Convert JSON to C# POCO
I have below JSON data
{
"appDesc": {
"description": "App description.",
"message": "Create and edit presentations "
},
"appName": {
"description": "App name.",
"message": "Slides"
}
}
I want to Deserialize into C# class object. I am using JsonConvert.DeserializeObject<>() to achieve this functionality. But some how it is not working.
string JsonData= System.IO.File.ReadAllText(msgJSONpath);
var moreInfo = JsonConvert.DeserializeObject<appName>(msg)
internal class appName
{
public string message { get; set; }
public string description { get; set; }
}
So moreInfo object will have 2 properties in message and description.
c# json
edited Nov 20 '18 at 7:58
JustLearning
1,13821637
asked Nov 20 '18 at 6:48
AMIT SHELKEAMIT SHELKE
2073923
|
show 4 more comments
I have below JSON data
{
"appDesc": {
"description": "App description.",
"message": "Create and edit presentations "
},
"appName": {
"description": "App name.",
"message": "Slides"
}
}
I want to Deserialize into C# class object. I am using JsonConvert.DeserializeObject<>() to achieve this functionality. But some how it is not working.
string JsonData= System.IO.File.ReadAllText(msgJSONpath);
var moreInfo = JsonConvert.DeserializeObject<appName>(msg)
internal class appName
{
public string message { get; set; }
public string description { get; set; }
}
So moreInfo object will have 2 properties in message and description.
c# json
edited Nov 20 '18 at 7:58
JustLearning
1,13821637
asked Nov 20 '18 at 6:48
AMIT SHELKEAMIT SHELKE
2073923
1
show your class you want to de-serialize into
– Just code
Nov 20 '18 at 6:49
internal class appName { public string message { get; set; } public string description { get; set; } }
– AMIT SHELKE
Nov 20 '18 at 6:50
edit your question and add your relevant code
– Just code
Nov 20 '18 at 6:51
Senario is, when I read a .json file which is having above data, I need to convert them into objects. Something like I have mentioned above.
– AMIT SHELKE
Nov 20 '18 at 6:52
Add your code which you have tried
– Rahul Neekhra
Nov 20 '18 at 6:55
|
show 4 more comments
I have below JSON data
{
"appDesc": {
"description": "App description.",
"message": "Create and edit presentations "
},
"appName": {
"description": "App name.",
"message": "Slides"
}
}
I want to Deserialize into C# class object. I am using JsonConvert.DeserializeObject<>() to achieve this functionality. But some how it is not working.
string JsonData= System.IO.File.ReadAllText(msgJSONpath);
var moreInfo = JsonConvert.DeserializeObject<appName>(msg)
internal class appName
{
public string message { get; set; }
public string description { get; set; }
}
So moreInfo object will have 2 properties in message and description.
c# json
edited Nov 20 '18 at 7:58
JustLearning
1,13821637
asked Nov 20 '18 at 6:48
AMIT SHELKEAMIT SHELKE
2073923
I have below JSON data
{
"appDesc": {
"description": "App description.",
"message": "Create and edit presentations "
},
"appName": {
"description": "App name.",
"message": "Slides"
}
}
I want to Deserialize into C# class object. I am using JsonConvert.DeserializeObject<>() to achieve this functionality. But some how it is not working.
string JsonData= System.IO.File.ReadAllText(msgJSONpath);
var moreInfo = JsonConvert.DeserializeObject<appName>(msg)
internal class appName
{
public string message { get; set; }
public string description { get; set; }
}
So moreInfo object will have 2 properties in message and description.
