【发布时间】：2018-03-06 21:05:46
【问题描述】：
我的请求如下所示：
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：
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
)
}
}
已被记录为：
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"
好吧，这是其中一个教程的复制和粘贴。它不工作。
我猜BeanProperty 指的是body 对象属性；这就是我可以从示例here 中猜到的。
如果我想要查询字符串？
尝试过：
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)
}
结果：
{
"message":"Hello default_name"
}
建议该类已使用空映射代替 queryStringParameters 进行初始化，但已正确提交
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：
我之所以走这条路，是因为我觉得将@BeanProperty queryStringParameters: Map[String, String] 中的Map 替换为T 类型会方便且富有表现力，例如
case class Person(@beanProperty val name: String, @beanProperty val surname: String)
但是，上面的代码将{"name":"ciao", "surname":"bonjour"} 视为String，但没有弄清楚它应该反序列化该字符串。
编辑
我也尝试用java.util.Map[String, String] 替换 scala 映射但没有成功
【问题讨论】：
-
你能以任何方式获得像
{"name":"ciao", "surname":"bonjour"}这样的字符串吗？在这么多（如果可能的话）你可以创建一个对象来保存（从字符串解析）数据？将字符串条目解析为 Java 对象对您来说是一种可能的解决方案吗？
标签： java scala aws-lambda aws-api-gateway serverless-framework