尽管在SELECT语句中使用了JSON标志,但来自Cassandra数据库的数据却以无效的JSON返回。
这是我收到的数据。
当我的前端接收到这个数据时,它当然被认为是无效的JSON。同时我也不知道为什么 "[json]"
是有的。
Future(Success(["[json]":'{"pasta_name": "conchiglie", "id": 2, "description": "description for conchiglie", "image": {"alt": "alt text", "src": "image.jpg"}}']))
这是我想收到的东西。
{
"pasta_name": "conchiglie",
"id": 2, "description":
"description for conchiglie",
"image":
{
"alt": "alt text",
"src": "image.jpg"
}
}
这是我的简化代码
object Server extends App {
implicit val system: ActorSystem = ActorSystem("helloworld")
implicit val executor: ExecutionContext = ExecutionContext.global
implicit val materializer: ActorMaterializer = ActorMaterializer()
val sessionSettings = CassandraSessionSettings()
implicit val cassandraSession: CassandraSession =
CassandraSessionRegistry.get(system).sessionFor(sessionSettings)
val recipes: Future[String] =
CassandraSource(s"SELECT JSON * FROM danlough.recipe_by_pasta").map(row => row.getFormattedContents()).runWith(Sink.head)
import akka.http.scaladsl.server.Directives._
def route = path("getRecipe") {
get {
respondWithHeaders(RawHeader("Access-Control-Allow-Origin", "http://localhost:3000"), RawHeader("Vary", "Origin")) {
complete(recipes)
}
}
}
Http().bindAndHandle(route, host, port)
你需要改变 row => row.getFormattedContents()
到 row => row.getString(0)
只提取包含Cassandra格式化的JSON的字段,而不是像您执行 row.getFormattedContents()
.