Question

我有一个（看似）简单的问题：我试图在子线性布局上设置onTouchListener，但我无法编译代码。我收到错误＆＃34;无法解析符号setOnTouchListener＆＃34;当我尝试在我选择的视图上使用setOnTouchListener（）时。

如何记录LinearLayout上的触摸？我做错了什么？

MainActivity.java

   public class MainActivity extends FragmentActivity {
    public static LinearLayout glView;
    public static OpenGL_GLSurface foo;
    public TouchController touchSurface;

    void configView(){    // used to configure an opengl view 
        foo = new OpenGL_GLSurface(this);
        setContentView(R.layout.activity_main);
        glView = (LinearLayout)findViewById(R.id.openglsurface);
        RelativeLayout.LayoutParams glParams = new RelativeLayout.LayoutParams(RelativeLayout.LayoutParams.MATCH_PARENT, RelativeLayout.LayoutParams.MATCH_PARENT);
        glView.addView(foo, glParams);
    }

    @Override
    protected void onCreate(Bundle savedInstanceState) {
        ...
        touchSurface = new TouchController(this);  //initialize touchable surface
    }}

TouchController.java

public class TouchController {
private Context mContext;

public TouchController(Context c) {   //constructor
    mContext = c;  
}

View.OnTouchListener touch = new View.OnTouchListener() {       //set OnTouchListener to OpenGL View
    @Override
    public boolean onTouch(View v, MotionEvent event) {
        switch (maskedAction) {
            //do stuff
        }
        return true;
    }
};
MainActivity.glView.setOnTouchListener(touch);   //Compilation Error here @ setOnTouchListener
}

Answer 1

问题出在TouchController中，当您设置触控侦听器时，此行：

MainActivity.glView.setOnTouchListener(touch);

这行代码是无效的java代码，因为它只是在类中闲逛。它必须位于构造函数之类的方法中。像这样：

修改

public class TouchController { private Context mContext; public TouchController(Context c) { //constructor mContext = c; View.OnTouchListener touch = new View.OnTouchListener() { @Override public boolean onTouch(View v, MotionEvent event) { switch (maskedAction) { //do stuff } return true; } }; //Register touch listener here in constructor or in another method CourseListActivity.glView.setOnTouchListener(touch); } }

您应该考虑移动成员变量的分配＆＃34;触摸＆＃34;在构造函数内部，就在设置触摸侦听器之前。它会更有条理。

Android捕获LinearLayout上的触摸：无法解析符号setOnTouchListener。

1 个答案: