value<string>();
But sadly this wouldn't work. Does anyone have a idea how to get the value?
c# json
add a comment |
I'm currently Deserializing a json string using the Newtonsoft.Json nuget packet using the following code:
var data = (JObject)JsonConvert.DeserializeObject(json);
Now I'm receiving an object in the following format:
{{ "meta": { "rap": 2098, "count": 5 }, "data": [ { "name": "Gold Tetramino of Mastery", "rap": 735, "uaid": "16601901", "link": "https://www.roblox.com/Gold-Tetramino-of-Mastery-item?id=5786047", "img": "https://t4.rbxcdn.com/081337d7ea86e6a406512aaa83bbcdeb", "serial": "---", "count": 1 }, { "name": "Silver Tetramino of Accomplishment", "rap": 385, "uaid": "16601900", "link": "https://www.roblox.com/Silver-Tetramino-of-Accomplishment-item?id=5786026", "img": "https://t1.rbxcdn.com/60da69cd76f8dad979326f63f4a5b657", "serial": "---", "count": 1 }, { "name": "Subzero Ski Specs", "rap": 370, "uaid": "155175547", "link": "https://www.roblox.com/Subzero-Ski-Specs-item?id=19644587", "img": "https://t4.rbxcdn.com/8ead2b0418ef418c7650d34103d39b6d", "serial": "---", "count": 1 }, { "name": "Rusty Tetramino of Competence", "rap": 319, "uaid": "16601899", "link": "https://www.roblox.com/Rusty-Tetramino-of-Competence-item?id=5785985", "img": "https://t2.rbxcdn.com/968ad11ee2f4ee0861ae511c419148c8", "serial": "---", "count": 1 }, { "name": "Bluesteel Egg of Genius", "rap": 289, "uaid": "16601902", "link": "https://www.roblox.com/Bluesteel-Egg-of-Genius-item?id=1533893", "img": "https://t7.rbxcdn.com/48bf59fe531dd1ff155e455367e52e73", "serial": "---", "count": 1 } ]}}
Now I'm trying to get the following value from it:
"rap": 2098,
I just need 2098 and I've been trying the following code:
string rap = data["rap"].Value<string>();
But sadly this wouldn't work. Does anyone have a idea how to get the value?
c# json
Parse it instead of deserializing if you only need one value
– Make StackOverflow Good Again
Jan 19 '17 at 22:33
add a comment |
I'm currently Deserializing a json string using the Newtonsoft.Json nuget packet using the following code:
var data = (JObject)JsonConvert.DeserializeObject(json);
Now I'm receiving an object in the following format:
{{ "meta": { "rap": 2098, "count": 5 }, "data": [ { "name": "Gold Tetramino of Mastery", "rap": 735, "uaid": "16601901", "link": "https://www.roblox.com/Gold-Tetramino-of-Mastery-item?id=5786047", "img": "https://t4.rbxcdn.com/081337d7ea86e6a406512aaa83bbcdeb", "serial": "---", "count": 1 }, { "name": "Silver Tetramino of Accomplishment", "rap": 385, "uaid": "16601900", "link": "https://www.roblox.com/Silver-Tetramino-of-Accomplishment-item?id=5786026", "img": "https://t1.rbxcdn.com/60da69cd76f8dad979326f63f4a5b657", "serial": "---", "count": 1 }, { "name": "Subzero Ski Specs", "rap": 370, "uaid": "155175547", "link": "https://www.roblox.com/Subzero-Ski-Specs-item?id=19644587", "img": "https://t4.rbxcdn.com/8ead2b0418ef418c7650d34103d39b6d", "serial": "---", "count": 1 }, { "name": "Rusty Tetramino of Competence", "rap": 319, "uaid": "16601899", "link": "https://www.roblox.com/Rusty-Tetramino-of-Competence-item?id=5785985", "img": "https://t2.rbxcdn.com/968ad11ee2f4ee0861ae511c419148c8", "serial": "---", "count": 1 }, { "name": "Bluesteel Egg of Genius", "rap": 289, "uaid": "16601902", "link": "https://www.roblox.com/Bluesteel-Egg-of-Genius-item?id=1533893", "img": "https://t7.rbxcdn.com/48bf59fe531dd1ff155e455367e52e73", "serial": "---", "count": 1 } ]}}
Now I'm trying to get the following value from it:
"rap": 2098,
I just need 2098 and I've been trying the following code:
string rap = data["rap"].Value<string>();
But sadly this wouldn't work. Does anyone have a idea how to get the value?
