Json 如何在模式数组中形成多个响应
我正试图以这种格式形成一个大摇大摆的文档。响应是下面200 http代码上的json。我无法像下面这样形成jsonJson 如何在模式数组中形成多个响应,json,rest,swagger,swagger-ui,swagger-2.0,Json,Rest,Swagger,Swagger Ui,Swagger 2.0,我正试图以这种格式形成一个大摇大摆的文档。响应是下面200 http代码上的json。我无法像下面这样形成json [ { "key":"test1", "val":"val1" }, { "key":"test2", "val":"val2" }, { "key":"test3", "val":"val3" } ] 到目前为止,我有: "responses": { "2
[
{
"key":"test1",
"val":"val1"
},
{
"key":"test2",
"val":"val2"
},
{
"key":"test3",
"val":"val3"
}
]
到目前为止,我有:
"responses": {
"200": {
"description": "response",
"schema": {
"type": "array",
"items": {
"$ref": "#/definitions/res1",
"$ref": "#/definitions/res2"
}
}
}
}
"definitions": {
"res1": {
"type": "object",
"properties": {
"key": {
"type": "string",
"example": "test1"
},
"keyURL": {
"type": "string",
"example": "val1"
}
}
},
"res2": {
"type": "object",
"properties": {
"key": {
"type": "string",
"example": "test2"
},
"val": {
"type": "string",
"example": "val2"
}
}
}
但是我根本看不到res2
块。我只看到res1
JSON指针($ref
)必须“单独”存在于对象中。因此,项
块中不能有多个指针:
"$ref": "#/definitions/res1",
"$ref": "#/definitions/res2"
将忽略第二个参考(res2
),只应用第一个参考
对于你想做的事情,你有很多选择。最简单的可能是这样的:
type: array
items:
$ref: '#/definitions/Pair'
并有这样的定义:
definitions:
Pair:
properties:
key:
type: string
value:
type: string