Conditional Deserialization of either a Dictionary or a List of Objects
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I am sending a GET request to a Rest API that returns some JSON in two different formats (based on some external setting that I can't influence).
I can either receive:
"content": {
"fields": [
{
"name": "test1",
"value": 1
},
{
"name": "test2",
"value": "test"
},
{
"name": "test3",
"value": "test",
"links": [...]
}
]
}
or
"content": {
"test1": 1,
"test2": "test",
"test3": "test"
}
You can see that I receive either a list of objects containing name
and value
properties (along with some other properties like links
), or I receive a single object containing key-value pairs as in a Dictionary. I now want to know if there is a way to conditionally deserialize the JSON into a class with Dictionary<string, string>
and List<Field>
properties like this:
[Serializable]
public class Content
{
/// <summary>
/// The Type of the Content
/// </summary>
public string _Type { get; set; }
public Dictionary<string, string> Dictionary { get; set; }
public List<Field> Fields { get; set; }
}
and fill either the dictionary or the list of fields depending on the JSON.
c# json json.net deserialization
|
show 2 more comments
I am sending a GET request to a Rest API that returns some JSON in two different formats (based on some external setting that I can't influence).
I can either receive:
"content": {
"fields": [
{
"name": "test1",
"value": 1
},
{
"name": "test2",
"value": "test"
},
{
"name": "test3",
"value": "test",
"links": [...]
}
]
}
or
"content": {
"test1": 1,
"test2": "test",
"test3": "test"
}
You can see that I receive either a list of objects containing name
and value
properties (along with some other properties like links
), or I receive a single object containing key-value pairs as in a Dictionary. I now want to know if there is a way to conditionally deserialize the JSON into a class with Dictionary<string, string>
and List<Field>
properties like this:
[Serializable]
public class Content
{
/// <summary>
/// The Type of the Content
/// </summary>
public string _Type { get; set; }
public Dictionary<string, string> Dictionary { get; set; }
public List<Field> Fields { get; set; }
}
and fill either the dictionary or the list of fields depending on the JSON.
c# json json.net deserialization
a custom jsonconverter that can inspect the json and extract the desired format
– Nkosi
Nov 23 '18 at 13:16
Im geusing it should be a public Dictionary<string, List<Fields>> Dictionary { get; set; }
– mahlatse
Nov 23 '18 at 13:16
No it should be Dictionary<string,string> as that is what I am expecting. The Fields Class has way more members than I need here. @Nkosi could you give me an Example or link that points me in the right direction
– Brezelmann
Nov 23 '18 at 13:50
@Brezelmann you can find details in documentation newtonsoft.com/json/help/html/CustomJsonConverter.htm
– Nkosi
Nov 23 '18 at 14:03
@Nkosi I checked your Idea. I am not sure how to do it in my case because i have two quite different data structures with either a Dictionary or a List and one of those will always be null because I can either get a Dictionary of Data or a List of Data but never both.
– Brezelmann
Nov 23 '18 at 14:45
|
show 2 more comments
I am sending a GET request to a Rest API that returns some JSON in two different formats (based on some external setting that I can't influence).
I can either receive:
"content": {
"fields": [
{
"name": "test1",
"value": 1
},
{
"name": "test2",
"value": "test"
},
{
"name": "test3",
"value": "test",
"links": [...]
}
]
}
or
"content": {
"test1": 1,
"test2": "test",
"test3": "test"
}
You can see that I receive either a list of objects containing name
and value
properties (along with some other properties like links
), or I receive a single object containing key-value pairs as in a Dictionary. I now want to know if there is a way to conditionally deserialize the JSON into a class with Dictionary<string, string>
and List<Field>
properties like this:
[Serializable]
public class Content
{
/// <summary>
/// The Type of the Content
/// </summary>
public string _Type { get; set; }
public Dictionary<string, string> Dictionary { get; set; }
public List<Field> Fields { get; set; }
}
and fill either the dictionary or the list of fields depending on the JSON.
c# json json.net deserialization
I am sending a GET request to a Rest API that returns some JSON in two different formats (based on some external setting that I can't influence).
I can either receive:
"content": {
"fields": [
{
"name": "test1",
"value": 1
},
{
"name": "test2",
"value": "test"
},
{
"name": "test3",
"value": "test",
"links": [...]
}
]
}
or
"content": {
"test1": 1,
"test2": "test",
"test3": "test"
}
You can see that I receive either a list of objects containing name
and value
properties (along with some other properties like links
), or I receive a single object containing key-value pairs as in a Dictionary. I now want to know if there is a way to conditionally deserialize the JSON into a class with Dictionary<string, string>
and List<Field>
properties like this:
[Serializable]
public class Content
{
/// <summary>
/// The Type of the Content
/// </summary>
public string _Type { get; set; }
public Dictionary<string, string> Dictionary { get; set; }
public List<Field> Fields { get; set; }
}
and fill either the dictionary or the list of fields depending on the JSON.
c# json json.net deserialization
c# json json.net deserialization
edited Nov 25 '18 at 7:08
Brian Rogers
78.8k18197211
78.8k18197211
asked Nov 23 '18 at 13:12
BrezelmannBrezelmann
678
678
a custom jsonconverter that can inspect the json and extract the desired format
– Nkosi
Nov 23 '18 at 13:16
Im geusing it should be a public Dictionary<string, List<Fields>> Dictionary { get; set; }
– mahlatse
Nov 23 '18 at 13:16
No it should be Dictionary<string,string> as that is what I am expecting. The Fields Class has way more members than I need here. @Nkosi could you give me an Example or link that points me in the right direction
– Brezelmann
Nov 23 '18 at 13:50
@Brezelmann you can find details in documentation newtonsoft.com/json/help/html/CustomJsonConverter.htm
– Nkosi
Nov 23 '18 at 14:03
@Nkosi I checked your Idea. I am not sure how to do it in my case because i have two quite different data structures with either a Dictionary or a List and one of those will always be null because I can either get a Dictionary of Data or a List of Data but never both.
– Brezelmann
Nov 23 '18 at 14:45
|
show 2 more comments
a custom jsonconverter that can inspect the json and extract the desired format
– Nkosi
Nov 23 '18 at 13:16
Im geusing it should be a public Dictionary<string, List<Fields>> Dictionary { get; set; }
– mahlatse
Nov 23 '18 at 13:16
No it should be Dictionary<string,string> as that is what I am expecting. The Fields Class has way more members than I need here. @Nkosi could you give me an Example or link that points me in the right direction
– Brezelmann
Nov 23 '18 at 13:50
@Brezelmann you can find details in documentation newtonsoft.com/json/help/html/CustomJsonConverter.htm
– Nkosi
Nov 23 '18 at 14:03
@Nkosi I checked your Idea. I am not sure how to do it in my case because i have two quite different data structures with either a Dictionary or a List and one of those will always be null because I can either get a Dictionary of Data or a List of Data but never both.
– Brezelmann
Nov 23 '18 at 14:45
a custom jsonconverter that can inspect the json and extract the desired format
– Nkosi
Nov 23 '18 at 13:16
a custom jsonconverter that can inspect the json and extract the desired format
– Nkosi
Nov 23 '18 at 13:16
Im geusing it should be a public Dictionary<string, List<Fields>> Dictionary { get; set; }
– mahlatse
Nov 23 '18 at 13:16
Im geusing it should be a public Dictionary<string, List<Fields>> Dictionary { get; set; }
– mahlatse
Nov 23 '18 at 13:16
No it should be Dictionary<string,string> as that is what I am expecting. The Fields Class has way more members than I need here. @Nkosi could you give me an Example or link that points me in the right direction
– Brezelmann
Nov 23 '18 at 13:50
No it should be Dictionary<string,string> as that is what I am expecting. The Fields Class has way more members than I need here. @Nkosi could you give me an Example or link that points me in the right direction
– Brezelmann
Nov 23 '18 at 13:50
@Brezelmann you can find details in documentation newtonsoft.com/json/help/html/CustomJsonConverter.htm
– Nkosi
Nov 23 '18 at 14:03
@Brezelmann you can find details in documentation newtonsoft.com/json/help/html/CustomJsonConverter.htm
– Nkosi
Nov 23 '18 at 14:03
@Nkosi I checked your Idea. I am not sure how to do it in my case because i have two quite different data structures with either a Dictionary or a List and one of those will always be null because I can either get a Dictionary of Data or a List of Data but never both.
– Brezelmann
Nov 23 '18 at 14:45
@Nkosi I checked your Idea. I am not sure how to do it in my case because i have two quite different data structures with either a Dictionary or a List and one of those will always be null because I can either get a Dictionary of Data or a List of Data but never both.
– Brezelmann
Nov 23 '18 at 14:45
|
show 2 more comments
1 Answer
1
active
oldest
votes
You can handle this situation by creating a custom JsonConverter
for your Content
class as shown below. It works by loading the content
portion of the JSON into a JObject
and checking for the presence of the fields
property to determine how to populate the Content
instance.
public class ContentConverter : JsonConverter
{
public override bool CanConvert(Type objectType)
{
return objectType == typeof(Content);
}
public override object ReadJson(JsonReader reader, Type objectType, object existingValue, JsonSerializer serializer)
{
JObject jo = JObject.Load(reader);
Content content = new Content();
if (jo["fields"] != null)
{
// if the fields property is present, we have a list of fields
content.Fields = jo["fields"].ToObject<List<Field>>(serializer);
content._Type = "Fields";
}
else
{
// fields property is not present so we have a simple dictionary
content.Dictionary = jo.Properties().ToDictionary(p => p.Name, p => (string)p.Value);
content._Type = "Dictionary";
}
return content;
}
public override bool CanWrite
{
get { return false; }
}
public override void WriteJson(JsonWriter writer, object value, JsonSerializer serializer)
{
throw new NotImplementedException();
}
}
I was not sure how you wanted to handle the _Type
property, so I just set it to either "Fields" or "Dictionary" to indicate which property was populated. Feel free to change it to suit your needs.
To use the converter, just add a [JsonConverter]
attribute to your Content
class like this:
[JsonConverter(typeof(ContentConverter))]
public class Content
{
...
}
Here is a working demo: https://dotnetfiddle.net/geg5fA
Works really nice. A good Example and a real good explanation. Thank you @Brian Rogers
– Brezelmann
Nov 30 '18 at 10:32
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
You can handle this situation by creating a custom JsonConverter
for your Content
class as shown below. It works by loading the content
portion of the JSON into a JObject
and checking for the presence of the fields
property to determine how to populate the Content
instance.
public class ContentConverter : JsonConverter
{
public override bool CanConvert(Type objectType)
{
return objectType == typeof(Content);
}
public override object ReadJson(JsonReader reader, Type objectType, object existingValue, JsonSerializer serializer)
{
JObject jo = JObject.Load(reader);
Content content = new Content();
if (jo["fields"] != null)
{
// if the fields property is present, we have a list of fields
content.Fields = jo["fields"].ToObject<List<Field>>(serializer);
content._Type = "Fields";
}
else
{
// fields property is not present so we have a simple dictionary
content.Dictionary = jo.Properties().ToDictionary(p => p.Name, p => (string)p.Value);
content._Type = "Dictionary";
}
return content;
}
public override bool CanWrite
{
get { return false; }
}
public override void WriteJson(JsonWriter writer, object value, JsonSerializer serializer)
{
throw new NotImplementedException();
}
}
I was not sure how you wanted to handle the _Type
property, so I just set it to either "Fields" or "Dictionary" to indicate which property was populated. Feel free to change it to suit your needs.
To use the converter, just add a [JsonConverter]
attribute to your Content
class like this:
[JsonConverter(typeof(ContentConverter))]
public class Content
{
...
}
Here is a working demo: https://dotnetfiddle.net/geg5fA
Works really nice. A good Example and a real good explanation. Thank you @Brian Rogers
– Brezelmann
Nov 30 '18 at 10:32
add a comment |
You can handle this situation by creating a custom JsonConverter
for your Content
class as shown below. It works by loading the content
portion of the JSON into a JObject
and checking for the presence of the fields
property to determine how to populate the Content
instance.
public class ContentConverter : JsonConverter
{
public override bool CanConvert(Type objectType)
{
return objectType == typeof(Content);
}
public override object ReadJson(JsonReader reader, Type objectType, object existingValue, JsonSerializer serializer)
{
JObject jo = JObject.Load(reader);
Content content = new Content();
if (jo["fields"] != null)
{
// if the fields property is present, we have a list of fields
content.Fields = jo["fields"].ToObject<List<Field>>(serializer);
content._Type = "Fields";
}
else
{
// fields property is not present so we have a simple dictionary
content.Dictionary = jo.Properties().ToDictionary(p => p.Name, p => (string)p.Value);
content._Type = "Dictionary";
}
return content;
}
public override bool CanWrite
{
get { return false; }
}
public override void WriteJson(JsonWriter writer, object value, JsonSerializer serializer)
{
throw new NotImplementedException();
}
}
I was not sure how you wanted to handle the _Type
property, so I just set it to either "Fields" or "Dictionary" to indicate which property was populated. Feel free to change it to suit your needs.
To use the converter, just add a [JsonConverter]
attribute to your Content
class like this:
[JsonConverter(typeof(ContentConverter))]
public class Content
{
...