c# json
I'm currently Deserializing a json string using the Newtonsoft.Json nuget packet using the following code:
var data = (JObject)JsonConvert.DeserializeObject(json);
Now I'm receiving an object in the following format:
{{ "meta": { "rap": 2098, "count": 5 }, "data": [ { "name": "Gold Tetramino of Mastery", "rap": 735, "uaid": "16601901", "link": "https://www.roblox.com/Gold-Tetramino-of-Mastery-item?id=5786047", "img": "https://t4.rbxcdn.com/081337d7ea86e6a406512aaa83bbcdeb", "serial": "---", "count": 1 }, { "name": "Silver Tetramino of Accomplishment", "rap": 385, "uaid": "16601900", "link": "https://www.roblox.com/Silver-Tetramino-of-Accomplishment-item?id=5786026", "img": "https://t1.rbxcdn.com/60da69cd76f8dad979326f63f4a5b657", "serial": "---", "count": 1 }, { "name": "Subzero Ski Specs", "rap": 370, "uaid": "155175547", "link": "https://www.roblox.com/Subzero-Ski-Specs-item?id=19644587", "img": "https://t4.rbxcdn.com/8ead2b0418ef418c7650d34103d39b6d", "serial": "---", "count": 1 }, { "name": "Rusty Tetramino of Competence", "rap": 319, "uaid": "16601899", "link": "https://www.roblox.com/Rusty-Tetramino-of-Competence-item?id=5785985", "img": "https://t2.rbxcdn.com/968ad11ee2f4ee0861ae511c419148c8", "serial": "---", "count": 1 }, { "name": "Bluesteel Egg of Genius", "rap": 289, "uaid": "16601902", "link": "https://www.roblox.com/Bluesteel-Egg-of-Genius-item?id=1533893", "img": "https://t7.rbxcdn.com/48bf59fe531dd1ff155e455367e52e73", "serial": "---", "count": 1 } ]}}
Now I'm trying to get the following value from it:
"rap": 2098,
I just need 2098 and I've been trying the following code:
string rap = data["rap"].Value<string>();
But sadly this wouldn't work. Does anyone have a idea how to get the value?
c# json
c# json
edited Jan 20 '17 at 0:06
Peter Ritchie
30.1k97192
30.1k97192
asked Jan 19 '17 at 22:31
d4ned4ne
1612
1612
Parse it instead of deserializing if you only need one value
– Make StackOverflow Good Again
Jan 19 '17 at 22:33
add a comment |
Parse it instead of deserializing if you only need one value
– Make StackOverflow Good Again
Jan 19 '17 at 22:33
Parse it instead of deserializing if you only need one value
– Make StackOverflow Good Again
Jan 19 '17 at 22:33
Parse it instead of deserializing if you only need one value
– Make StackOverflow Good Again
Jan 19 '17 at 22:33
add a comment |
8 Answers
8
active
oldest
votes
Try:
var result = data["meta"]["rap"].Value<int>();
or
var result = data.SelectToken("meta.rap").ToString();
or if you don't want to want to pass the whole path in, you could just search for the property like this:
var result = data.Descendants()
.OfType<JProperty>()
.FirstOrDefault(x => x.Name == "rap")
?.Value;
add a comment |
Instead of declaring as type var and letting the compiler sort it, declare as a dynamic and using the Parse method.
dynamic data = JArray.Parse(json);
Then try
data.meta.rap
To get the internal rap object.
I edited from using the deserializeobject method as i incorrectly thought that had the dynamic return type. See here on json.net documentation for more details: http://www.newtonsoft.com/json/help/html/QueryJsonDynamic.htm
There is noParse
method onJsonConvert
.
– William
Jan 19 '17 at 23:08
@billisphere thanks, should have triple checked..!
– Chris Watts
Jan 20 '17 at 6:18
1
As suggested by stackoverflow.com/questions/47134936/…, I'm usingdynamic data = JObject.Parse(json);
, which seems to work without exceptions.
– Jari Turkia
Nov 20 '18 at 9:40
add a comment |
var jobject = (JObject)JsonConvert.DeserializeObject(json);
var jvalue = (JValue)jobject["meta"]["rap"];
Console.WriteLine(jvalue.Value); // 2098
add a comment |
The value is actually an int
type. Try:
int rap = data["rap"].Value<int>();
add a comment |
string rap = JsonConvert.DeserializeObject<dynamic>(json).meta.rap;
Console.WriteLine(rap); // 2098
If you're not into dynamic
(or aren't using .NET 4+), I like how this other answer relies solely on Json.NET's API.
