java - 将 json 值 key 对解压为 SharedPreferences 的变量

Question

我有一个 json，我需要一些帮助来格式化它，以便我可以将值保存在 SharedPreferences 中，这是困扰我的代码。

public class MainActivity extends AppCompatActivity {
public static final String EXTRA_MESSAGE = "com.example.cumaskp.mercfood";
private SharedPreferences mPreferences;
private SharedPreferences.Editor mEditor;
EditText username;
EditText password;
Button loginbutton;
Button signupBtn;


@Override
protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.activity_main);

    mPreferences = PreferenceManager.getDefaultSharedPreferences(this);
    mEditor = mPreferences.edit();

    username = (EditText) findViewById(R.id.usernameEditText);
    password = (EditText) findViewById(R.id.mailEditText);
    signupBtn = (Button) findViewById(R.id.signUpButton);
    loginbutton = (Button) findViewById(R.id.loginBtn);

    signupBtn.setOnClickListener(new View.OnClickListener() {
        @Override
        public void onClick(View view){

            Intent myIntent = new Intent(MainActivity.this, getData.class);
            MainActivity.this.startActivity(myIntent);
        }
    });

    loginbutton.setOnClickListener(new View.OnClickListener() {
        @Override
        public void onClick(View view) {
            new SendPostRequest().execute();
        }
    });
}



public class SendPostRequest extends AsyncTask<String, Void, String> {

    protected void onPreExecute(){}

    protected String doInBackground(String... arg0) {

        try {

            URL url = new URL("http://192.168.111.42/api/login"); //TODO here is your URL path

            JSONObject postDataParams = new JSONObject();
            postDataParams.put("username", username.getText().toString());
            postDataParams.put("password", password.getText().toString());
            Log.e("params",postDataParams.toString());

            HttpURLConnection conn = (HttpURLConnection) url.openConnection();
            conn.setReadTimeout(15000 /* milliseconds */);
            conn.setConnectTimeout(15000 /* milliseconds */);
            conn.setRequestMethod("POST");
            conn.setDoInput(true);
            conn.setDoOutput(true);

            OutputStream os = conn.getOutputStream();
            BufferedWriter writer = new BufferedWriter(new OutputStreamWriter(os, "UTF-8"));
            writer.write(getPostDataString(postDataParams));

            writer.flush();
            writer.close();
            os.close();

            int responseCode=conn.getResponseCode();

            if (responseCode == HttpsURLConnection.HTTP_OK) {

                BufferedReader in=new BufferedReader(new
                        InputStreamReader(
                        conn.getInputStream()));

                StringBuffer sb = new StringBuffer("");
                String line="";

                while((line = in.readLine()) != null) {

                    sb.append(line);
                    break;
                }

                in.close();
                return sb.toString();

            }
            else {
                return new String("false : "+responseCode);
            }
        }
        catch(Exception e){
            return new String("Exception: " + e.getMessage());
        }
    }

@Override
protected void onPostExecute(String result) {

 // it is right here the problem is! -----------------------------------------------------------------------------

    try {
        for (int i = 0; i < result.length(); i++) {
            JSONObject jsonObj = result.getJSONObject(i);
            String k = jsonObj.keys().next();
            Log.i("Info", "Key: " + k + ", value: " + jsonObj.getString(k));
        }

    } catch (JSONException ex) {
        ex.printStackTrace();
    }


        Toast.makeText(getApplicationContext(), result, Toast.LENGTH_LONG).show();
        mEditor.putString("token",result);

        Toast.makeText(getApplicationContext(), result, Toast.LENGTH_LONG).show();
        String token = mPreferences.getString("key", "default");
        Log.d("myTag", "onPostExecute: "+ token);
    }
}

public String getPostDataString(JSONObject params) throws Exception {

    StringBuilder result = new StringBuilder();
    boolean first = true;

    Iterator<String> itr = params.keys();

    while(itr.hasNext()){

        String key= itr.next();
        Object value = params.get(key);

        if (first)
            first = false;
        else
            result.append("&");

        result.append(URLEncoder.encode(key, "UTF-8"));
        result.append("=");
        result.append(URLEncoder.encode(value.toString(), "UTF-8"));

    }
    return result.toString();
 }
}

我认为这个问题经常被问到，但我尝试在谷歌上搜索答案，但非常不清楚我应该如何将字符串格式化为变量。

我的应用程序的字符串输出如下所示:{"api_token":"eyJ0eXAiOiJKV1QiLCJhbGciOiJIUzI1NiJ9.eyJpc3MiOiJsdW1lbi1qd3QiLCJzdWIiOjEsImlhdCI6MTU0MTQyNjYxMiwiZXhwIjoxNTQxNTEzMDEy LCJuYW0iOiIxMjM0In0.7TR1ueOB-xqAkI6XgStUnn7HZxBTvxx3wzjKGHDnD4I","user_id":1}

有人告诉我，代码越多越好，中间有一个指针。

Answer 1

您不需要创建变量。 由于响应是字符串， 保存 json 本身， 我还看到一个语法错误 OnPostExecute() result 参数是字符串，而字符串没有名为 getJsonObject() 的方法，所以你需要执行以下操作

JSONObject resultAsJsonObject=new  JSONObject(result);

上面的方法可以让你使用变量作为对象，而不是字符串

我建议您将整个 json 保存到共享首选项中，这样您此时就不需要转换它 您可以创建一个模型并使用 GSON这样你就不需要将其转换为 JSONObject，但结果将是 json 结构，Gson 库会将你的 json 结果转换为 java 对象