The specified child already has a parent. You must call removeView() on the child's parent first分析

  • 异常
java.lang.IllegalStateException: The specified child already has a parent. You must call removeView() on the child's parent first.
        at android.view.ViewGroup.addViewInner(ViewGroup.java:4465)
        at android.view.ViewGroup.addView(ViewGroup.java:4301)
        at android.view.ViewGroup.addView(ViewGroup.java:4242)
        at android.view.ViewGroup.addView(ViewGroup.java:4215)
  • 异常的原因child.getParent()!=null
//ViewGroup中
private void addViewInner(View child, int index, LayoutParams params,
            boolean preventRequestLayout) {
        if (mTransition != null) {
            // Don't prevent other add transitions from completing, but cancel remove
            // transitions to let them complete the process before we add to the container
            mTransition.cancel(LayoutTransition.DISAPPEARING);
        }

        if (child.getParent() != null) {
            throw new IllegalStateException("The specified child already has a parent. " +
                    "You must call removeView() on the child's parent first.");
        }
       //省略代码。。。。
    }
  • view.getParent初始化及重置

    • 初始化

      //无法通过View重置child.mParent,只能使用ViewGroup中的removeXXX方法进行重置,下面是View.mParent初始化方法,位于View中
          /*
          * Caller is responsible for calling requestLayout if necessary.
          * (This allows addViewInLayout to not request a new layout.)
          */
          void assignParent(ViewParent parent) {
              if (mParent == null) {
                  mParent = parent;
              } else if (parent == null) {
                  mParent = null;
              } else {
                  throw new RuntimeException("view " + this + " being added, but"
                          + " it already has a parent");
              }
          }
      
      
    • 重置

      //位于ViewGroup中,具体调用逻辑不再展开
      
      // This method also sets the child's mParent to null
          private void removeFromArray(int index) {
              final View[] children = mChildren;
              if (!(mTransitioningViews != null && mTransitioningViews.contains(children[index]))) {
                  children[index].mParent = null;
              }
              final int count = mChildrenCount;
              if (index == count - 1) {
                  children[--mChildrenCount] = null;
              } else if (index >= 0 && index < count) {
                  System.arraycopy(children, index + 1, children, index, count - index - 1);
                  children[--mChildrenCount] = null;
              } else {
                  throw new IndexOutOfBoundsException();
              }
              if (mLastTouchDownIndex == index) {
                  mLastTouchDownTime = 0;
                  mLastTouchDownIndex = -1;
              } else if (mLastTouchDownIndex > index) {
                  mLastTouchDownIndex--;
              }
          }
      
          // This method also sets the children's mParent to null
          private void removeFromArray(int start, int count) {
              final View[] children = mChildren;
              final int childrenCount = mChildrenCount;
      
              start = Math.max(0, start);
              final int end = Math.min(childrenCount, start + count);
      
              if (start == end) {
                  return;
              }
      
              if (end == childrenCount) {
                  for (int i = start; i < end; i++) {
                      children[i].mParent = null;
                      children[i] = null;
                  }
              } else {
                  for (int i = start; i < end; i++) {
                      children[i].mParent = null;
                  }
      
                  // Since we're looping above, we might as well do the copy, but is arraycopy()
                  // faster than the extra 2 bounds checks we would do in the loop?
                  System.arraycopy(children, end, children, start, childrenCount - end);
      
                  for (int i = childrenCount - (end - start); i < childrenCount; i++) {
                      children[i] = null;
                  }
              }
      
              mChildrenCount -= (end - start);
          }
      
      
  • 解决方法

    View childView=...;
    ViewGroup parent=....;
    /**
    * 抛出异常的错误代码
    * parent.removeView(child);
    * parent.addView(child);
    **/
     parent.removeView(childView);
     //parent和childView.getParent可能不是同一对象
     if(childView.getParent instanceof ViewGroup){
        ((ViewGroup)child.getParent).removeView(childView);
     }
     parent.addView(childView);
    

你可能感兴趣的:(The specified child already has a parent. You must call removeView() on the child's parent first分析)