Deserializar JSON

UiPath.WebAPI.Activities.DeserializeJson

Descripción

Deserializes the JSON string or sample to a JSON object. By default, the result is a JObject, but you can choose a different output type using the Output type property.

Compatibilidad de proyectos

Windows - Heredado | Windows | Multiplataforma

Requisitos previos

To configure the Serializer settings property using data mapping, use Studio 25.10 or later. In earlier Studio versions, this property uses the Expression Editor window instead.

Windows, configuración multiplataforma

• JSON String - The string containing the JSON to be deserialized. The JSON should be written without indents, and between double quotes. Strings inside the JSON should be written with single quotes. For example, the following JSON is valid: "{'TenancyName': 'Documentation', 'UsernameOrEmail': 'Documentation', 'Password': 'Documentation'}".
• JSON Sample - A sample of the JSON used to generate the output object properties. The sample must be in JSON format. The activity converts this sample into a similar object that you can use during design time.

Opciones avanzadas

Entrada

• Serializer settings - Custom deserializer settings to be used when deserializing the JSON string. Use this property when you want to fully control how the JSON string is deserialized.
• Output type - A drop-down list that lets you choose the argument type of the object to deserialize into.
  Nota:

  In Studio Web, generating the output structure from the JSON Sample property makes the Output type property read-only.

Salida

• JSON object - The deserialized JSON, returned as an object of the type selected in the Output type property. When the default type is used, the result is a Newtonsoft.Json.Linq.JObject variable. Read more information on Newtonsoft.Json.Linq.JObject objects.

Example of using the JSON Sample

The JSON Sample property generates the output object structure at design time from a representative JSON snippet. After you provide a sample, the activity infers a matching type and assigns it to the Output type property, so you can reference the deserialized values by name in downstream activities instead of querying them with SelectToken.

Nota:

The JSON Sample property is a design-time helper. It is available in Studio Web and in Studio versions that support custom type inference. When these features are not available, the property is hidden.

To exemplify how to use the JSON Sample, we extract a person's name and city from a JSON string:

1. Add a Deserialize JSON activity to your workflow.

2. In the JSON String field, enter the JSON to deserialize. For example: "{'name': 'John Doe', 'age': 30, 'address': {'city': 'New York'}}".

3. In the JSON Sample field, paste a representative sample with the same structure:

   {
     "name": "John Doe",
     "age": 30,
     "address": {
       "city": "New York"
     }
   }
   

   The activity reads the sample and generates an output object that mirrors this structure. The Output type property updates to the generated type.

4. In a subsequent activity, reference the deserialized values directly through the property names defined in your sample, for example JsonObject.name and JsonObject.address.city, instead of using JsonObject.SelectToken("name").

Resultado

The output object exposes the properties from your sample, so JsonObject.name returns John Doe and JsonObject.address.city returns New York without querying the JSON with SelectToken.

Example of using Serializer settings

When no data mapping is available for the Serializer settings property, configure it in the Expression Editor using one of the following code snippets. For more information, see the Newtonsoft documentation.

VB.NET

New JsonDeserializationSettings With {
.MissingMemberHandling = JsonMissingMemberHandling.Ignore,
.NullValueHandling = JsonNullValueHandling.Ignore,
.DateParseHandling = JsonDateParseHandling.DateTime,
.DateTimeZoneHandling = JsonDateTimeZoneHandling.Utc,
.FloatParseHandling = JsonFloatParseHandling.Double
}

C#

new JsonDeserializationSettings
{
MissingMemberHandling = JsonMissingMemberHandling.Ignore,
NullValueHandling = JsonNullValueHandling.Ignore,
DateParseHandling = JsonDateParseHandling.DateTime,
DateTimeZoneHandling = JsonDateTimeZoneHandling.Utc,
FloatParseHandling = JsonFloatParseHandling.Double
}

Windows - Configuración heredada

Panel de propiedades

Común

• Nombre para mostrar: el nombre para mostrar de la actividad.

Entrada

• JSON Sample - The JSON sample for the output structure. The activity converts this sample into a similar object which you can use during design time.
• CadenaJson : la cadena que contiene el JSON que se va a deserializar. Escribe el JSON sin sangrías y entre comillas dobles. Las cadenas dentro del JSON deben escribirse con comillas simples. Por ejemplo, el siguiente JSON es válido: "{'TenancyName': 'Documentation', 'UsernameOrEmail': 'Documentation', 'Password': 'Documentation'}".

Otros

• Privado : si se selecciona, los valores de variables y argumentos ya no se registran en el nivel Detallado.
• TipoDeArgumento : una lista desplegable que te permite seleccionar el tipo de salida devuelta. La cadena JSON proporcionada se convertirá en un objeto del tipo seleccionado.

Salida

• JsonObject - The deserialized JSON, as a Newtonsoft.Json.Linq.JObject variable. Read more information on Newtonsoft.Json.Linq.JObject objects.

Ejemplo de uso de la actividad Deserializar Json

La actividad Deserializar Json se utiliza para extraer datos de archivos JSON y cadenas.

Para ejemplificar cómo utilizar esta actividad, hemos creado un ejemplo que utiliza la actividad Deserializar Json para extraer la edad de una persona de un archivo JSON.

Así es como se puede construir el proceso de automatización:

1. Abre Studio y crea un nuevo Proceso.
2. Drag a Sequence to the Workflow Designer.

   • Crea las siguientes variables:

     Nombre de la variable	Tipo de variable	Valor predeterminado
     JsonText	Cadena
     JsonObj	JObject
     Age	ValorGenérico

3. Drag a Read Text File activity inside the sequence container.
   • In the Properties panel, add the value "JsonString.txt" in the FileName field.
   • Add the variable JsonText in the Content field.
4. Añade una actividad Deserializar JSON después de la actividad Leer archivo de texto.
   • In the Properties panel, add the variable JsonText in the JsonString field and the variable JsonObj in the JsonObject field.
   • Select the value Newtonsoft.Json.Linq.JObject from the TypeArgument drop-down list.
5. Place an Assign activity under the Deserialize JSON activity.
   • Add the variable Age in the To field and the expression JsonObj.SelectToken("age").ToString in the Value field.
6. Arrastra una actividad Escribir línea después de la actividad Asignar.
   • Agrega la variable Age en el campo Texto.
7. Run the process. The robot extracts the age value from the .json file and displays it in the Output panel.

Aquí puedes descargar un ejemplo.