本文介绍了如何在Kotlin中与Gson一起使用TypeToken +泛型的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
val turnsType = TypeToken< p> ; List< Turns>>(){} .type
val turns = Gson()。fromJson(pref.turns,turnsType)
它表示:
无法访问'< init>'它是'公开的/ *包* /'in'TypeToken'
解决方案
创建这个内联乐趣:
内联趣味< reified T> Gson.fromJson(json:String)= this.fromJson< T>(json,object:TypeToken< T>(){} .type)
然后你可以这样调用它:
val turns = Gson() .fromJson< Turns>(pref.turns)
//或
val turn:Turns = Gson()。fromJson(pref.turns)
注意:这种方法在旧的kotlin插件版本中是不可能的,但现在您可以使用它。
以前的替代品:
替代品1:
val turnsType = object:TypeToken< List< Turns>>(){} .type
val turns = Gson ).fromJson< List< Turns>>(pref.turns,turnsType)
object:
以及 fromJson的具体类型< List< Turns>>
其他选择2:
它也可以通过这种方式实现:
inline fun< reified T> useType:genericType()= object:TypeToken< T>(){} .type
p>
val turnsType = genericType< List< Turns>>()
I'm unable to get a List of generic type from a custom class (Turns):
val turnsType = TypeToken<List<Turns>>() {}.type
val turns = Gson().fromJson(pref.turns, turnsType)
it said:
cannot access '<init>' it is 'public /*package*/' in 'TypeToken'
解决方案
Create this inline fun:
inline fun <reified T> Gson.fromJson(json: String) = this.fromJson<T>(json, object: TypeToken<T>() {}.type)
and then you can call it in this way:
val turns = Gson().fromJson<Turns>(pref.turns)
// or
val turns: Turns = Gson().fromJson(pref.turns)
NOTE: This approach was not possible before in old kotlin plugin versions but now you are able to use it.
Previous Alternatives:
ALTERNATIVE 1:
val turnsType = object : TypeToken<List<Turns>>() {}.type
val turns = Gson().fromJson<List<Turns>>(pref.turns, turnsType)
You have to put object :
and the specific type in fromJson<List<Turns>>
ALTERNATIVE 2:
As @cypressious mention it can be achieved also in this way:
inline fun <reified T> genericType() = object: TypeToken<T>() {}.type
use as:
val turnsType = genericType<List<Turns>>()
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