c# json
c# json
edited Nov 20 '18 at 7:58
JustLearning
1,13821637
asked Nov 20 '18 at 6:48
AMIT SHELKEAMIT SHELKE
2073923
edited Nov 20 '18 at 7:58
JustLearning
1,13821637
asked Nov 20 '18 at 6:48
AMIT SHELKEAMIT SHELKE
2073923
edited Nov 20 '18 at 7:58
JustLearning
1,13821637
edited Nov 20 '18 at 7:58
JustLearning
1,13821637
edited Nov 20 '18 at 7:58
JustLearning
1,13821637
1,13821637
asked Nov 20 '18 at 6:48
AMIT SHELKEAMIT SHELKE
2073923
asked Nov 20 '18 at 6:48
AMIT SHELKEAMIT SHELKE
2073923
asked Nov 20 '18 at 6:48
AMIT SHELKEAMIT SHELKE
2073923
2073923
1
show your class you want to de-serialize into
– Just code
Nov 20 '18 at 6:49
internal class appName { public string message { get; set; } public string description { get; set; } }
– AMIT SHELKE
Nov 20 '18 at 6:50
edit your question and add your relevant code
– Just code
Nov 20 '18 at 6:51
Senario is, when I read a .json file which is having above data, I need to convert them into objects. Something like I have mentioned above.
– AMIT SHELKE
Nov 20 '18 at 6:52
Add your code which you have tried
– Rahul Neekhra
Nov 20 '18 at 6:55
|
show 4 more comments
1
show your class you want to de-serialize into
– Just code
Nov 20 '18 at 6:49
internal class appName { public string message { get; set; } public string description { get; set; } }
– AMIT SHELKE
Nov 20 '18 at 6:50
edit your question and add your relevant code
– Just code
Nov 20 '18 at 6:51
Senario is, when I read a .json file which is having above data, I need to convert them into objects. Something like I have mentioned above.
– AMIT SHELKE
Nov 20 '18 at 6:52
Add your code which you have tried
– Rahul Neekhra
Nov 20 '18 at 6:55
1
1
show your class you want to de-serialize into
– Just code
Nov 20 '18 at 6:49
show your class you want to de-serialize into
– Just code
Nov 20 '18 at 6:49
internal class appName { public string message { get; set; } public string description { get; set; } }
– AMIT SHELKE
Nov 20 '18 at 6:50
internal class appName { public string message { get; set; } public string description { get; set; } }
– AMIT SHELKE
Nov 20 '18 at 6:50
edit your question and add your relevant code
– Just code
Nov 20 '18 at 6:51
edit your question and add your relevant code
– Just code
Nov 20 '18 at 6:51
Senario is, when I read a .json file which is having above data, I need to convert them into objects. Something like I have mentioned above.
– AMIT SHELKE
Nov 20 '18 at 6:52
Senario is, when I read a .json file which is having above data, I need to convert them into objects. Something like I have mentioned above.
– AMIT SHELKE
Nov 20 '18 at 6:52
Add your code which you have tried
– Rahul Neekhra
Nov 20 '18 at 6:55
Add your code which you have tried
– Rahul Neekhra
Nov 20 '18 at 6:55
|
show 4 more comments
3 Answers
3
active
oldest
votes
JObject defines method Parse for this:
JObject json = JObject.Parse(str);
or try for a typed object try:
Foo json = JsonConvert.DeserializeObject<Foo>(str)
answered Nov 20 '18 at 6:58
A.M. PatelA.M. Patel
689
add a comment |
you need 2 C# classes since the properties of appName and appDesc are exactly the same.
