com.google.gson.JsonPrimitive无法转换为com.google.gson.JsonObject [英] com.google.gson.JsonPrimitive cannot be cast to com.google.gson.JsonObject

问题描述

我有一个来自PHP的json列表:

I have a json list like this from PHP:

$Json = '[{"test":"1", "message":"try it"}, 
{"test":"2", "message":"try it"}, {"test":"3", "message":"try it"} ...]';
$final = [ 'error' => '1', 'json' => json_encode($Json)];
die(json_encode($final));

在Android上，我可以显示如下结果:

From Android i can show the result like this:

JsonParser jsonParser = new JsonParser();
JsonObject res = (JsonObject) jsonParser.parse(new Gson().toJson(response.body()));
Log.w("Return", response.body().toString());

到目前为止一切正常，但是当我尝试从返回的结果中创建新的Json Object时，出现以下错误消息:

All works fine until now, but when i try to make a new Json Object from the returned results, i get this error message:

com.google.gson.JsonPrimitive cannot be cast to com.google.gson.JsonObject

这是我所做的:

JsonObject json = (JsonObject) jsonParser.parse(new Gson().toJson(res.get("json").toString()));
Log.w("JSON", json.toString());

请修复吗?

推荐答案

首先，修复您的PHP.

First, fix your PHP.

$Json = array(
    array('test' => '1', 'message' => 'try it'), 
    array('test' => '2', 'message' => 'try it')
  );
$final = array(
  'error' => '1', 
  'json' => $Json
);
die(json_encode($final, JSON_FORCE_OBJECT));

您不需要编码一个已经有效的JSON字符串.

You don't need to encode an already valid JSON string.

一切正常

您从未在那里使用过res.您按原样打印了响应正文.

You never used res there. You printed the response body as-is.

Log.w("Return", response.body().toString());

这看起来是错误的.响应已经是一个JSON字符串，因此toJson不是必需的.

This looks wrong. The response is already a JSON string, so toJson wouldn't be necessary.

jsonParser.parse(new Gson().toJson(response.body()));

您的错误是"JsonPrimitive"，在这种情况下是字符串，它不是JSON对象.

Your error is the "JsonPrimitive" in this case is a String, which is not a JSON object.

您应该这样做

final String body = response.body().toString(); // Or use response.raw(), if want raw response string
JsonParser jsonParser = new JsonParser();
JsonObject res = jsonParser.parse(body).getAsJsonObject();
Log.w("Return", res.toString());

如果您想要数据，那么可以拥有

If you want the data, then you can have

JsonArray data = res.getAsJsonArray("data");

---

或者，您将需要进行POJO并反序列化数据列表.

---

Alternatively, you will need to make a POJO and deserialize your data list.

class Data {
    String test, message;
}

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