Serialize Object To Json
- Serialize Object To Json Python
- Serialize Object To Json File C#
- Serialize Object To Json C#
- Serialize Object To Json String
Nowadays, we are dealing with JSON data mostly when receiving data in JSON format from a web service and getting data from it. To accomplish getting data from JSON or creating JSON text from a custom object we will use JSON serialization and deserialization in C#. Serialize an Object. Serialize a Collection. Serialize a Dictionary. Serialize JSON to a file. This sample serializes an object to JSON. Serialize an Object. Serialize a Collection. Serialize a Dictionary. Serialize JSON to a file. This sample deserializes JSON to an object. This sample serializes an object to JSON. Serialize an Object. Serialize a Collection. Serialize a Dictionary. Serialize JSON to a file. Serialize with JsonConverters. Serialize a DataSet. Serialize Raw JSON value. Serialize Unindented JSON. Serialize Conditional Property. Deserialize an Object.
Important
The JSON serialization documentation is under construction. This article doesn't cover all scenarios. For more information, examine System.Text.Json issues in the dotnet/corefx repository on GitHub, especially those labeled json-functionality-doc.
This article shows how to use the System.Text.Json namespace to serialize and deserialize to and from JavaScript Object Notation (JSON). The directions and sample code use the library directly, not through a framework such as ASP.NET Core.
Namespaces
The System.Text.Json namespace contains all the entry points and the main types. The System.Text.Json.Serialization namespace contains attributes and APIs for advanced scenarios and customization specific to serialization and deserialization. Therefore, the code examples shown in this article require one or both of the following using directives:
Attributes from the System.Runtime.Serialization namespace aren't currently supported in System.Text.Json.
How to write .NET objects to JSON (serialize)
To write JSON to a string, call the JsonSerializer.Serialize method. The following example uses an overload with a generic type parameter:
You can omit the generic type parameter and use generic type inference instead:
Here's an example type to be serialized, which contains collections and nested classes:
The JSON output is minified by default:
The following example shows the same JSON, formatted (that is, pretty-printed with whitespace and indentation):
Overloads of Serialize let you serialize to a Stream. Async versions of the Stream overloads are available.
Serialize to UTF-8
To serialize to UTF-8, call the JsonSerializer.SerializeToUtf8Bytes method:
As an alternative, a Serialize overload that takes a Utf8JsonWriter is available.
Serializing to UTF-8 is about 5-10% faster than using the string-based methods. The difference is because the bytes (as UTF-8) don't need to be converted to strings (UTF-16).
Serialization behavior
- By default, all public properties are serialized. You can specify properties to exclude.
- The default encoder escapes non-ASCII characters, HTML-sensitive characters within the ASCII-range, and characters that must be escaped according to the JSON spec.
- By default, JSON is minified. You can pretty-print the JSON.
- By default, casing of JSON names matches the .NET names. You can customize JSON name casing.
- Circular references are detected and exceptions thrown. For more information, see the issue on circular references in the dotnet/corefx repository on GitHub.
- Currently, fields are excluded.
Supported types include:
- .NET primitives that map to JavaScript primitives, such as numeric types, strings, and Boolean.
- User-defined Plain Old CLR Objects (POCOs).
- One-dimensional and jagged arrays (
ArrayName[][]). Dictionary<string,TValue>whereTValueisobject,JsonElement, or a POCO.- Collections from the following namespaces. For more information, see the issue on collection support in the dotnet/corefx repository on GitHub.
How to read JSON into .NET objects (deserialize)
To deserialize from a string, call the JsonSerializer.Deserialize method, as shown in the following example:
For an example, see the serialize section. The JSON and .NET object are the same, but the direction is reversed.
Overloads of Deserialize let you deserialize from a Stream. Async versions of the Stream overloads are available.
Deserialize from UTF-8
To deserialize from UTF-8, call a JsonSerializer.Deserialize overload that takes a Utf8JsonReader or a ReadOnlySpan<byte>, as shown in the following examples:
Deserialization behavior
- By default, property name matching is case-sensitive. You can specify case-insensitivity.
- If the JSON contains a value for a read-only property, the value is ignored and no exception is thrown.
- Deserialization to reference types without a parameterless constructor isn't supported.
