Spring Boot REST API 以格式显示响应结果
Spring Boot REST API display response results in the format
我想以以下格式显示响应结果:
[
"author":"Author",
"status_code":"200"
"message":"GET",
"description":"Description..."
"data":{
"id":"123",
"name":""My Name",
... : ...
}
]
但是我只显示每条数据结果如下:
[
{
"id":"123",
"name":""My Name",
... : ...
}
]
这是显示响应结果的代码片段
- 控制器
@GetMapping(value = "categorie")
fun getAllFoodCategories(@Param(value = "key") key:String): ResponseEntity<List<Category>> {
if (key==AppUtils.APY_KEY){
var foodCategories: List<Category> = foodCategoryRespository.findAll()
return if(!foodCategories.isEmpty()){
foodCategories.forEach { v ->
run {
logger.info(v.toString())
}
}
ResponseEntity(foodCategories, HttpStatus.OK)
}else{
ResponseEntity(HttpStatus.NO_CONTENT)
}
}else{
return ResponseEntity(HttpStatus.UNAUTHORIZED)
}
}
- 型号class
@Entity(name = "categories")
data class Category(
@Id
@Column(name = "id")
val id: Int,
@get: NotBlank
@Column(name = "name")
val name: String
)
谁能帮我解答一下,谢谢。
你不能return不同的元素作为一个列表,但是你可以return作为一个对象,它包含一个类别的列表。像这样:
{
"author":"Author",
"status_code":"200"
"message":"GET",
"description":"Description..."
"data":[
{
"id":"123",
"name":""My Name",
... : ...
}
]
}
为此,创建如下模型:
data class MyResponse(
val author: String,
val statusCode: String,
val message: String,
val description: String,
val data: List<Category>
)
然后 return 与:
fun getAllFoodCategories(@Param(value = "key") key:String): ResponseEntity<MyResponse> {...}
我想以以下格式显示响应结果:
[
"author":"Author",
"status_code":"200"
"message":"GET",
"description":"Description..."
"data":{
"id":"123",
"name":""My Name",
... : ...
}
]
但是我只显示每条数据结果如下:
[
{
"id":"123",
"name":""My Name",
... : ...
}
]
这是显示响应结果的代码片段
- 控制器
@GetMapping(value = "categorie")
fun getAllFoodCategories(@Param(value = "key") key:String): ResponseEntity<List<Category>> {
if (key==AppUtils.APY_KEY){
var foodCategories: List<Category> = foodCategoryRespository.findAll()
return if(!foodCategories.isEmpty()){
foodCategories.forEach { v ->
run {
logger.info(v.toString())
}
}
ResponseEntity(foodCategories, HttpStatus.OK)
}else{
ResponseEntity(HttpStatus.NO_CONTENT)
}
}else{
return ResponseEntity(HttpStatus.UNAUTHORIZED)
}
}
- 型号class
@Entity(name = "categories")
data class Category(
@Id
@Column(name = "id")
val id: Int,
@get: NotBlank
@Column(name = "name")
val name: String
)
谁能帮我解答一下,谢谢。
你不能return不同的元素作为一个列表,但是你可以return作为一个对象,它包含一个类别的列表。像这样:
{
"author":"Author",
"status_code":"200"
"message":"GET",
"description":"Description..."
"data":[
{
"id":"123",
"name":""My Name",
... : ...
}
]
}
为此,创建如下模型:
data class MyResponse(
val author: String,
val statusCode: String,
val message: String,
val description: String,
val data: List<Category>
)
然后 return 与:
fun getAllFoodCategories(@Param(value = "key") key:String): ResponseEntity<MyResponse> {...}