我希望能够使用GSON存储和获取序列化的对象列表。
但是,当我试图获取List时,我得到了LinkedTreeMaps的ArrayList的奇怪结构(每个对象属性)
这是我存储prefs存储的方法:
fun <T> putAsJson(key: String, valueObject: T?) {
if (valueObject != null) {
put(key, gson.toJson(valueObject))
} else {
put(key, null)
}
}
以下是我如何获得它:
inline fun <reified T: Any> getFromJson(key: String): T? {
val jsonValue = get<String>(key)
return try {
gson.fromJson(jsonValue, T::class.java)
} catch (ex: Exception) {
Timber.e(ex, "Error when parsing JSON representing ${T::class.java} class")
null
}
}
所以我只是将它存储为:
fun saveSomeList(list: List<SomeObject>?) {
someStorage.putAsJson(KEY, list)
}
然后我尝试使用此方法获取它:
fun getSomeList(): List<SomeObject> {
return someStorage.getFromJson<List<SomeObject>>(KEY) ?: emptyList()
}
我做错了什么?
对于集合(和其他通用)类型,GSON无法自动提取正确的类型。您需要通过提供“类型令牌”来帮助它:
// Known list
val ints = listOf(1, 2, 3)
val collectionType = object : TypeToken<Collection<Int>>() {}.type
val json = gson.toJson(ints, collectionType)
val ints2: List<Int> = gson.fromJson(json, collectionType)
// Known generic
val foo = Foo<Bar>()
val fooType = object : TypeToken<Foo<Bar>>() {}.type
val json = gson.toJson(foo, fooType)
val foo2: Foo<Bar> = gson.fromJson(json, fooType)
如果你想保持内部类型的通用性,即List<T>
中的T,你可以这样做:
// Unknown generic
val modelType = T::class.java
val listType = TypeToken.getParameterized(List::class.java, modelType).type
val list = gson.fromJson(jsonValue, listType)
我认为您最好的选择是列表的特定方法:
inline fun <reified T: Any> getListFromJson(key: String): T? {
val jsonValue = get<String>(key)
return try {
val listType = TypeToken.getParameterized(List::class.java, T::class.java).type
gson.fromJson(jsonValue, listType)
} catch (ex: Exception) {
Timber.e(ex, "Error when parsing JSON representing List<${T::class.java}> class")
null
}
}
资料来源: - https://github.com/google/gson/blob/master/UserGuide.md#TOC-Collections-Examples - https://stackoverflow.com/a/44303909/2957169