- Deserialization to immutable objects or read-only properties isn't supported. For more information, see the GitHub issue on immutable object support and the issue on read-only property support in the dotnet/corefx repository on GitHub.
- By default, enums are supported as numbers.
- Fields aren't supported.
- By default, comments or trailing commas in the JSON throw exceptions. You can allow comments and trailing commas if needed.
- The default maximum depth is 64.
Serialize to formatted JSON
To pretty-print the JSON output, set JsonSerializerOptions.WriteIndented to true:
Here's an example type to be serialized and JSON output:
Allow comments and trailing commas
By default comments and trailing commas are not allowed in JSON. To allow comments in the JSON, set the JsonSerializerOptions.ReadCommentHandling property to JsonCommentHandling.Skip. And to allow trailing commas, set the JsonSerializerOptions.AllowTrailingCommas property to true. The following example shows how to allow both:
Here's example JSON with comments and a trailing comma:
Customize JSON names
By default, property names and dictionary keys are unchanged in the JSON output, including case. This section explains how to:
- Customize individual property names
- Convert all property names to camel case
- Implement a custom property naming policy
- Convert dictionary keys to camel case
Currently, there's no support for automatically converting enums to camel case. For more information, see the issue on enum camel case support in the dotnet/corefx repository on GitHub.
Customize individual property names
To set the name of individual properties, use the [JsonPropertyName] attribute.
Here's an example type to serialize and resulting JSON:
The property name set by this attribute:
- Applies in both directions, for serialization and deserialization.
- Takes precedence over property naming policies.
Use camel case for all JSON property names
To use camel case for all JSON property names, set JsonSerializerOptions.PropertyNamingPolicy to JsonNamingPolicy.CamelCase, as shown in the following example:
Here's an example class to serialize and JSON output:
The camel case property naming policy:
- Applies to serialization and deserialization.
- Is overridden by
[JsonPropertyName]attributes.
Use a custom JSON property naming policy
To use a custom JSON property naming policy, create a class that derives from JsonNamingPolicy and override the ConvertName method, as shown in the following example:
Then set the JsonSerializerOptions.PropertyNamingPolicy property to an instance of your naming policy class:
Here's an example class to serialize and JSON output:
The JSON property naming policy:
- Applies to serialization and deserialization.
- Is overridden by
[JsonPropertyName]attributes.
Camel case dictionary keys
If a property of an object to be serialized is of type Dictionary<string,TValue>, the string keys can be converted to camel case. To do that, set DictionaryKeyPolicy to JsonNamingPolicy.CamelCase, as shown in the following example:
Here's an example object to serialize and JSON output:
| Property | Value |
|---|---|
| Date | 8/1/2019 12:00:00 AM -07:00 |
| TemperatureC | 25 |
| Summary | Hot |
| TemperatureRanges | Cold, 20 Hot, 40 |
The camel case naming policy applies to serialization only.
Exclude properties
By default, all public properties are serialized. If you don't want some of them to appear in the JSON output, you have several options. This section explains how to exclude:
- Individual properties
- All read-only properties
- All null-value properties
Exclude individual properties
To ignore individual properties, use the [JsonIgnore] attribute.
Here's an example type to serialize and JSON output:
Exclude all read-only properties
To exclude all read-only properties, set the JsonSerializerOptions.IgnoreReadOnlyProperties to true, as shown in the following example:
Here's an example type to serialize and JSON output:
This option applies only to serialization. During deserialization, read-only properties are ignored by default. A property is read-only if it contains a public getter but not a public setter.
Exclude all null value properties
To exclude all null value properties, set the IgnoreNullValues property to true, as shown in the following example:
Here's an example object to serialize and JSON output:
| Property | Value |
|---|---|
| Date | 8/1/2019 12:00:00 AM -07:00 |
| TemperatureC | 25 |
| Summary | null |
This setting applies to serialization and deserialization. During deserialization, null values in the JSON are ignored only if they are valid. Null values for non-nullable value types cause exceptions. For more information, see the issue on non-nullable value types in the dotnet/corefx repository on GitHub.