To store appname
public class appName {
public string description { get; set; }
public string message { get; set; }
}
A class to have both above classes as properties
public class appResult {
public appName appDesc { get; set; }
public appName appName { get; set; }
public appResult() {
appDesc = new appName();
appName = new appName();
}
}
}
desirialize the json
var result = JsonConvert.DeserializeObject<appResult>(msg);
Once you have the result object you can get your
appName
var appName = result.appName;
answered Nov 20 '18 at 7:04
JustLearningJustLearning
1,13821637
1
appNameandappDescare the same. There's no reason to use separate classes for them
– Panagiotis Kanavos
Nov 20 '18 at 7:50
i see your point your are right
– JustLearning
Nov 20 '18 at 7:59
@BackSlash i just saw it now, why it makes sense
– JustLearning
Nov 20 '18 at 8:00
add a comment |
First ,you need to create some classes based on your JSON , if you,re using visual studio you can copy the JSON string to your clipboard and than go to
Edit > Paste Special > Paste JSON As Classes
otherwise you can use This Online Tool
after that your code should be something like this :
string JsonData= System.IO.File.ReadAllText(msgJSONpath);
var moreInfo = JsonConvert.DeserializeObject<RootObject>(msg);
Generated Classes Based On Your JSON :
public class AppDesc
{
public string description { get; set; }
public string message { get; set; }
}
public class AppName
{
public string description { get; set; }
public string message { get; set; }
}
public class RootObject
{
public AppDesc appDesc { get; set; }
public AppName appName { get; set; }
}
answered Nov 20 '18 at 7:43
FarhadMohseniFarhadMohseni
215
1
This would be a good answer if you pasted the generated classes as well
– Panagiotis Kanavos
Nov 20 '18 at 7:50
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
JObject defines method Parse for this:
JObject json = JObject.Parse(str);
or try for a typed object try:
Foo json = JsonConvert.DeserializeObject<Foo>(str)
answered Nov 20 '18 at 6:58
A.M. PatelA.M. Patel
689
add a comment |
JObject defines method Parse for this:
JObject json = JObject.Parse(str);
or try for a typed object try:
Foo json = JsonConvert.DeserializeObject<Foo>(str)
answered Nov 20 '18 at 6:58
A.M. PatelA.M. Patel
689
add a comment |
JObject defines method Parse for this:
JObject json = JObject.Parse(str);
or try for a typed object try:
Foo json = JsonConvert.DeserializeObject<Foo>(str)
answered Nov 20 '18 at 6:58
A.M. PatelA.M. Patel
689
JObject defines method Parse for this:
JObject json = JObject.Parse(str);
or try for a typed object try:
Foo json = JsonConvert.DeserializeObject<Foo>(str)
answered Nov 20 '18 at 6:58
A.M. PatelA.M. Patel
689
answered Nov 20 '18 at 6:58
A.M. PatelA.M. Patel
689
answered Nov 20 '18 at 6:58
A.M. PatelA.M. Patel
689
answered Nov 20 '18 at 6:58
A.M. PatelA.M. Patel
689
689
add a comment |
add a comment |
you need 2 C# classes since the properties of appName and appDesc are exactly the same.
To store appname
public class appName {
public string description { get; set; }
public string message { get; set; }
}
A class to have both above classes as properties
public class appResult {
public appName appDesc { get; set; }
public appName appName { get; set; }
public appResult() {
appDesc = new appName();
appName = new appName();
}
}
}
desirialize the json
var result = JsonConvert.DeserializeObject<appResult>(msg);
Once you have the result object you can get your
appName
var appName = result.appName;
answered Nov 20 '18 at 7:04
JustLearningJustLearning
1,13821637
1
appNameandappDescare the same. There's no reason to use separate classes for them
– Panagiotis Kanavos
Nov 20 '18 at 7:50
i see your point your are right
– JustLearning
Nov 20 '18 at 7:59
@BackSlash i just saw it now, why it makes sense
– JustLearning
Nov 20 '18 at 8:00
add a comment |
you need 2 C# classes since the properties of appName and appDesc are exactly the same.
To store appname
public class appName {
public string description { get; set; }
public string message { get; set; }
}
A class to have both above classes as properties
public class appResult {
public appName appDesc { get; set; }
public appName appName { get; set; }
public appResult() {
appDesc = new appName();
appName = new appName();
}
}
}
desirialize the json
var result = JsonConvert.DeserializeObject<appResult>(msg);
Once you have the result object you can get your
appName
var appName = result.appName;
answered Nov 20 '18 at 7:04
JustLearningJustLearning
1,13821637
1
appNameandappDescare the same. There's no reason to use separate classes for them
– Panagiotis Kanavos
Nov 20 '18 at 7:50
i see your point your are right
– JustLearning
Nov 20 '18 at 7:59
@BackSlash i just saw it now, why it makes sense
– JustLearning
Nov 20 '18 at 8:00
add a comment |
you need 2 C# classes since the properties of appName and appDesc are exactly the same.
