本文介绍了指定的孩子已经有一个父母.您必须首先在孩子的父母上调用 removeView() (Android)的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着跟版网的小编来一起学习吧!
问题描述
我必须经常在两种布局之间切换.错误发生在下面发布的布局中.
I have to switch between two layouts frequently. The error is happening in the layout posted below.
当我的布局第一次被调用时,没有发生任何错误,一切都很好.然后当我调用不同的布局(空白的)然后再次调用我的布局时,它会引发以下错误:
When my layout is called the first time, there doesn't occur any error and everything's fine. When I then call a different layout (a blank one) and afterwards call my layout a second time, it throws the following error:
> FATAL EXCEPTION: main
> java.lang.IllegalStateException: The specified child already has a parent. You must call removeView() on the child's parent first.
我的布局代码如下所示:
My layout-code looks like this:
tv = new TextView(getApplicationContext()); // are initialized somewhere else
et = new EditText(getApplicationContext()); // in the code
private void ConsoleWindow(){
runOnUiThread(new Runnable(){
@Override
public void run(){
// MY LAYOUT:
setContentView(R.layout.activity_console);
// LINEAR LAYOUT
LinearLayout layout=new LinearLayout(getApplicationContext());
layout.setOrientation(LinearLayout.VERTICAL);
setContentView(layout);
// TEXTVIEW
layout.addView(tv); // <========== ERROR IN THIS LINE DURING 2ND RUN
// EDITTEXT
et.setHint("Enter Command");
layout.addView(et);
}
}
}
我知道以前有人问过这个问题,但对我的情况没有帮助.
I know this question has been asked before, but it didn't help in my case.
推荐答案
错误信息说明了你应该做什么.
The error message says what You should do.
// TEXTVIEW
if(tv.getParent() != null) {
((ViewGroup)tv.getParent()).removeView(tv); // <- fix
}
layout.addView(tv); // <========== ERROR IN THIS LINE DURING 2ND RUN
// EDITTEXT
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