使用Scala中的Play Framework转换原始Json(Transform original Json using Play Framework in Scala)

使用Scala中的Play Framework转换原始Json(Transform original Json using Play Framework in Scala)

我有以下Json：

{"id": 1, "url":[ {"format":""}, {"creator":""}, {"value":"http://..."} ] }

如何将其转换为"url": "http://..." 。

我怎样才能做到这一点？ 我尝试了以下，但它似乎没有工作。

(json \ "value").as[JsString].value

I have the following Json:

{"id": 1, "url":[ {"format":""}, {"creator":""}, {"value":"http://..."} ] }

How can I transform it into "url": "http://...".

How can I do this? I tried the following, but it does not seem to be working.

(json \ "value").as[JsString].value

最满意答案

因为“url”具有数组值，所以需要对其进行适当的索引以获得所需的值，因此在您的情况下，它将类似于：

scala> val url = (json \ "url")(2).get url: play.api.libs.json.JsValue = {"value":"http://..."} scala> (url \ "value").as[String] res22: String = http://...

如果您对处理JSON的更实用的方法感到满意，我还建议您查看Argonaut 。 它远远优于Play提供的JSON设施。

Because "url" has an array value, you need to index it appropriately to get the value you require, so in your case it would be something like:

scala> val url = (json \ "url")(2).get url: play.api.libs.json.JsValue = {"value":"http://..."} scala> (url \ "value").as[String] res22: String = http://...

I also recommend looking at Argonaut if you are comfortable with a more functional approach for dealing with JSON. It is much superior to the JSON facilities provided by Play.