Case-insensitive property matching
By default, deserialization looks for case-sensitive property name matches between JSON and the target object properties. To change that behavior, set the JsonSerializerOptions.PropertyNameCaseInsensitive to true:
Here's example JSON with camel case property names. It can be deserialized into the following type that has Pascal case property names.
Include properties of derived classes
Polymorphic serialization isn't supported when you specify at compile time the type to be serialized. For example, suppose you have a WeatherForecast class and a derived class WeatherForecastWithWind:
And suppose the type passed to, or inferred by, the Serialize method at compile time is WeatherForecast:
In this scenario, the WindSpeed property is not serialized even if the weatherForecast object is actually a WeatherForecastWithWind object. Only the base class properties are serialized:
This behavior is intended to help prevent accidental exposure of data in a derived runtime-created type.
To serialize the properties of the derived type, use one of the following approaches:
Call an overload of Serialize that lets you specify the type at runtime:
Declare the object to be serialized as
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In the preceding example scenario, both approaches cause the WindSpeed property to be included in the JSON output:
Handle overflow JSON
While deserializing, you might receive data in the JSON that is not represented by properties of the target type. For example, suppose your target type is this:
And the JSON to be deserialized is this:
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If you deserialize the JSON shown into the type shown, the DatesAvailable and SummaryWords properties have nowhere to go and are lost. To capture extra data such as these properties, apply the JsonExtensionData attribute to a property of type Dictionary<string,object> or Dictionary<string,JsonElement>:
When you deserialize the JSON shown earlier into this sample type, the extra data becomes key-value pairs of the ExtensionData property:
| Property | Value | Notes |
|---|---|---|
| Date | 8/1/2019 12:00:00 AM -07:00 | |
| TemperatureC | 0 | Case-sensitive mismatch (temperatureC in the JSON), so the property isn't set. |
| Summary | Hot | |
| ExtensionData | temperatureC: 25 | Since the case didn't match, this JSON property is an extra and becomes a key-value pair in the dictionary. |
| DatesAvailable: 8/1/2019 12:00:00 AM -07:00 8/2/2019 12:00:00 AM -07:00 | Extra property from the JSON becomes a key-value pair, with an array as the value object. | |
| SummaryWords: Cool Windy Humid | Extra property from the JSON becomes a key-value pair, with an array as the value object. |
When the target object is serialized, the extension data key value pairs become JSON properties just as they were in the incoming JSON:
Notice that the ExtensionData property name doesn't appear in the JSON. This behavior lets the JSON make a round trip without losing any extra data that otherwise wouldn't be deserialized.
Use Utf8JsonWriter directly
The following example shows how to use the Utf8JsonWriter class directly.
Use Utf8JsonReader directly
The following example shows how to use the Utf8JsonReader class directly. The code assumes that the jsonUtf8 variable is a byte array that contains valid JSON, encoded as UTF-8.
Additional resources
I have a JSON string and I need some help to deserialize it.
Nothing worked for me.. This is the JSON:
I have an example of the classes, but I don't have to use those classes. I don't mind using some other classes.
These are the classes:
I want to mention that I have Newtonsoft.Json already, so I can use some functions from there.
How can I do this?
Peter Mortensen11 Answers
I am using like this in my code and it's working fine
below is a piece of code which you need to write
Should just be this:
You can paste the json string to here: http://json2csharp.com/ to check your classes are correct.
stevepkr84stevepkr84If you use C# 2010 or newer, you can use dynamic type:
Then you can access attributes and arrays in dynamic object using dot notation:
First install newtonsoft.json package to Visual Studio using NuGet Package Manager then add the following code:
I had a scenario, and this one helped me
JObject objParserd = JObject.Parse(jsonString);
JObject arrayObject1 = (JObject)objParserd['d'];
D myOutput= JsonConvert.DeserializeObject<D>(arrayObject1.ToString());
I solved this problem to add a public setter for all properties, which should be deserialized.
You can solve your problem like below bunch of codes
Serialize Object To Json Python
I also had the issue of parsing and using JSON objects in C#. I checked the dynamic type with some libraries, but the issue was always checking if a property exists.
In the end, I stumbled upon this web page, which saved me a lot of time. It automatically creates a strongly typed class based on your JSON data, that you will use with the Newtonsoft library, and it works perfectly. It also works with languages other than C#.
Peter Mortensen