Question

我受到以下代码的启发：

@Test
public void onType() throws Exception {
    ...
    onView(isAssignableFrom(EditText.class)).perform(typeText("a"));
    ...
}

如何在活动中使用视图类查看视图，我认为控制第3个自定义视图很有用

Answer 1

您所说的代码是Espresso testing framework的一部分。它用于仪器UI测试，但它不是简单应用程序的常见用法。

您甚至可以为您的应用程序（不是测试模块）设置此库的依赖关系，如下所示：

dependencies {
       compile 'com.android.support.test.espresso:espresso-core:2.2.2'
}

但我认为这些方法对性能有不良影响（可能使用简单的视图迭代）。

Answer 2

如果您有某个类的对象引用，例如从View.class派生的示例类

View someView <- by  findViewById(int),  getView(), getRootView() etc // any method that return view

您可以检查是否可以通过向上演员将其分配给其他类 - 通过这种方式从右向左移动（下面的一些示例）：

EditText <- TextView <- View 
FrameLayout <- ViewGroup  <- View
NumberPicker <- LinearLayou <- ViewGroup <- View

示例应用程序：

// get View class reference by call to find view by id 
View view = findViewById(R.id.someView);
if(view!=null) {
     Class viewClass = view.getClass()
     //  check if we can assign edit text class  to this class ???  
     if(EditText.class.isAssingableFrom(viewClass)) {
           // u can cast it to edit text and do some stuff with edit text  class as we did up casting  while in this case it is allowed:
           EditText editText = (EditText) view;
            editText.setText(String);

     }
}

这样我们就不会出现异常，例如我们在没有任何检查的情况下进行此类演员：

 ((EditText) view).setText(String);

如果视图类是其他类型的EditText，或者不是从编辑文本类派生的，则会抛出异常。

来自类实现：

public final class Class<T> implements Serializable, AnnotatedElement, GenericDeclaration, Type {

...

 /**
 * Can {@code c}  be assigned to this class? For example, String can be assigned to Object
 * (by an upcast), however, an Object cannot be assigned to a String as a potentially exception
 * throwing downcast would be necessary. Similarly for interfaces, a class that implements (or
 * an interface that extends) another can be assigned to its parent, but not vice-versa. All
 * Classes may assign to themselves. Classes for primitive types may not assign to each other.
 *
 * @param c the class to check.
 * @return {@code true} if {@code c} can be assigned to the class
 *         represented by this {@code Class}; {@code false} otherwise.
 * @throws NullPointerException if {@code c} is {@code null}.
 */
public boolean isAssignableFrom(Class<?> c) {

另见：

The Type Comparison Operator instanceof

Answer 3

以下代码应该有效：

static class FindV {
    Context ctx;
    ViewGroup vg;
    Class<? extends View> targetK;


    public FindV(ViewGroup vg, Class<? extends View> targetK) {
        ctx = vg.getContext();
        this.vg = vg;
        this.targetK = targetK;
    }

    ArrayList<View> vs = new ArrayList<View>();

    //        public static <T extends View> ArrayList<T> findV(ViewGroup vg, Class<T> targetK) {
    ArrayList<View> st() {
        st(vg);
        return vs;
    }

    void st(ViewGroup vg) {
        int cc = vg.getChildCount();
        if (cc > 0) {
            for (int i = 0; i < cc; i++) {
                View v = vg.getChildAt(i);
                if (v.getClass() == targetK) {
                    vs.add(v);
                }
            }

            if (vs.isEmpty()) {
                for (int i = 0; i < cc; i++) {
                    View v = vg.getChildAt(i);
                    if (ancestorKs(v).contains(targetK)) {
                        vs.add(v);
                    }
                }
            }

            for (int i = 0; i < cc; i++) {
                View v = vg.getChildAt(i);
                if (v instanceof ViewGroup) {
                    st((ViewGroup) v);
                }
            }
        }
    }
}

public static <T extends View> ArrayList<T> findV(ViewGroup vg, Class<T> targetK) {
    return (ArrayList<T>) new FindV(vg, targetK).st();
}

如何通过android中的视图类查找视图

3 个答案: