如何解析Scala中复杂的json

Question

我是一个真正的Scala新手（使用Play框架）并试图找出解析复杂JSON的最佳方法。

这是一个例子，我有： { “ID”： “测试1”， “的t max”：270， “AT”：2， “BCAT”：[ “IAB26”， “IAB25 ” “IAB24” ]，

"imp":[ 
     { 
     "id":"1", 
     "banner":{ 
      "w":320, 
      "h":480, 
      "battr":[ 
       10 
      ], 
      "api":[ 
      ] 
     }, 
     "bidfloor":0.69 
     } 
    ], 
    "app":{ 
     "id":"1234", 
     "name":"Video Games", 
     "bundle":"www.testapp.com", 
     "cat":[ 
     "IAB1" 
     ], 
     "publisher":{ 
     "id":"1111" 
     } 
    }, 
    "device":{ 
     "dnt":0, 
     "connectiontype":2, 
     "carrier":"Ellijay Telephone Company", 
     "dpidsha1":"e5f61ae0597d8abee94860d66f7d512aa68d0985", 
     "dpidmd5":"c1827fe90bae819017dfdb30db7e84fa", 
     "didmd5":"1f6cb9dc519db8e48cf9592b31cce04e", 
     "didsha1":"2e6a5d7f5fd1b2b5dea56a80f2b9dc24902a0ca7", 
     "ifa":"422aeb3a-6507-40b5-9e6f-42a0e14b51be", 
     "osv":"4.4", 
     "os":"Android", 
     "ua":"Mozilla/5.0 (Linux; Android 4.4.2; DL1010Q Build/KOT49H) AppleWebKit/537.36 (KHTML, like Gecko) Version/4.0 Chrome/30.0.0.0 Safari/537.36", 
     "ip":"24.75.160.74", 
     "devicetype":1, 
     "geo":{ 
     "type":1, 
     "country":"USA", 
     "city":"Jasper" 
     } 
    }, 
    "user":{ 
     "id":"9c2be7c2a3dfe19070f193910e92b2e0" 
    }, 
    "cur":[ 
     "USD" 
    ] 
    } 

非常感谢您！

Answer 1

我会按照http://json4s.org/#extracting-values（例如，

scala> import org.json4s._ 
scala> import org.json4s.jackson.JsonMethods._ 
scala> implicit val formats = DefaultFormats // Brings in default date formats etc. 
scala> case class Child(name: String, age: Int, birthdate: Option[java.util.Date]) 
scala> case class Address(street: String, city: String) 
scala> case class Person(name: String, address: Address, children: List[Child]) 
scala> val json = parse(""" 
     { "name": "joe", 
      "address": { 
      "street": "Bulevard", 
      "city": "Helsinki" 
      }, 
      "children": [ 
      { 
       "name": "Mary", 
       "age": 5, 
       "birthdate": "2004-09-04T18:06:22Z" 
      }, 
      { 
       "name": "Mazy", 
       "age": 3 
      } 
      ] 
     } 
     """) 

scala> json.extract[Person] 
res0: Person = Person(joe,Address(Bulevard,Helsinki),List(Child(Mary,5,Some(Sat Sep 04 18:06:22 EEST 2004)), Child(Mazy,3,None))) 

Answer 2

不知道它是否可以单独使用，但Play Framework有很好的工具可以使用JSON。看一看here

Answer 3

使用隐式自动获取具有相同名称的json值。

，如果你有一个像

val jsonString = 
    { "company": { 
      "name": "google", 
      "department": [ 
       { 
        "name": "IT", 
        "members": [ 
        { 
         "firstName": "Max", 
         "lastName": "Joe", 
         "age" : 25 
        }, 
        { 
         "firstName": "Jean", 
         "lastName": "Nick", 
         "age" : 55, 
         "salary": 100,000 
        } 
        ] 
       }, 
       { 
        "name": "Marketing" 
        "members": [ 
         { 
         "firstName": "Mike", 
         "lastName": "Lucas", 
         "age" : 43 
         } 
        ] 
       }, 
      ] 
     } 
     } 

一个字符串，则我们应该做如下解析它变成阶对象

val json: JsValue = Json.parse(jsonString) // string to JsValue 
case class Person(firstName: String, lastName: String, age: Int, salary: Option[Int]) // keep the lower level object above since you will use them below 
implicit personR = Json.reads[Person] 
// implicit will read the json attribute if you use the same name as it is in json String. For example "name" : "google" will be read if you have Company(name: String). See the "name" and name:String . 

case class Department(name: String, members: Seq[Person]) 
implicit departmentR = Json.reads[Department] 

case class Company(name: String, department: Seq[Department]) 
implicit val companyR = Json.reads[Company] 

val result = json.validate[Company] match { 
    case s: JsSuccess[Company] => s.get // s.get will return a Company object 
    case e: JsError => println("error") 
}