有条件的显示按钮 [英] Conditionally displaying button

问题描述

我知道如何启用/找到它后禁止在布局XML定义的按钮：

I know how to enable/disable a button defined in a layout XML after finding it:

  testBtn = (Button)findViewById(R.id.test);

但是，除了有条件地装载布局等，有没有办法在我的code告诉使用该布局XML，但不加载其中定义的按钮？

But other than conditionally loading layouts, is there a way to tell in my code "Use that layout XML, but don't load the button defined there"?

推荐答案

在XML中设置视图的可见性，使用android：visibility属性。

To set a View's visibility in Xml, use the android:visibility attribute.

下面的设置按钮能见度走了。当设置为离去的Andr​​oid不会显示该按钮，而不是布局计算时包括它的大小。

The following sets the button visibility to gone. When set to gone Android will not show the button and not include it's size during layout calculation.

<Button android:id="@+id/mybutton"
            android:layout_width="wrap_content"
            android:layout_height="wrap_content"
            android:text="Hello" 
            android:visibility="gone"/>

设置机器人：知名度=看不见不会显示按钮，但布局计算中包含它

Setting android:visibility="invisible" will not show the button, but include it during layout calculation.

<Button android:id="@+id/mybutton"
            android:layout_width="wrap_content"
            android:layout_height="wrap_content"
            android:text="Hello" 
            android:visibility="invisible"/>

要编程显示$ C $的按钮C调用setVisibility（）方法。

To programmatically show the button in code you call the setVisibility() method.

Button btn = (Button)findViewById(R.id.thebuttonid);
btn.setVisibility(View.VISIBLE); //View.GONE, View.INVISIBLE are available too.

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