问题描述

addView同一个对象两次会出错

        //首页按钮
        ImageView home=new ImageView(context);
        home.setImageResource(R.drawable.icon_03);
        home.setLayoutParams(iconparams);
        home.setPadding(10, 10, 10, 10);
        addView(home);

        //分隔符
        ImageView seprator = new ImageView(context);
        seprator.setImageResource(R.drawable.seprator_03);
        seprator.setLayoutParams(sepratorparams);
        addView(seprator);

        //2
        ImageView icon2=new ImageView(context);
        icon2.setImageResource(R.drawable.icon_05);
        icon2.setLayoutParams(iconparams);
        icon2.setPadding(10, 10, 10, 10);
        addView(icon2);
        addView(seprator);

上边是我在android项目中写的一个用户控件的部分代码，调试后发现再第二次addView(seprator);时候会报错，然后我试了一下其他的，结果是只要是第二次addView的参数对象之前已经addView过一次，就会报错，请问是因为什么？有什么解决办法？

解决方案

你可以去看看源码，ViewGroup→addView()→addViewInner()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.");
    }

    if (mTransition != null) {
        mTransition.addChild(this, child);
    }

    if (!checkLayoutParams(params)) {
        params = generateLayoutParams(params);
    }

    if (preventRequestLayout) {
        child.mLayoutParams = params;
    } else {
        child.setLayoutParams(params);
    }

    if (index < 0) {
        index = mChildrenCount;
    }

    addInArray(child, index);

    // tell our children
    if (preventRequestLayout) {
        child.assignParent(this);
    } else {
        child.mParent = this;
    }

    if (child.hasFocus()) {
        requestChildFocus(child, child.findFocus());
    }

    AttachInfo ai = mAttachInfo;
    if (ai != null && (mGroupFlags & FLAG_PREVENT_DISPATCH_ATTACHED_TO_WINDOW) == 0) {
        boolean lastKeepOn = ai.mKeepScreenOn;
        ai.mKeepScreenOn = false;
        child.dispatchAttachedToWindow(mAttachInfo, (mViewFlags&VISIBILITY_MASK));
        if (ai.mKeepScreenOn) {
            needGlobalAttributesUpdate(true);
        }
        ai.mKeepScreenOn = lastKeepOn;
    }

    if (child.isLayoutDirectionInherited()) {
        child.resetRtlProperties();
    }

    onViewAdded(child);

    if ((child.mViewFlags & DUPLICATE_PARENT_STATE) == DUPLICATE_PARENT_STATE) {
        mGroupFlags |= FLAG_NOTIFY_CHILDREN_ON_DRAWABLE_STATE_CHANGE;
    }

    if (child.hasTransientState()) {
        childHasTransientStateChanged(child, true);
    }

    if (child.getVisibility() != View.GONE) {
        notifySubtreeAccessibilityStateChangedIfNeeded();
    }
}

解决方案二：

没错，因为它已经有parent了，你要先removeallviews，然后再添加

解决方案三：

同一个对象添加两次当然会崩溃了，会提示已经有父view

如果你想添加两次都构造两个seperator对象！！！new 两次就可以了

解决方案四：

每次都有new一个新的view来add

解决方案五：

已经存在的view 具有一个parent,会报错的。或者你需要移除他的parentView

解决方案六：

这个问题前面几位已经回答很详细了。我的建议：遇到奇怪的问题，如API文档中没有找到原因。这个时候就应该看看源代码了。