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我在 Android Scala 应用程序（使用"com.hanhuy.sbt" % "android-sdk-plugin" % "1.3.5"）中将案例类图序列化为 JSON。

我正在使用"org.json4s" %% "json4s-native" % "3.2.10"，即使使用如下所示的简单案例类，它也会失败：

package test
case class Test(text: String)

实际序列化的代码如下所示：

import org.json4s._
import org.json4s.native.Serialization
import org.json4s.native.Serialization.{read, write}

// ...

implicit val formats = Serialization.formats(NoTypeHints)

val test = Test("test")
val serialized = write(test)
info(s"Serialized to '$serialized'")

输出是：

Serialized to '{}'

我怀疑是 ProGuard 问题，我的 ProGuard 设置build.sbt如下：

proguardScala in Android := true

proguardOptions in Android ++= Seq(
  "-dontobfuscate",
  "-dontoptimize",
  "-keepattributes Signature",
  "-dontwarn scala.collection.**", // required from Scala 2.11.3
  "-dontwarn scala.collection.mutable.**", // required from Scala 2.11.0
  "-ignorewarnings",
  "-keep class scala.Dynamic",
  "-keep class test.**"
)

我也尝试过，json4s-jackson但没有任何区别。

ProGuard 日志如下所示：

Warning: com.thoughtworks.paranamer.AnnotationParanamer$Jsr330Helper: can't find referenced class javax.inject.Named
Warning: com.thoughtworks.paranamer.AnnotationParanamer$Jsr330Helper: can't find referenced class javax.inject.Named
Warning: org.joda.convert.JDKStringConverter$9: can't find referenced class javax.xml.bind.DatatypeConverter
Warning: org.joda.convert.JDKStringConverter$9: can't find referenced class javax.xml.bind.DatatypeConverter
Warning: org.joda.convert.JDKStringConverter$9: can't find referenced class javax.xml.bind.DatatypeConverter
Note: com.google.android.gms.internal.av calls '(com.google.ads.mediation.MediationAdapter)Class.forName(variable).newInstance()'
Note: com.google.android.gms.maps.internal.q: can't find dynamically referenced class com.google.android.gms.maps.internal.CreatorImpl
Note: org.joda.time.DateTimeZone calls '(org.joda.time.tz.Provider)Class.forName(variable).newInstance()'
Note: org.joda.time.DateTimeZone calls '(org.joda.time.tz.NameProvider)Class.forName(variable).newInstance()'
Note: there were 1 unresolved dynamic references to classes or interfaces.
  You should check if you need to specify additional program jars.
  (http://proguard.sourceforge.net/manual/troubleshooting.html#dynamicalclass)
Note: there were 3 class casts of dynamically created class instances.
  You might consider explicitly keeping the mentioned classes and/or
  their implementations (using '-keep').
  (http://proguard.sourceforge.net/manual/troubleshooting.html#dynamicalclasscast)
Warning: there were 5 unresolved references to classes or interfaces.
     You may need to add missing library jars or update their versions.
     If your code works fine without the missing classes, you can suppress
     the warnings with '-dontwarn' options.
     (http://proguard.sourceforge.net/manual/troubleshooting.html#unresolvedclass)
Note: You're ignoring all warnings!

有任何想法吗？

4

1 回答 1

-1

You can create a JSONObject from a String using the constructor:

 JSONObject json = new JSONObject(myString);

And to convert your JSONObject to a String, just use the toString() method:

 String myString = json.toString();

Additionally, if you are trying to get a specific String value from the JSONObject, you can do this:

 if (json.has("content"))
{
    String content = json.getString("content");
    //do something with content string
}

Finally, if you aren't very comfortable using JSONObject, I recommend using the tools provided by droidQuery to help you parse, such as:

Object[] array = $.toArray(myJSONArray);

and

Map<String, ?> map = $.map(myJSONObject);
于 2014-11-13T07:58:22.730 回答