add a comment |
Just use dynamic representation of object:
dynamic obj = JsonConvert.DeserializeObject(json)
var value = obj.meta.rap;
JObject easily convertable to dynamic type itself. You can either get string or int from this value:
var ivalue = (int)obj.meta.rap;
var svalue = (string)obj.meta.rap;
add a comment |
Try to use as following
var test = JsonConvert.DeserializeObject<dynamic>(param);
var testDTO = new TPRDTO();
testDTO.TPR_ID = test.TPR_ID.Value;
Note: For using of JsonConvert
class you have to install Newton-Soft from your package-manager
add a comment |
the problem is that you are casting the deserialized json into a JObject. if you want to have the JObject then simple do this:
JObject.Parse(json);
then you have the JObject and you can access a specific path (for extracting value see this )
you have also another option which is to deserialize your json into a class that you have in your code like this:
var instanceOFTheClass = JsonConvert.DeserializeObject<YourClassName>(json);
with the above code you can access any property and values you want.
add a comment |
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8 Answers
8
active
oldest
votes
8 Answers
8
active
oldest
votes
active
oldest
votes
active
oldest
votes
Try:
var result = data["meta"]["rap"].Value<int>();
or
var result = data.SelectToken("meta.rap").ToString();
or if you don't want to want to pass the whole path in, you could just search for the property like this:
var result = data.Descendants()
.OfType<JProperty>()
.FirstOrDefault(x => x.Name == "rap")
?.Value;
add a comment |
Try:
var result = data["meta"]["rap"].Value<int>();
or
var result = data.SelectToken("meta.rap").ToString();
or if you don't want to want to pass the whole path in, you could just search for the property like this:
var result = data.Descendants()
.OfType<JProperty>()
.FirstOrDefault(x => x.Name == "rap")
?.Value;
add a comment |
Try:
var result = data["meta"]["rap"].Value<int>();
or
var result = data.SelectToken("meta.rap").ToString();
or if you don't want to want to pass the whole path in, you could just search for the property like this:
var result = data.Descendants()
.OfType<JProperty>()
.FirstOrDefault(x => x.Name == "rap")
?.Value;
Try:
var result = data["meta"]["rap"].Value<int>();
or
var result = data.SelectToken("meta.rap").ToString();
or if you don't want to want to pass the whole path in, you could just search for the property like this:
var result = data.Descendants()
.OfType<JProperty>()
.FirstOrDefault(x => x.Name == "rap")
?.Value;
edited Jan 19 '17 at 23:12
answered Jan 19 '17 at 22:35
StuartStuart
2,6711016
2,6711016
add a comment |
add a comment |
Instead of declaring as type var and letting the compiler sort it, declare as a dynamic and using the Parse method.
dynamic data = JArray.Parse(json);
Then try
data.meta.rap
To get the internal rap object.
I edited from using the deserializeobject method as i incorrectly thought that had the dynamic return type. See here on json.net documentation for more details: http://www.newtonsoft.com/json/help/html/QueryJsonDynamic.htm
There is noParse
method onJsonConvert
.
– William
Jan 19 '17 at 23:08
@billisphere thanks, should have triple checked..!
– Chris Watts
Jan 20 '17 at 6:18
1
As suggested by stackoverflow.com/questions/47134936/…, I'm usingdynamic data = JObject.Parse(json);
, which seems to work without exceptions.
– Jari Turkia
Nov 20 '18 at 9:40
add a comment |
Instead of declaring as type var and letting the compiler sort it, declare as a dynamic and using the Parse method.
dynamic data = JArray.Parse(json);
Then try
data.meta.rap
To get the internal rap object.
I edited from using the deserializeobject method as i incorrectly thought that had the dynamic return type. See here on json.net documentation for more details: http://www.newtonsoft.com/json/help/html/QueryJsonDynamic.htm
There is noParse
method onJsonConvert
.
– William
Jan 19 '17 at 23:08
@billisphere thanks, should have triple checked..!
– Chris Watts
Jan 20 '17 at 6:18
1
As suggested by stackoverflow.com/questions/47134936/…, I'm usingdynamic data = JObject.Parse(json);
, which seems to work without exceptions.
– Jari Turkia
Nov 20 '18 at 9:40
add a comment |
Instead of declaring as type var and letting the compiler sort it, declare as a dynamic and using the Parse method.
dynamic data = JArray.Parse(json);
Then try
data.meta.rap
To get the internal rap object.
I edited from using the deserializeobject method as i incorrectly thought that had the dynamic return type. See here on json.net documentation for more details: http://www.newtonsoft.com/json/help/html/QueryJsonDynamic.htm
Instead of declaring as type var and letting the compiler sort it, declare as a dynamic and using the Parse method.
dynamic data = JArray.Parse(json);
Then try
data.meta.rap
To get the internal rap object.
