scala 如何在 Play 中替换 JSON 值
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How to replace a JSON value in Play
提问by Farmor
How do I replace a value in a JSON value in Play?
Code to illustrate:
如何在 Play 中替换 JSON 值中的值？
代码说明：
def newReport() = Action(parse.json) { request =>
var json = request.body
if((json \ "customerId").as[Int] == -1){
// replace customerId after some logic to find the new value
}
json.validate[Report](Reports.readsWithoutUser).map {
case _: Report =>
回答by Zeimyth
According to the Play Documentation, JsObjects have a method ++that will merge two JsObjects. So, when you have your new integer value, you simply need:
根据Play 文档，JsObjects 有一个方法++可以合并两个 JsObjects。因此，当您拥有新的整数值时，您只需要：
val updatedJson = json.as[JsObject] ++ Json.obj("customerId" -> newValue)
As of Play 2.4.x you can use +:
从 Play 2.4.x 开始，您可以使用+：
val updatedJson = json.as[JsObject] + ("customerId" -> newValue)
(NOTE: the +method was added already in 2.1.x but adds a duplicate field to the object instead of replacing the existing value in versions prior to 2.4.x)
（注意：该+方法已在 2.1.x 中添加，但向对象添加了重复字段，而不是替换 2.4.x 之前版本中的现有值）
回答by JMess
One approach is, as Marc B says, convert the JSON to something like a case class, manipulate that, and then create a new JSON.
一种方法是，正如 Marc B 所说，将 JSON 转换为类似于 case 类的东西，对其进行操作，然后创建一个新的 JSON。
However you can also use JSON 'transformers'. Effectively what you do is build a Reads[SomeThing] object. This object is passed to the transform method which you call on your JSON object. It will change the JSON object and return a Success(result) or Failure(error) where result is the new modified JSON. Here's a (comparatively)verysimple example:
但是，您也可以使用 JSON“转换器”。您所做的实际上是构建一个 Reads[SomeThing] 对象。该对象被传递给您在 JSON 对象上调用的转换方法。它将更改 JSON 对象并返回 Success(result) 或 Failure(error)，其中 result 是新修改的 JSON。这是一个（相对）非常简单的例子：
using json of format: {key -> value}
使用 json 格式：{key -> value}
def jsonXForm(value: String) = (__ \ "customerId").json.update(
(__ \ "customerId").json.put(JsString(value))
)
json.transform(jsonXForm(yourNewValue)) match {...}`
There is a decentguide here
有一个体面的指南这里
回答by scalapeno
Something along the lines of:
类似的东西：
val updatedJson = if((request.body \ "customerId").as[Int] == -1){
val newId = JsObject(Seq(("customerId",JsString("ID12345"))))
(request.body ++ newId).as[JsValue]
} else request.body
updatedJson.validate[Report](Reports.readsWithoutUser).map {
case _: Report =>
回答by Wilfred Springer
I'm considering moving away of all of those immutable "JSON" solutions. It's just making the code a horrible mess. This is how it would look in SON of JSON:
我正在考虑放弃所有那些不可变的“JSON”解决方案。它只是让代码变得一团糟。这是它在JSON 的 SON 中的样子：
import nl.typeset.sonofjson._
val json = …
if (json.customerId.as[Int] == -1) {
json.customerId = 987938
}
回答by rogue-one
a glorified version of Zeimyth's answer that uses implicit conversion
使用隐式转换的 Zeimyth 答案的美化版本
implicit class JsObjectEnhancer(jsObject: JsObject) {
def update(args: (String, Json.JsValueWrapper)*): JsObject = {
jsObject ++ Json.obj(args: _*)
}
}
usage(json must be of type JsObject and the implicit class should be in scope)
用法（json 必须是 JsObject 类型并且隐式类应该在范围内）
json.update("customerId", 1000)
