Question

我使用JSON.parseFull解析Json。在解析之前，Json就像这样

{
"response":
    {
    "status":"ok",
    "userTier":"developer",
    "total":1,
    "content":
        {
        "id":"technology/2014/feb/18/doge-such-questions-very-answered",
        "type":"article",
        "sectionId":"technology",
        "sectionName":"Technology",
        "webPublicationDate":"2014-02-18T10:25:30Z",
        "webTitle":"What is Doge?",
        "webUrl":"https://www.theguardian.com/technology/2014/feb/18/doge-such-questions-very-answered",
        "apiUrl":"https://content.guardianapis.com/technology/2014/feb/18/doge-such-questions-very-answered",
        "isHosted":false
        }
    } 
}

解析后，它就变成了这个，

Map(response -> 
    Map(status -> ok, 
    userTier -> developer, 
    total -> 1.0, 
    content -> 
    Map(webUrl -> 
        https://www.theguardian.com/technology/2014/feb/18/doge-such-questions-very-answered, 
        webPublicationDate -> 2014-02-18T10:25:30Z, 
        webTitle -> What is Doge?, 
        sectionName -> Technology, 
        apiUrl -> https://content.guardianapis.com/technology/2014/feb/18/doge-such-questions-very-answered, 
        id -> technology/2014/feb/18/doge-such-questions-very-answered, 
        isHosted -> false, 
        sectionId -> technology, 
        type -> article
        )
    )
)

我需要获取像webUrl和webtitle这样的值。

任何人都知道如何实现这一目标？

Answer 1

Scala Json库有点脆弱，从某种意义上说它并不强烈地键入你解析的Json。因此，我的猜测是在这里和那里放弃asInstanceOf个电话。以下示例

  type JsonMap = Map[String, Any]

  val maybeParsedMap: Option[JsonMap] = JSON.parseFull(jsonString).map(_.asInstanceOf[JsonMap])

  val content: Option[JsonMap] = for {
    parsedMap <- maybeParsedMap
    response <- parsedMap.get("response")
    content <- response.asInstanceOf[JsonMap].get("content")
  } yield content.asInstanceOf[JsonMap]

  val webUrl: Option[String] = content.asInstanceOf[JsonMap].get("webUrl").map(_.asInstanceOf[String])

从JSON.parseAll提供的嵌套Map获取值

1 个答案: