我有一个类似以下内容的请求：
package pricing
import scala.beans.BeanProperty
class Request(@BeanProperty var name: String, @BeanProperty var surname: String) {
def this() = this(name="defName", surname="defSurname")
}
处理程序如下所示：
package pricing
import com.amazonaws.services.lambda.runtime.{Context, RequestHandler}
import scala.collection.JavaConverters
import spray.json._
class ApiGatewayHandler extends RequestHandler[Request, ApiGatewayResponse] {
import DefaultJsonProtocol._
def handleRequest(input: Request, context: Context): ApiGatewayResponse = {
val headers = Map("x-foo" -> "coucou")
val msg = "Hello " + input.name
val message = Map[String, String]("message" -> msg )
ApiGatewayResponse(
200,
message.toJson.toString(),
JavaConverters.mapAsJavaMap[String, Object](headers),
true
)
}
}
这被记录为:
functions:
pricing:
handler: pricing.ApiGatewayHandler
events:
- http:
path: pricing/test
method: get
documentation:
summary: "submit your name and surname, the API says hi"
description: ".. well, the summary is pretty exhaustive"
requestBody:
description: "Send over name and surname"
queryParams:
- name: "name"
description: "your 1st name"
- name: "surname"
description: ".. guess .. "
methodResponses:
- statusCode: "200"
responseHeaders:
- name: "x-foo"
description: "you can foo in here"
responseBody:
description: "You'll see a funny message here"
responseModels:
"application/json": "HelloWorldResponse"
好的，这是从教程中复制粘贴的内容。但它并没有起作用。我猜测BeanProperty是指body对象属性；这是我从这里的示例中所猜测的。
如果我想要查询字符串怎么办？
尝试如下：
package pricing
import scala.beans.BeanProperty
import spray.json._
abstract class ApiGatewayGetRequest(
@BeanProperty httpMethod: String,
@BeanProperty headers: Map[String, String],
@BeanProperty queryStringParameters: Map[String, String])
abstract class ApiGatewayPostRequest(
@BeanProperty httpMethod: String,
@BeanProperty headers: Map[String, String],
@BeanProperty queryStringParameters: Map[String, String])
class HelloWorldRequest(
@BeanProperty httpMethod: String,
@BeanProperty headers: Map[String, String],
@BeanProperty queryStringParameters: Map[String, String]
) extends ApiGatewayGetRequest(httpMethod, headers, queryStringParameters) {
private def getParam(param: String): String =
queryStringParameters get param match {
case Some(s) => s
case None => "default_" + param
}
def name: String = getParam("name")
def surname: String = getParam("surname")
def this() = this("GET", Map.empty, Map.empty)
}
这导致：
{
"message":"Hello default_name"
}
提示该类已使用空映射初始化，代替正确提交的queryStringParameters
Mon Sep 25 20:45:22 UTC 2017 : Endpoint request body after
transformations:
{"resource":"/pricing/test","path":"/pricing/test","httpMethod":"GET","headers":null,"queryStringParameters":{"name":"ciao", "surname":"bonjour"},"pathParameters":null,"stageVariables":null,
...
注意： 我选择这个路径是因为我认为用类型T（例如）替换
@BeanProperty queryStringParameters: Map[String, String] 中的 Map 更加方便和表达清晰。case class Person(@beanProperty val name: String, @beanProperty val surname: String)
然而，上面的代码将
{"name":"ciao", "surname":"bonjour"}视为一个String，而没有弄清楚它应该反序列化那个字符串。
编辑
我还尝试将scala map替换为java.util.Map[String, String]，但没有成功。
{"name":"ciao", "surname":"bonjour"}这样的字符串返回吗？如果可能的话，你能否创建一个对象来保存（从字符串解析出来的）数据？对于你来说，将字符串条目解析为Java对象是一种可行的解决方案吗？ - VC.One