}
Here is a working demo: https://dotnetfiddle.net/geg5fA
Works really nice. A good Example and a real good explanation. Thank you @Brian Rogers
– Brezelmann
Nov 30 '18 at 10:32
add a comment |
You can handle this situation by creating a custom JsonConverter
for your Content
class as shown below. It works by loading the content
portion of the JSON into a JObject
and checking for the presence of the fields
property to determine how to populate the Content
instance.
public class ContentConverter : JsonConverter
{
public override bool CanConvert(Type objectType)
{
return objectType == typeof(Content);
}
public override object ReadJson(JsonReader reader, Type objectType, object existingValue, JsonSerializer serializer)
{
JObject jo = JObject.Load(reader);
Content content = new Content();
if (jo["fields"] != null)
{
// if the fields property is present, we have a list of fields
content.Fields = jo["fields"].ToObject<List<Field>>(serializer);
content._Type = "Fields";
}
else
{
// fields property is not present so we have a simple dictionary
content.Dictionary = jo.Properties().ToDictionary(p => p.Name, p => (string)p.Value);
content._Type = "Dictionary";
}
return content;
}
public override bool CanWrite
{
get { return false; }
}
public override void WriteJson(JsonWriter writer, object value, JsonSerializer serializer)
{
throw new NotImplementedException();
}
}
I was not sure how you wanted to handle the _Type
property, so I just set it to either "Fields" or "Dictionary" to indicate which property was populated. Feel free to change it to suit your needs.
To use the converter, just add a [JsonConverter]
attribute to your Content
class like this:
[JsonConverter(typeof(ContentConverter))]
public class Content
{
...
}
Here is a working demo: https://dotnetfiddle.net/geg5fA
You can handle this situation by creating a custom JsonConverter
for your Content
class as shown below. It works by loading the content
portion of the JSON into a JObject
and checking for the presence of the fields
property to determine how to populate the Content
instance.
public class ContentConverter : JsonConverter
{
public override bool CanConvert(Type objectType)
{
return objectType == typeof(Content);
}
public override object ReadJson(JsonReader reader, Type objectType, object existingValue, JsonSerializer serializer)
{
JObject jo = JObject.Load(reader);
Content content = new Content();
if (jo["fields"] != null)
{
// if the fields property is present, we have a list of fields
content.Fields = jo["fields"].ToObject<List<Field>>(serializer);
content._Type = "Fields";
}
else
{
// fields property is not present so we have a simple dictionary
content.Dictionary = jo.Properties().ToDictionary(p => p.Name, p => (string)p.Value);
content._Type = "Dictionary";
}
return content;
}
public override bool CanWrite
{
get { return false; }
}
public override void WriteJson(JsonWriter writer, object value, JsonSerializer serializer)
{
throw new NotImplementedException();
}
}
I was not sure how you wanted to handle the _Type
property, so I just set it to either "Fields" or "Dictionary" to indicate which property was populated. Feel free to change it to suit your needs.
To use the converter, just add a [JsonConverter]
attribute to your Content
class like this:
[JsonConverter(typeof(ContentConverter))]
public class Content
{
...
}
Here is a working demo: https://dotnetfiddle.net/geg5fA
answered Nov 25 '18 at 6:48
Brian RogersBrian Rogers
78.8k18197211
78.8k18197211
Works really nice. A good Example and a real good explanation. Thank you @Brian Rogers
– Brezelmann
Nov 30 '18 at 10:32
add a comment |
Works really nice. A good Example and a real good explanation. Thank you @Brian Rogers
– Brezelmann
Nov 30 '18 at 10:32
Works really nice. A good Example and a real good explanation. Thank you @Brian Rogers
– Brezelmann
Nov 30 '18 at 10:32
Works really nice. A good Example and a real good explanation. Thank you @Brian Rogers
– Brezelmann
Nov 30 '18 at 10:32
add a comment |
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a custom jsonconverter that can inspect the json and extract the desired format
– Nkosi
Nov 23 '18 at 13:16
Im geusing it should be a public Dictionary<string, List<Fields>> Dictionary { get; set; }
– mahlatse
Nov 23 '18 at 13:16
No it should be Dictionary<string,string> as that is what I am expecting. The Fields Class has way more members than I need here. @Nkosi could you give me an Example or link that points me in the right direction
– Brezelmann
Nov 23 '18 at 13:50
@Brezelmann you can find details in documentation newtonsoft.com/json/help/html/CustomJsonConverter.htm
– Nkosi
Nov 23 '18 at 14:03
@Nkosi I checked your Idea. I am not sure how to do it in my case because i have two quite different data structures with either a Dictionary or a List and one of those will always be null because I can either get a Dictionary of Data or a List of Data but never both.
– Brezelmann
Nov 23 '18 at 14:45