Question

在尝试了许多不同的方式表达以下程序后，我无法将活动中的值传递给另一个活动。由于某种原因，捆绑似乎保持为空。我不认为这可能是一个许可问题。

MainActivity：

public class MainActivity extends AppCompatActivity {

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

    public void ButtonRAWclick(View view)
    {
        Intent intent = new Intent(MainActivity.this
                , RestClient.class);
        intent.putExtra("type", "RAW_TYPE");
        startActivity(intent);
    }

    public void ButtonHTTPclick(View view)
    {
        Intent intent = new Intent(MainActivity.this
                , RestClient.class);
        intent.putExtra("type", "HTTP_TYPE");
        startActivity(intent);
    }

    public void ButtonJSONclick(View view)
    {
        Intent intent = new Intent(MainActivity.this
                , RestClient.class);
        intent.putExtra("type", "JSON_TYPE");
        startActivity(intent);
    }

}

RestClient：

public class RestClient extends AppCompatActivity implements SensorListener {
    RawHttpSensor rhs1;
    TextSensor rhs2;
    TextView temperature;
    String output;
    String type;

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

        Bundle bundle = getIntent().getExtras();
        if (bundle != null) {
            type = bundle.getString("type");
        }

        if (type == "RAW_TYPE") {
            rhs1 = new RawHttpSensor();
            temperature = (TextView) findViewById(R.id.temperature);
            rhs1.registerListener(this);
            rhs1.getTemperature();
        }

        if (type == "HTTP_TYPE")
        {
            rhs2 = new TextSensor();
            temperature = (TextView) findViewById(R.id.temperature);
            rhs2.registerListener(this);
            rhs2.getTemperature();
        }

        if (type == "JSON_TYPE")
        {
           ...
        }

      ...
    }

有人可以帮我发现错误吗？

Answer 1

<强>不

 if(type ==

<强>不要

if(type.equals("YOUR_STRING"))

您应该使用equals()

将此字符串与指定对象进行比较。如果，结果是真的并且仅当参数不为null并且是String对象时表示与此对象相同的字符序列。

if (type.equals("RAW_TYPE")) {
        rhs1 = new RawHttpSensor();
        temperature = (TextView) findViewById(R.id.temperature);
        rhs1.registerListener(this);
        rhs1.getTemperature();
    }

无法从always null bundle获取值

1 个答案: