Question

给出以下JSON ......

{
  "firstName": "Joe",
  "lastName": "Grey",
  ...
  "addresses":
  [
    {
      "name": "Default",
      "street": "...",
      ...,
      "isDefault": true
    },
    {
      "name": "Home",
      "street": "...",
      ...,
      "isDefault": false
    },
    {
      "name": "Office",
      "street": "...",
      ...,
      "isDefault": false
    }
  ]
}

...如何让我们说name等于Home的项目？

    {
      "name": "Home",
      "street": "...",
      ...,
      "isDefault": false
    }

感谢。

Answer 1

user1502304建议的链接不提供有关此主题的任何信息。此外，根据this discussion，当前的JSON库有一些限制......如上所述here，良好的替代方案是JsZipper。也就是说，下面是如何按值获取项目：

scala> import play.api.libs.json._
import play.api.libs.json._

scala> import play.api.libs.json.extensions._
import play.api.libs.json.extensions._

scala> val user = Json.parse("""
     | {
     |   "firstName": "Joe",
     |   "lastName": "Grey",
     |   "isDefault": false,
     |   "addresses":
     |   [
     |     {
     |       "name": "Default",
     |       "street": "one",
     |       "isDefault": true
     |     },
     |     {
     |       "name": "Home",
     |       "street": "two",
     |       "isDefault": false
     |     },
     |     {
     |       "name": "Office",
     |       "street": "three",
     |       "isDefault": false
     |     }
     |   ]
     | }""")
json: play.api.libs.json.JsValue = {"firstName":"Joe","lastName":"Grey","isDefault":false,"addresses":[{"name":"Default","street":"one","isDefault":true},{"name":"Home","street":"two","isDefault":false},{"name":"Office","street":"three","isDefault":false}]}

scala> JsZipper(json).findByValue(_ \ "name" == JsString("Default")).value
res8: play.api.libs.json.JsValue = {"name":"Default","street":"one","isDefault":true}

我希望它有所帮助。

Play Framework＆amp; JSON：如何按值获取数组项

1 个答案: