科特林的Gson toJson和ArrayList

Question

{
  "metadata": {
    "count": 0,
    "type": "string"
  },
  "results": [
    {
      "obj1": {
        "param1": "s1",
        "param2": "s2"
      },
      "obj2": {
        "param3": 0,
        "param4": 0,
        "param5": 0
      },
      "obj3": 0,
      "obj4": "27/12/2017"
    }
  ]
}

var contentResponse: String = ""
try{
    this.contentResponse.plus(Gson().toJson(response))
} catch (e: Exception){
    Log.e("Gson error", e.toString())
}

Answer 1

字符串在JVM中是不可变的。呼叫

this.contentResponse.plus(Gson().toJson(response))

相当于

this.contentResponse + (Gson().toJson(response))

这样，您可以更好地看到，您没有将结果分配给任何东西。将其更改为

this.contentResponse = this.contentResponse.plus(Gson().toJson(response))