I edited from using the deserializeobject method as i incorrectly thought that had the dynamic return type. See here on json.net documentation for more details: http://www.newtonsoft.com/json/help/html/QueryJsonDynamic.htm
edited Jan 20 '17 at 6:19
answered Jan 19 '17 at 22:39
Chris WattsChris Watts
3151218
3151218
There is noParse
method onJsonConvert
.
– William
Jan 19 '17 at 23:08
@billisphere thanks, should have triple checked..!
– Chris Watts
Jan 20 '17 at 6:18
1
As suggested by stackoverflow.com/questions/47134936/…, I'm usingdynamic data = JObject.Parse(json);
, which seems to work without exceptions.
– Jari Turkia
Nov 20 '18 at 9:40
add a comment |
There is noParse
method onJsonConvert
.
– William
Jan 19 '17 at 23:08
@billisphere thanks, should have triple checked..!
– Chris Watts
Jan 20 '17 at 6:18
1
As suggested by stackoverflow.com/questions/47134936/…, I'm usingdynamic data = JObject.Parse(json);
, which seems to work without exceptions.
– Jari Turkia
Nov 20 '18 at 9:40
There is no
Parse
method on JsonConvert
.– William
Jan 19 '17 at 23:08
There is no
Parse
method on JsonConvert
.– William
Jan 19 '17 at 23:08
@billisphere thanks, should have triple checked..!
– Chris Watts
Jan 20 '17 at 6:18
@billisphere thanks, should have triple checked..!
– Chris Watts
Jan 20 '17 at 6:18
1
1
As suggested by stackoverflow.com/questions/47134936/…, I'm using
dynamic data = JObject.Parse(json);
, which seems to work without exceptions.– Jari Turkia
Nov 20 '18 at 9:40
As suggested by stackoverflow.com/questions/47134936/…, I'm using
dynamic data = JObject.Parse(json);
, which seems to work without exceptions.– Jari Turkia
Nov 20 '18 at 9:40
add a comment |
var jobject = (JObject)JsonConvert.DeserializeObject(json);
var jvalue = (JValue)jobject["meta"]["rap"];
Console.WriteLine(jvalue.Value); // 2098
add a comment |
var jobject = (JObject)JsonConvert.DeserializeObject(json);
var jvalue = (JValue)jobject["meta"]["rap"];
Console.WriteLine(jvalue.Value); // 2098
add a comment |
var jobject = (JObject)JsonConvert.DeserializeObject(json);
var jvalue = (JValue)jobject["meta"]["rap"];
Console.WriteLine(jvalue.Value); // 2098
var jobject = (JObject)JsonConvert.DeserializeObject(json);
var jvalue = (JValue)jobject["meta"]["rap"];
Console.WriteLine(jvalue.Value); // 2098
answered Jan 19 '17 at 22:46
EricEric
1,1321915
1,1321915
add a comment |
add a comment |
The value is actually an int
type. Try:
int rap = data["rap"].Value<int>();
add a comment |
The value is actually an int
type. Try:
int rap = data["rap"].Value<int>();
add a comment |
The value is actually an int
type. Try:
int rap = data["rap"].Value<int>();
The value is actually an int
type. Try:
int rap = data["rap"].Value<int>();
answered Jan 19 '17 at 22:46
simonlchildssimonlchilds
2,47011222
2,47011222
add a comment |
add a comment |
string rap = JsonConvert.DeserializeObject<dynamic>(json).meta.rap;
Console.WriteLine(rap); // 2098
If you're not into dynamic
(or aren't using .NET 4+), I like how this other answer relies solely on Json.NET's API.
add a comment |
string rap = JsonConvert.DeserializeObject<dynamic>(json).meta.rap;
Console.WriteLine(rap); // 2098
If you're not into dynamic
(or aren't using .NET 4+), I like how this other answer relies solely on Json.NET's API.
add a comment |
string rap = JsonConvert.DeserializeObject<dynamic>(json).meta.rap;
Console.WriteLine(rap); // 2098
If you're not into dynamic
(or aren't using .NET 4+), I like how this other answer relies solely on Json.NET's API.
string rap = JsonConvert.DeserializeObject<dynamic>(json).meta.rap;
Console.WriteLine(rap); // 2098
If you're not into dynamic
(or aren't using .NET 4+), I like how this other answer relies solely on Json.NET's API.
