回答

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LinearLayout row = new LinearLayout(this); 
    row.setLayoutParams(new LayoutParams(LayoutParams.FILL_PARENT, LayoutParams.WRAP_CONTENT)); 
Button btnTag = new Button(this); 
     btnTag.setLayoutParams(new LayoutParams(LayoutParams.WRAP_CONTENT, LayoutParams.WRAP_CONTENT)); 
     btnTag.setText("Button 1"); 
     btnTag.setId(1); 
     row.addView(btnTag); 
layout.addView(row); 
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显示错误在SETID(“BTN”) –

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IDS将永远是一个整数。它不能是一个字符串 –

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请现在检查它 –

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是的,你可以动态地添加按钮到线性布局。

对于如:

LinearLayout layout = new LinearLayout(this); 
layout.setLayoutParams(new AbsListView.LayoutParams(AbsListView.LayoutParams.WRAP_CONTENT,AbsListView.LayoutParams.WRAP_CONTENT)); 
layout.setId(5000); 

Random random = new Random(); 
//Five button 
for(int i=0;i<5;i++) 
{ 
Button button = new Button(this); 
button.setLayoutParams(new LayoutParams(LayoutParams.WRAP_CONTENT, LayoutParams.WRAP_CONTENT)); 
button.setText("Button"); 
button.setId(random.nextInt(5)); //generate unique number between 0 to 4 and set it as id 
layout.addView(button); 
} 
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但在这种情况下,所有创建的新按钮将具有相同的ID。是否有任何方式可以动态生成id –

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@ user3172071.use随机类生成一个随机ID ..将其转换为字符串形式,并将其分配给setId参数 –

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@ user3172071..updated代码根据您的随机唯一ID要求 –