Android Studio 预计 "class" 或 "interface"

Question

我正在按照 http://developer.android.com/training/location/retrieve-current.html 中提供的说明显示用户位置。但导致上述编译错误。

这是我的代码

public class MainActivity extends ActionBarActivity implements GoogleApiClient.ConnectionCallbacks, GoogleApiClient.OnConnectionFailedListener {

    private GoogleApiClient mGoogleApiClient;
    private Location mLastLocation;
    TextView tv1 = (TextView) findViewById(R.id.lat);
    TextView tv2 = (TextView) findViewById(R.id.lon);

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

        protected synchronized void buildGoogleApiClient() {
            mGoogleApiClient = new GoogleApiClient.Builder(this)
                    .addConnectionCallbacks(this)
                    .addOnConnectionFailedListener(this)
                    .addApi(LocationServices.API)
                    .build();
        }
    } // ERROR 'class' or 'interface' expected


    @Override
    public boolean onCreateOptionsMenu(Menu menu) {
        // Inflate the menu; this adds items to the action bar if it is present.
        getMenuInflater().inflate(R.menu.menu_main, menu);
        return true;
    }

    @Override
    public boolean onOptionsItemSelected(MenuItem item) {
        // Handle action bar item clicks here. The action bar will
        // automatically handle clicks on the Home/Up button, so long
        // as you specify a parent activity in AndroidManifest.xml.
        int id = item.getItemId();

        //noinspection SimplifiableIfStatement
        if (id == R.id.action_settings) {
            return true;
        }

        return super.onOptionsItemSelected(item);
    }

    @Override
    public void onConnected(Bundle connectionHint) {
        mLastLocation = LocationServices.FusedLocationApi.getLastLocation(
                mGoogleApiClient);
        if (mLastLocation != null) {
            tv1.setText(String.valueOf(mLastLocation.getLatitude()));
            tv1.setText(String.valueOf(mLastLocation.getLongitude()));
        }
    }

    @Override
    public void onConnectionSuspended(int i) {

    }

    @Override
    public void onConnectionFailed(ConnectionResult connectionResult) {

    }
}

我检查了这个错误，发现它是一个语法错误。谁能告诉我编译失败的原因？

Answer 1

您的 buildGoogleApiClient 方法不能在您的 onCreate 方法中。

将其更改为：

protected synchronized void buildGoogleApiClient() {
    mGoogleApiClient = new GoogleApiClient.Builder(this)
                .addConnectionCallbacks(this)
                .addOnConnectionFailedListener(this)
                .addApi(LocationServices.API)
                .build();
}

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

或：

@Override
protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.activity_main);
    mGoogleApiClient = new GoogleApiClient.Builder(this)
                .addConnectionCallbacks(this)
                .addOnConnectionFailedListener(this)
                .addApi(LocationServices.API)
                .build();
}

Answer 2

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

    protected synchronized void buildGoogleApiClient() {
        mGoogleApiClient = new GoogleApiClient.Builder(this)
                .addConnectionCallbacks(this)
                .addOnConnectionFailedListener(this)
                .addApi(LocationServices.API)
                .build();
    }
} // ERROR 'class' or 'interface' expected

应该是：

@Override
protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.activity_main); } // curly brace to close method and clean up error

    protected synchronized void buildGoogleApiClient() {
        mGoogleApiClient = new GoogleApiClient.Builder(this)
                .addConnectionCallbacks(this)
                .addOnConnectionFailedListener(this)
                .addApi(LocationServices.API)
                .build();
    }
// removing this curly brace should clear up the error here

由于花括号放错位置，您的语法错误。注意小事。

Answer 3

您的代码存在的问题是您试图在另一个函数 ( onCreate ) 中定义一个函数 ( buildGoogleApiClient )，这在 Java.

中是不可能的

protected void onCreate(Bundle savedInstanceState) { 
    //
    // Body of this function
    //
}

所以基本上在 Java 中，花括号标记了代码块的边界。代码块可以是 if 块、while 块或功能块等。Java 不允许功能块内部有功能块。只有 class-块可以包含功能块。

因此，您需要直接在 class 块下定义函数。

public class Blah extends Activity implements BlahInterface {

    private BlahApiClient mBlahApiClient;

    protected synchronized void buildBlahApiClient() {
        mBlahApiClient = new BlahApiClient.Builder(this)
            .addConnectionCallbacks(this)
            .addOnConnectionFailedListener(this)
            .addApi(LocationServices.API)
            .build();
    }

    protected void onCreate( Bundel savedInstanceState ) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.activity_main);

        // You can call (execute) any defined function inside this function
        buildBlahApiClient();

   }

}

Android Studio 预计 "class" 或 "interface"

Android Studio "class" or "interface" expected

java

android

android-studio