To store appname
public class appName {
public string description { get; set; }
public string message { get; set; }
}
A class to have both above classes as properties
public class appResult {
public appName appDesc { get; set; }
public appName appName { get; set; }
public appResult() {
appDesc = new appName();
appName = new appName();
}
}
}
desirialize the json
var result = JsonConvert.DeserializeObject<appResult>(msg);
Once you have the result object you can get your
appName
var appName = result.appName;
answered Nov 20 '18 at 7:04
JustLearningJustLearning
1,13821637
you need 2 C# classes since the properties of appName and appDesc are exactly the same.
To store appname
public class appName {
public string description { get; set; }
public string message { get; set; }
}
A class to have both above classes as properties
public class appResult {
public appName appDesc { get; set; }
public appName appName { get; set; }
public appResult() {
appDesc = new appName();
appName = new appName();
}
}
}
desirialize the json
var result = JsonConvert.DeserializeObject<appResult>(msg);
Once you have the result object you can get your
appName
var appName = result.appName;
answered Nov 20 '18 at 7:04
JustLearningJustLearning
1,13821637
edited Nov 20 '18 at 8:01
answered Nov 20 '18 at 7:04
JustLearningJustLearning
1,13821637
answered Nov 20 '18 at 7:04
JustLearningJustLearning
1,13821637
answered Nov 20 '18 at 7:04
JustLearningJustLearning
1,13821637
1,13821637
1
appNameandappDescare the same. There's no reason to use separate classes for them
– Panagiotis Kanavos
Nov 20 '18 at 7:50
i see your point your are right
– JustLearning
Nov 20 '18 at 7:59
@BackSlash i just saw it now, why it makes sense
– JustLearning
Nov 20 '18 at 8:00
add a comment |
1
appNameandappDescare the same. There's no reason to use separate classes for them
– Panagiotis Kanavos
Nov 20 '18 at 7:50
i see your point your are right
– JustLearning
Nov 20 '18 at 7:59
@BackSlash i just saw it now, why it makes sense
– JustLearning
Nov 20 '18 at 8:00
1
1
appName and appDesc are the same. There's no reason to use separate classes for them– Panagiotis Kanavos
Nov 20 '18 at 7:50
appName and appDesc are the same. There's no reason to use separate classes for them– Panagiotis Kanavos
Nov 20 '18 at 7:50
i see your point your are right
– JustLearning
Nov 20 '18 at 7:59
i see your point your are right
– JustLearning
Nov 20 '18 at 7:59
@BackSlash i just saw it now, why it makes sense
– JustLearning
Nov 20 '18 at 8:00
@BackSlash i just saw it now, why it makes sense
– JustLearning
Nov 20 '18 at 8:00
add a comment |
First ,you need to create some classes based on your JSON , if you,re using visual studio you can copy the JSON string to your clipboard and than go to
Edit > Paste Special > Paste JSON As Classes
otherwise you can use This Online Tool
after that your code should be something like this :
string JsonData= System.IO.File.ReadAllText(msgJSONpath);
var moreInfo = JsonConvert.DeserializeObject<RootObject>(msg);
Generated Classes Based On Your JSON :
public class AppDesc
{
public string description { get; set; }
public string message { get; set; }
}
public class AppName
{
public string description { get; set; }
public string message { get; set; }
}
public class RootObject
{
public AppDesc appDesc { get; set; }
public AppName appName { get; set; }
}
answered Nov 20 '18 at 7:43
FarhadMohseniFarhadMohseni
215
1
This would be a good answer if you pasted the generated classes as well
– Panagiotis Kanavos
Nov 20 '18 at 7:50
add a comment |
First ,you need to create some classes based on your JSON , if you,re using visual studio you can copy the JSON string to your clipboard and than go to
Edit > Paste Special > Paste JSON As Classes
otherwise you can use This Online Tool
after that your code should be something like this :
string JsonData= System.IO.File.ReadAllText(msgJSONpath);
var moreInfo = JsonConvert.DeserializeObject<RootObject>(msg);
Generated Classes Based On Your JSON :
public class AppDesc
{
public string description { get; set; }
public string message { get; set; }
}
public class AppName
{
public string description { get; set; }
public string message { get; set; }
}
public class RootObject
{
public AppDesc appDesc { get; set; }
public AppName appName { get; set; }
}
answered Nov 20 '18 at 7:43
FarhadMohseniFarhadMohseni
215
1
This would be a good answer if you pasted the generated classes as well
– Panagiotis Kanavos
Nov 20 '18 at 7:50
add a comment |
First ,you need to create some classes based on your JSON , if you,re using visual studio you can copy the JSON string to your clipboard and than go to
Edit > Paste Special > Paste JSON As Classes
otherwise you can use This Online Tool
after that your code should be something like this :
string JsonData= System.IO.File.ReadAllText(msgJSONpath);
var moreInfo = JsonConvert.DeserializeObject<RootObject>(msg);
Generated Classes Based On Your JSON :
public class AppDesc
{
public string description { get; set; }
public string message { get; set; }
}
public class AppName
{
public string description { get; set; }
public string message { get; set; }
}
public class RootObject
{
public AppDesc appDesc { get; set; }
public AppName appName { get; set; }
}
answered Nov 20 '18 at 7:43
FarhadMohseniFarhadMohseni
215
First ,you need to create some classes based on your JSON , if you,re using visual studio you can copy the JSON string to your clipboard and than go to
Edit > Paste Special > Paste JSON As Classes
otherwise you can use This Online Tool
after that your code should be something like this :
string JsonData= System.IO.File.ReadAllText(msgJSONpath);
var moreInfo = JsonConvert.DeserializeObject<RootObject>(msg);
Generated Classes Based On Your JSON :
public class AppDesc
{
public string description { get; set; }
public string message { get; set; }
}
public class AppName
{
public string description { get; set; }
public string message { get; set; }
}
public class RootObject
{
public AppDesc appDesc { get; set; }
public AppName appName { get; set; }
}
answered Nov 20 '18 at 7:43
FarhadMohseniFarhadMohseni
215
edited Nov 20 '18 at 8:01
answered Nov 20 '18 at 7:43
FarhadMohseniFarhadMohseni
215
answered Nov 20 '18 at 7:43
FarhadMohseniFarhadMohseni
215
answered Nov 20 '18 at 7:43
FarhadMohseniFarhadMohseni
215
215
1
This would be a good answer if you pasted the generated classes as well
– Panagiotis Kanavos
Nov 20 '18 at 7:50
add a comment |
1
This would be a good answer if you pasted the generated classes as well
– Panagiotis Kanavos
Nov 20 '18 at 7:50
1
1
This would be a good answer if you pasted the generated classes as well
– Panagiotis Kanavos
Nov 20 '18 at 7:50
This would be a good answer if you pasted the generated classes as well
– Panagiotis Kanavos
Nov 20 '18 at 7:50
add a comment |
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1
show your class you want to de-serialize into
– Just code
Nov 20 '18 at 6:49
internal class appName { public string message { get; set; } public string description { get; set; } }
– AMIT SHELKE
Nov 20 '18 at 6:50
edit your question and add your relevant code
– Just code
Nov 20 '18 at 6:51
Senario is, when I read a .json file which is having above data, I need to convert them into objects. Something like I have mentioned above.
– AMIT SHELKE
Nov 20 '18 at 6:52
Add your code which you have tried
– Rahul Neekhra
Nov 20 '18 at 6:55