blocks|key|851740|text|是否先尝试从父视图中删除scrollChildLayout？|type|unstyled|depth|inlineStyleRanges|entityRanges|data|851741|scrollview.removeView(scrollChildLayout)|code-block|syntax|javascript|851742|或者从父视图中删除所有子视图，然后重新添加它们。|851743|scrollview.removeAllViews()|851744|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

Try remove scrollChildLayout from its parent view first?

<pre><code>scrollview.removeView(scrollChildLayout)
</code></pre>

Or remove all the child from the parent view, and add them again.

<pre><code>scrollview.removeAllViews()
</code></pre>

blocks|key|635474|text|您所要做的就是post()一个执行addView()的Runnable。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|635475|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

All you have to do is post() a Runnable that does the addView().

blocks|key|851879|text|解决方案：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|851880|((ViewGroup)scrollChildLayout.getParent()).removeView(scrollChildLayout);
//scrollView.removeView(scrollChildLayout);|code-block|syntax|javascript|851881|使用子元素获取对父元素的引用。将父对象强制转换为ViewGroup，这样您就可以访问并使用removeView方法。|851882|感谢@东升to的解决方案|851883|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|P|8|@]|9|@]|A|$]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

Solution:

<pre><code>((ViewGroup)scrollChildLayout.getParent()).removeView(scrollChildLayout);
//scrollView.removeView(scrollChildLayout);
</code></pre>

Use the child element to get a reference to the parent. Cast the parent to a ViewGroup so that you get access to the removeView method and use that.

Thanks to @Dongshengcn for the solution

blocks|key|851960|text|你也可以通过检查视图的indexOfView方法来实现，如果indexOfView方法返回-1，那么我们可以使用。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|851961|ViewGroup的detachViewFromParent(v)；紧跟ViewGroup的removeDetachedView(v，true/false)；|851962|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|F|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|G|8|@]|9|@]|A|$]]|$1|D|3|-4|5|6|7|H|8|@]|9|@]|A|$]]]|E|$]]

You can also do it by checking if View's indexOfView method 
if indexOfView method returns -1 then we can use.

ViewGroup's detachViewFromParent(v);
followed by 
ViewGroup's removeDetachedView(v, true/false);

blocks|key|635628|text|好吧，叫我偏执狂，但我建议：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|635629|++final+android.view.ViewParent+parent+=+view.getParent+();

++if+(parent+instanceof+android.view.ViewManager)
++{
+++++final+android.view.ViewManager+viewManager+=+(android.view.ViewManager)+parent;

+++++viewManager.removeView+(view);
++}+//+if|code-block|syntax|javascript|635630|在没有instanceof的情况下进行造型似乎是错误的。而且(感谢IntelliJ+IDEA告诉我)+removeView是ViewManager接口的一部分。当有一个非常合适的接口可用时，不应该强制转换到一个具体的类。|offset|length|style|CODE|635631|entityMap^0|0|0|3|A|1F|A|1Q|B|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@$I|R|J|S|K|L]|$I|T|J|U|K|L]|$I|V|J|W|K|L]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|X|8|@]|9|@]|A|$]]]|N|$]]

Ok, call me paranoid but I suggest:

<pre><code> final android.view.ViewParent parent = view.getParent ();

 if (parent instanceof android.view.ViewManager)
 {
 final android.view.ViewManager viewManager = (android.view.ViewManager) parent;

 viewManager.removeView (view);
 } // if
</code></pre>

casting without <code>instanceof</code> just seems wrong. And (thanks IntelliJ IDEA for telling me) <code>removeView</code> is part of the <code>ViewManager</code> interface. And one should not cast to a concrete class when a perfectly suitable interface is available.

blocks|key|635684|text|我打电话给parentView.removeView(+childView+)，childView还在显示。我最终意识到一个方法以某种方式被触发了两次，并将childView添加到parentView中两次。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|635685|因此，在添加视图之前和之后，使用parentView.getChildCount()来确定父对象有多少子对象。如果孩子被添加了太多次，那么最上面的孩子将被删除，而拷贝childView仍然存在-这看起来removeView即使是工作的也是有效的。|635686|此外，您不应该使用View.GONE来删除视图。如果它真的被移除了，那么你就不需要隐藏它了，否则它仍然在那里，你只是对自己隐藏了它：|635687|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|H|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|I|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|J|8|@]|9|@]|A|$]]|$1|F|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|G|$]]

I was calling parentView.removeView(childView) and childView was still showing. I eventually realized that a method was somehow being triggered twice and added the childView to the parentView twice.

So, use parentView.getChildCount() to determine how many children the parent has before you add a view and afterwards. If the child is added too many times then the top most child is being removed and the copy childView remains-which looks like removeView is working even when it is.

Also, you shouldn't use View.GONE to remove a view. If it's truly removed then you won't need to hide it, otherwise it's still there and you're just hiding it from yourself :(

blocks|key|635712|text|在带有activity的onCreate或带有片段的onCreateView中。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|635713|+if+(view+!=+null)+{
++++ViewGroup+parent+=+(ViewGroup)+view.getParent();
++++if+(parent+!=+null)+{
++++++++parent.removeView(view);
++++}
}
try+{
++++view+=+inflater.inflate(R.layout.fragment_main,+container,+false);
}+catch+(InflateException+e)+{

}|code-block|syntax|javascript|635714|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

In onCreate with activity or in onCreateView with fragment.

<pre><code> if (view != null) {
 ViewGroup parent = (ViewGroup) view.getParent();
 if (parent != null) {
 parent.removeView(view);
 }
}
try {
 view = inflater.inflate(R.layout.fragment_main, container, false);
} catch (InflateException e) {

}
</code></pre>

blocks|key|635734|text|我所做的错误是我没有实例化动态布局并向其添加childs，所以我得到了这个错误|type|unstyled|depth|inlineStyleRanges|entityRanges|data|635735|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

What I was doing wrong so I got this error is I wasn't instantiating dynamic layout and adding childs to it so got this error

blocks|key|852203|text|在我的例子中，我有BaseFragment和所有其他从这里继承的片段。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|852204|因此我解决方案是在OnDestroyView()方法中添加以下代码行|offset|length|style|CODE|852205|@Override
public+final+View+onCreateView(LayoutInflater+inflater,+ViewGroup+container,+Bundle+savedInstanceState)
{
++++if+(mRootView+==+null)
++++{
++++++++mRootView+=+(inflater+==+null+?+getActivity().getLayoutInflater()+:+inflater).inflate(mContentViewResourceId,+container,+false);
++++}
....////
}

@Override
public+void+onDestroyView()
{
++++if+(mRootView+!=+null)
++++{
++++++++ViewGroup+parentViewGroup+=+(ViewGroup)+mRootView.getParent();

++++++++if+(parentViewGroup+!=+null)
++++++++{
++++++++++++parentViewGroup.removeAllViews();
++++++++}
++++}

++++super.onDestroyView();
}|code-block|syntax|javascript|852206|entityMap^0|0|9|F|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|P|8|@$D|Q|E|R|F|G]]|9|@]|A|$]]|$1|H|3|I|5|J|7|S|8|@]|9|@]|A|$K|L]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

In my case , I have BaseFragment and all other fragment inherits from this.

So my solytion was add this lines in <code>OnDestroyView()</code> method

<pre><code>@Override
public final View onCreateView(LayoutInflater inflater, ViewGroup container, Bundle savedInstanceState)
{
 if (mRootView == null)
 {
 mRootView = (inflater == null ? getActivity().getLayoutInflater() : inflater).inflate(mContentViewResourceId, container, false);
 }
....////
}

@Override
public void onDestroyView()
{
 if (mRootView != null)
 {
 ViewGroup parentViewGroup = (ViewGroup) mRootView.getParent();

 if (parentViewGroup != null)
 {
 parentViewGroup.removeAllViews();
 }
 }

 super.onDestroyView();
}
</code></pre>

blocks|key|852251|text|这是我的解决方案。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|852252|假设您有两个TextViews，并将它们放在一个名为ll的LinearLayout上。您将把这个LinerLayout放在另一个LinerLayout上。|offset|length|style|CODE|852253|<+lm+Linear+Layout>+
+++++++<+ll+Linear+Layout>+
+++++++++++++<name+Text+view>
+++++++++++++</name+Text+view>
+++++++++++++<surname+Text+view>
+++++++++++++</surname+Text+view>
+++++++</ll+Linear+Layout>+
</lm+Linear+Layout>+|code-block|syntax|javascript|852254|当您想要创建此结构时，您需要将parent作为继承。|852255|如果你想在onCreate方法中使用它，this就足够了。|852256|否则这就是孤单：|852257|LinerLayout+lm+=+new+LinearLayout(this);+//+You+can+use+getApplicationContext()+also
LinerLayout+ll+=+new+LinearLayout(lm.getContext());
TextView+name+=+new+TextView(ll.getContext());
TextView+surname+=+new+TextView(ll.getContext());|852258|entityMap^0|0|6|9|Q|2|T|C|1C|B|1S|B|0|0|0|5|8|K|4|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|X|8|@$D|Y|E|Z|F|G]|$D|10|E|11|F|G]|$D|12|E|13|F|G]|$D|14|E|15|F|G]|$D|16|E|17|F|G]]|9|@]|A|$]]|$1|H|3|I|5|J|7|18|8|@]|9|@]|A|$K|L]]|$1|M|3|N|5|6|7|19|8|@]|9|@]|A|$]]|$1|O|3|P|5|6|7|1A|8|@$D|1B|E|1C|F|G]|$D|1D|E|1E|F|G]]|9|@]|A|$]]|$1|Q|3|R|5|6|7|1F|8|@]|9|@]|A|$]]|$1|S|3|T|5|J|7|1G|8|@]|9|@]|A|$K|L]]|$1|U|3|-4|5|6|7|1H|8|@]|9|@]|A|$]]]|V|$]]

Here is my solution.

Lets say you have two <code>TextViews</code> and put them on a <code>LinearLayout</code> (named <code>ll</code>). You'll put this <code>LinerLayout</code> on another <code>LinerLayout</code>.

<pre><code>&lt; lm Linear Layout&gt; 
 &lt; ll Linear Layout&gt; 
 &lt;name Text view&gt;
 &lt;/name Text view&gt;
 &lt;surname Text view&gt;
 &lt;/surname Text view&gt;
 &lt;/ll Linear Layout&gt; 
&lt;/lm Linear Layout&gt; 
</code></pre>

When you want to create this structure you need to give parent as inheritance. 

If you want use it in an <code>onCreate</code> method <code>this</code> will enough.

Otherwise here is solition:

<pre><code>LinerLayout lm = new LinearLayout(this); // You can use getApplicationContext() also
LinerLayout ll = new LinearLayout(lm.getContext());
TextView name = new TextView(ll.getContext());
TextView surname = new TextView(ll.getContext());
</code></pre>

First a little background:

I have a layout inside a scrollview. At first, when the user scrolls on the screen, the scrollview scrolls. However, after a certain amount of scroll, I was to disable the scroll on the scroll view the move the "scroll focus" onto a webview inside the child layout. This way, the scrollview sticks and all the scroll events go to the webview inside it.

So, for a solution, when the scroll threshold is reached, I remove the child layout from the scrollview and put it in scrollview's parent.(And make the scrollview invisible).

<pre><code>// Remove the child view from the scroll view
scrollView.removeView(scrollChildLayout);

// Get scroll view out of the way
scrollView.setVisibility(View.GONE);

// Put the child view into scrollview's parent view
parentLayout.addView(scrollChildLayout);
</code></pre>

General Idea: (-> means contains)

Before: parentlayout -> scrollview -> scrollChildLayout

After : parentLayout -> scrollChildLayout

The above code is giving me this exception:

<pre><code>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:1976)
 at android.view.ViewGroup.addView(ViewGroup.java:1871)
 at android.view.ViewGroup.addView(ViewGroup.java:1828)
 at android.view.ViewGroup.addView(ViewGroup.java:1808)
</code></pre>

Do you know what's going on? I am clearly calling removeView on the parent.

Call removeView() on the child's parent first

IntelliJ IDEA

首先是一些背景知识：我在滚动视图中有一个布局。首先，当用户在屏幕上滚动时，滚动视图也会滚动。然而，经过一定数量的滚动后，我将禁用滚动视图上的滚动，并将“滚动焦点”移动到子布局内的was视图上。这样，scrollview就会固定下来，所有的scroll事件都会转到其中的webview中。因此，作为一种解决方案，当达到滚动...

问首先在子级的父级上调用removeView()
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问首先在子级的父级上调用removeView()
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