edited May 23 '17 at 12:24
Community♦
11
11
answered Jan 19 '17 at 23:05
WilliamWilliam
1,3101732
1,3101732
add a comment |
add a comment |
Just use dynamic representation of object:
dynamic obj = JsonConvert.DeserializeObject(json)
var value = obj.meta.rap;
JObject easily convertable to dynamic type itself. You can either get string or int from this value:
var ivalue = (int)obj.meta.rap;
var svalue = (string)obj.meta.rap;
add a comment |
Just use dynamic representation of object:
dynamic obj = JsonConvert.DeserializeObject(json)
var value = obj.meta.rap;
JObject easily convertable to dynamic type itself. You can either get string or int from this value:
var ivalue = (int)obj.meta.rap;
var svalue = (string)obj.meta.rap;
add a comment |
Just use dynamic representation of object:
dynamic obj = JsonConvert.DeserializeObject(json)
var value = obj.meta.rap;
JObject easily convertable to dynamic type itself. You can either get string or int from this value:
var ivalue = (int)obj.meta.rap;
var svalue = (string)obj.meta.rap;
Just use dynamic representation of object:
dynamic obj = JsonConvert.DeserializeObject(json)
var value = obj.meta.rap;
JObject easily convertable to dynamic type itself. You can either get string or int from this value:
var ivalue = (int)obj.meta.rap;
var svalue = (string)obj.meta.rap;
edited Jan 20 '17 at 6:35
answered Jan 20 '17 at 6:22
eocroneocron
3,896835
3,896835
add a comment |
add a comment |
Try to use as following
var test = JsonConvert.DeserializeObject<dynamic>(param);
var testDTO = new TPRDTO();
testDTO.TPR_ID = test.TPR_ID.Value;
Note: For using of JsonConvert
class you have to install Newton-Soft from your package-manager
add a comment |
Try to use as following
var test = JsonConvert.DeserializeObject<dynamic>(param);
var testDTO = new TPRDTO();
testDTO.TPR_ID = test.TPR_ID.Value;
Note: For using of JsonConvert
class you have to install Newton-Soft from your package-manager
add a comment |
Try to use as following
var test = JsonConvert.DeserializeObject<dynamic>(param);
var testDTO = new TPRDTO();
testDTO.TPR_ID = test.TPR_ID.Value;
Note: For using of JsonConvert
class you have to install Newton-Soft from your package-manager
Try to use as following
var test = JsonConvert.DeserializeObject<dynamic>(param);
var testDTO = new TPRDTO();
testDTO.TPR_ID = test.TPR_ID.Value;
Note: For using of JsonConvert
class you have to install Newton-Soft from your package-manager
edited Oct 10 '18 at 11:06
AmirReza-Farahlagha
594617
594617
answered Oct 10 '18 at 9:24
Srimitha VarshaSrimitha Varsha
1
1
add a comment |
add a comment |
the problem is that you are casting the deserialized json into a JObject. if you want to have the JObject then simple do this:
JObject.Parse(json);
then you have the JObject and you can access a specific path (for extracting value see this )
you have also another option which is to deserialize your json into a class that you have in your code like this:
var instanceOFTheClass = JsonConvert.DeserializeObject<YourClassName>(json);
with the above code you can access any property and values you want.
add a comment |
the problem is that you are casting the deserialized json into a JObject. if you want to have the JObject then simple do this:
JObject.Parse(json);
then you have the JObject and you can access a specific path (for extracting value see this )
you have also another option which is to deserialize your json into a class that you have in your code like this:
var instanceOFTheClass = JsonConvert.DeserializeObject<YourClassName>(json);
with the above code you can access any property and values you want.
add a comment |
the problem is that you are casting the deserialized json into a JObject. if you want to have the JObject then simple do this:
JObject.Parse(json);
then you have the JObject and you can access a specific path (for extracting value see this )
you have also another option which is to deserialize your json into a class that you have in your code like this:
var instanceOFTheClass = JsonConvert.DeserializeObject<YourClassName>(json);
with the above code you can access any property and values you want.
the problem is that you are casting the deserialized json into a JObject. if you want to have the JObject then simple do this:
JObject.Parse(json);
then you have the JObject and you can access a specific path (for extracting value see this )
you have also another option which is to deserialize your json into a class that you have in your code like this:
var instanceOFTheClass = JsonConvert.DeserializeObject<YourClassName>(json);
with the above code you can access any property and values you want.
answered Nov 23 '18 at 14:41
MeysamMeysam
180110
180110
add a comment |
add a comment |
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Parse it instead of deserializing if you only need one value
– Make StackOverflow Good Again
Jan 19 '17 at 22:33