In my src/resources folder, I have a JSON file like this :
{
"name": "${person.id}"
}
I don't want the value of name key to be hard coded. I want it to be dynamic (person.id value is defined in application.properties)
I'm using ObjectMapper to turn this JSON into Java Object
But it doesn't work, ${person.id} is interpreted as real value, not a placeholder
Related
how can write condition 'if' type is no details other two fields are hide in json response. if possible i want to do it in pojo or bean ?
Using spring boot
spring data rest and hal.
pojo
MongoDB Repository
I want to show accountNo and Accountdetails if type="details"
{
"Name":"Json",
"lastName":"Amazon",
"type":"Details",
"accountNo":"12123",
"AccountdetailsDetails":[ some details]
}
If type="noDetails" just show mentioned response.
{
"Name":"Json",
"lastName":"Amazon",
"type":"NoDetails"
}
I guess you need #JsonFilter.
You can use this annotation to exclude some properties in your entity response.
What you need to do is Add this annotation with unique name in your entity file.
Then serialize this entity values using filter class SimpleFilterProvider.
Take a look on
https://www.logicbig.com/tutorials/misc/jackson/json-filter-annotation.html
I need to decompose my incoming JSON by fields in me REST Controller with Spring Boot.
My request body:
{
"text": "my text",
"myEnum": "VALUE1"
}
And my controller:
#PatchMapping("/{id}")
Object updateEntity(#PathVariable Long id, String text, MyEnum myEnum) {
/* ... */
}
#RequestParam doesn't work because it's just for query string params, #RequestBody doesn't work too because it handle whole body. But I need decompose incoming body by fields and inject into controller. I know what I can use Map <String, String> for this, but I would like validate my incoming fields, and I have the fields with difference types. And I don't want to create one class by incoming body for each controller.
If I haven't misunderstood your requirement, the usual way to deal with incoming JSON is to define a class that reflects your expected input, and make that the controller method parameter annotated as RequestBody.
Spring Boot, by default, uses Jackson to deserialize to your class, and so if you use matching property names then you won't need any special annotations or setup. I think enums will be handled by default, as are other types (though you may need to provide some guidance for strings representing dates or timestamps). Any bad value will fail deserialisation, which I think you can handle in ControllerAdvice (though you'll want to double check that)
I'm using Java 11.0.2, Spring Boot 2.2.6 and Spring OpenApi Core 1.1.49 to create an OpenApi documentation using annotations.
During the request for creating a merchant in the Controller I need to have a custom header property, but which needs to be optional. Based on Swagger documentation for Parameter Object, the field "required" (Determines whether this parameter is mandatory. If the parameter location is "path", this property is REQUIRED and its value MUST be true. Otherwise, the property MAY be included and its default value is false.) by default for header is false, but below you can see that for some reason is true (nevertheless that I configured this option to "false").
Java - part ot Controller method
public ResponseDto create(#Parameter(in = ParameterIn.HEADER, required = false, schema = #Schema(type = "string", format = "uuid"), name = "X-Request-Correlation-Id", #RequestHeader("X-Request-Correlation-Id") #Nullable String headerXRequestId, ... <
This result in OpenApi yaml file - autogenerated with the information from the annotations
parameters:
- name: X-Request-Correlation-Id
in: header
required: true
schema:
type: string
format: uuid
Can you point out the problem, because I can't find a solution in the documentation or anywhere else?!
Found the solution - the problem wasn't in OpenApi annotation #Parameter, but in Spring binding annotation #RequestHeader, which binds the header field to the method parameter. #RequestHeader has also field "required" and by default, it's set on "true" and it overrides the one in #Parameter. So the solution was the following syntax - #RequestHeader(name = "X-Request-Correlation-Id", required = false).
I'm trying retrofit 2 for the first time and I have no idea how to tell it to get "Category" objects from an jsonarray named "data".
Method 1
If I do it like this it fails:
#GET("category")
Call<List<Category>> listCategories();
Method 2
But when I make a new model, called "Categories", which holds a List and is annotated with #SerializedName("data"), it works flawlessly.
#GET("category")
Call<Categories> listCategories();
My Question
Should I annotate something in the interface, like this
#GET("category")
#Annotation to look inside "data"
Call<List<Category>> listCategories();
Or should I annotate my "Category" model to tell Retrofit (or GSON)
that it lives inside the json array "data"?
JSON
{"data":[{"id":1,"name":"Fist Name","parent":0},{"id":2,"name":"Second Name","parent":1}]}
Method 2 Is correct and we use it when we dont want to use/define the json response object/arrays key names(field names). instead provide our own. Eg. In below code List object name is items but while Serialization and Deserialization it uses, what you have defined in #SerializedName annotation that is data.
public class Categories {
//solution 1
List<Category> data;//object name must match with the json response
//solution 2
#SerializedName("data")
List<Category> items;
}
Should I annotate something in the interface
No. There is no such annotation available and everything you can do is only in Response type class.
I have the following entity, that references another entity.
class Foo {
String id;
String name supplierName;
**#DBRef** TemplateSchema templateSchema;
...
}
I want to be able to use the following JSON (or similar) to create a new entity.
{
"supplierName": "Stormkind",
"templateSchema": "572878138b749120341e6cbf"
}
...but it looks like Spring forces you to use a URI like this:
{
"supplierName": "Stormkind",
"templateSchema": "/template-schema/572878138b749120341e6cbf"
}
Is there a way to create the DBRef by posting an ID instead of a URI?
Thanks!
In REST, the only form of ID's that exist are URIs (hence the name Unique Resource Identifier). Something like 572878138b749120341e6cbf does not identify a resource, /template-schema/572878138b749120341e6cbf does.
On the HTTP level, entities do not exist, only resources identified by URIs. That's why Spring Data REST expects you to use URIs as identifiers.