立即获取LinearLayout中的所有子视图

Question

我有LinearLayout，其中包含多个子TextViews。如何使用循环获取该LinerLayout的子视图？

Answer 1

使用getChildCount()和getChildAt(int index)。

示例：

LinearLayout ll = …
final int childCount = ll.getChildCount();
for (int i = 0; i < childCount; i++) {
      View v = ll.getChildAt(i);
      // Do something with v.
      // …
}

Answer 2

   ( (ViewGroup) findViewById(android.R.id.content));// you can use this in an Activity to get your layout root view, then pass it to findAllEdittexts() method below.

这里我只迭代EdiTexts，如果你想要所有的视图，你可以用View替换EditText。

SparseArray<Edittext> array = new SparseArray<Edittext>();

private void findAllEdittexts(ViewGroup viewGroup) {

    int count = viewGroup.getChildCount();
    for (int i = 0; i < count; i++) {
        View view = viewGroup.getChildAt(i);
        if (view instanceof ViewGroup)
            findAllEdittexts((ViewGroup) view);
        else if (view instanceof Edittext) {
            Edittext edittext = (Edittext) view;
            array.put(edittext.getId(), edittext);
        }
    }

}

Answer 3

使用此

    final int childCount = mainL.getChildCount();
    for (int i = 0; i < childCount; i++) {
          View element = mainL.getChildAt(i);

        // EditText
        if (element instanceof EditText) {
            EditText editText = (EditText)element;
            System.out.println("ELEMENTS EditText getId=>"+editText.getId()+ " getTag=>"+element.getTag()+
            " getText=>"+editText.getText());
        }

        // CheckBox
        if (element instanceof CheckBox) {
            CheckBox checkBox = (CheckBox)element;
            System.out.println("ELEMENTS CheckBox getId=>"+checkBox.getId()+ " getTag=>"+checkBox.getTag()+
            " getText=>"+checkBox.getText()+" isChecked=>"+checkBox.isChecked());
        }

        // DatePicker
        if (element instanceof DatePicker) {
            DatePicker datePicker = (DatePicker)element;
            System.out.println("ELEMENTS DatePicker getId=>"+datePicker.getId()+ " getTag=>"+datePicker.getTag()+
            " getDayOfMonth=>"+datePicker.getDayOfMonth());
        }

        // Spinner
        if (element instanceof Spinner) {
            Spinner spinner = (Spinner)element;
            System.out.println("ELEMENTS Spinner getId=>"+spinner.getId()+ " getTag=>"+spinner.getTag()+
            " getSelectedItemId=>"+spinner.getSelectedItemId()+
            " getSelectedItemPosition=>"+spinner.getSelectedItemPosition()+
            " getTag(key)=>"+spinner.getTag(spinner.getSelectedItemPosition()));
        }

    }

Answer 4

从任何类型的布局获取所有视图

public List<View> getAllViews(ViewGroup layout){
        List<View> views = new ArrayList<>();
        for(int i =0; i< layout.getChildCount(); i++){
            views.add(layout.getChildAt(i));
        }
        return views;
}

从任何类型的布局获取所有TextView

public List<TextView> getAllTextViews(ViewGroup layout){
        List<TextView> views = new ArrayList<>();
        for(int i =0; i< layout.getChildCount(); i++){
            View v =layout.getChildAt(i);
            if(v instanceof TextView){
                views.add((TextView)v);
            }
        }
        return views;
}

Answer 5

使用Kotlin更容易：

for (child in ll.children) {
     //child is a View         
}

此处ll是用XML定义的LinearLayout。

Answer 6

在Kotlin中递归获取视图及其子视图的所有视图：

private fun View.getAllViews(): List<View> {
    if (this !is ViewGroup || childCount == 0) return listOf(this)

    return children
            .toList()
            .flatMap { it.getAllViews() }
            .plus(this as View)
}