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)

时间:2022-11-12
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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