问题描述
在 Play Framework 的主页上,他们声称JSON 是一等公民".我还没有看到证据.
在我的项目中,我正在处理一些非常复杂的 JSON 结构.这只是一个非常简单的例子:
<代码>{key1":{子键1":{k1":值1"k2":["val1",val2"val3"]}}key2":[{"j1": "v1",j2":v2"},{"j1": "x1",j2":x2"}]}
现在我知道 Play 正在使用 Jackson 来解析 JSON.我在我的 Java 项目中使用 Jackson,我会做一些简单的事情:
ObjectMapper mapper = new ObjectMapper();映射<字符串,对象>obj = mapper.readValue(jsonString, Map.class);
这可以很好地将我的 JSON 解析为我想要的 Map 对象 - 字符串和对象对的映射,并且可以让我轻松地将数组转换为 ArrayList
.
Scala/Play 中的相同示例如下所示:
val obj: JsValue = Json.parse(jsonString)
这反而给了我一个专有的 JsObject
type我在追求什么.
我的问题是:我能否像在 Java 中一样轻松地将 Scala/Play 中的 JSON 字符串解析为 Map
而不是 JsObject
?
附带问题:在 Scala/Play 中使用 JsObject
而不是 Map
有什么原因吗?
我的堆栈:Play Framework 2.2.1/Scala 2.10.3/Java 8 64bit/Ubuntu 13.10 64bit
更新:我可以看到 Travis 的回答得到了赞成,所以我想这对每个人都有意义,但我仍然看不到如何应用它来解决我的问题.假设我们有这个例子(jsonString):
<预><代码>[{"key1": "v1",key2":v2"},{"key1": "x1",key2":x2"}]好吧,根据所有说明,我现在应该放入所有我不明白其目的的样板:
case class MyJson(key1: String, key2: String)隐式 val MyJsonReads = Json.reads[MyJson]val 结果 = Json.parse(jsonString).as[List[MyJson]]
看起来不错,是吧?但是等一下,数组中有另一个元素完全破坏了这种方法:
<预><代码>[{"key1": "v1",key2":v2"},{"key1": "x1",key2":x2"},{"key1": "y1",key2":{"subkey1": "subval1",subkey2":subval2"}}]第三个元素不再与我定义的案例类匹配 - 我又回到了第一个方格.我每天都可以在 Java 中使用这种更复杂的 JSON 结构,Scala 是否建议我应该简化我的 JSON 以适应它的类型安全"策略?如果我错了,请纠正我,但我认为该语言应该为数据服务,而不是相反?
UPDATE2: 解决方案是将 Jackson 模块用于 Scala(我的回答中的示例).
解决方案我选择使用 Scala 的杰克逊模块.
import com.fasterxml.jackson.databind.ObjectMapper导入 com.fasterxml.jackson.module.scala.DefaultScalaModule导入 com.fasterxml.jackson.module.scala.experimental.ScalaObjectMapperval mapper = new ObjectMapper() 与 ScalaObjectMappermapper.registerModule(DefaultScalaModule)val obj = mapper.readValue[Map[String, Object]](jsonString)
On Play Framework's homepage they claim that "JSON is a first class citizen". I have yet to see the proof of that.
In my project I'm dealing with some pretty complex JSON structures. This is just a very simple example:
{
"key1": {
"subkey1": {
"k1": "value1"
"k2": [
"val1",
"val2"
"val3"
]
}
}
"key2": [
{
"j1": "v1",
"j2": "v2"
},
{
"j1": "x1",
"j2": "x2"
}
]
}
Now I understand that Play is using Jackson for parsing JSON. I use Jackson in my Java projects and I would do something simple like this:
ObjectMapper mapper = new ObjectMapper();
Map<String, Object> obj = mapper.readValue(jsonString, Map.class);
This would nicely parse my JSON into Map object which is what I want - Map of string and object pairs and would allow me easily to cast array to ArrayList
.
The same example in Scala/Play would look like this:
val obj: JsValue = Json.parse(jsonString)
This instead gives me a proprietary JsObject
type which is not really what I'm after.
My question is: can I parse JSON string in Scala/Play to Map
instead of JsObject
just as easily as I would do it in Java?
Side question: is there a reason why JsObject
is used instead of Map
in Scala/Play?
My stack: Play Framework 2.2.1 / Scala 2.10.3 / Java 8 64bit / Ubuntu 13.10 64bit
UPDATE: I can see that Travis' answer is upvoted, so I guess it makes sense to everybody, but I still fail to see how that can be applied to solve my problem. Say we have this example (jsonString):
[
{
"key1": "v1",
"key2": "v2"
},
{
"key1": "x1",
"key2": "x2"
}
]
Well, according to all the directions, I now should put in all that boilerplate that I otherwise don't understand the purpose of:
case class MyJson(key1: String, key2: String)
implicit val MyJsonReads = Json.reads[MyJson]
val result = Json.parse(jsonString).as[List[MyJson]]
Looks good to go, huh? But wait a minute, there comes another element into the array which totally ruins this approach:
[
{
"key1": "v1",
"key2": "v2"
},
{
"key1": "x1",
"key2": "x2"
},
{
"key1": "y1",
"key2": {
"subkey1": "subval1",
"subkey2": "subval2"
}
}
]
The third element no longer matches my defined case class - I'm at square one again. I am able to use such and much more complicated JSON structures in Java everyday, does Scala suggest that I should simplify my JSONs in order to fit it's "type safe" policy? Correct me if I'm wrong, but I though that language should serve the data, not the other way around?
UPDATE2: Solution is to use Jackson module for scala (example in my answer).
解决方案I've chosen to use Jackson module for scala.
import com.fasterxml.jackson.databind.ObjectMapper
import com.fasterxml.jackson.module.scala.DefaultScalaModule
import com.fasterxml.jackson.module.scala.experimental.ScalaObjectMapper
val mapper = new ObjectMapper() with ScalaObjectMapper
mapper.registerModule(DefaultScalaModule)
val obj = mapper.readValue[Map[String, Object]](jsonString)
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