java 如何一次只允许 3 个中的 1 个 android 切换按钮打开
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/6282702/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
how to allow only 1 android toggle button out of 3 to be on at once
提问by molleman
i am using 3 toggle buttons. In my android application i would like that only 1 of these toggle buttons can be selected at once. How would i go about doing this?
我正在使用 3 个切换按钮。在我的 android 应用程序中，我希望一次只能选择这些切换按钮中的 1 个。我该怎么做呢？
采纳答案by Haphazard
You could use radio buttons. If you don't want that, check out this link- it shows you how to listen for changes to the button state. If you find that one of your buttons is changed, change the other 2 to the off state.
您可以使用单选按钮。如果您不想这样做，请查看此链接- 它向您展示了如何监听按钮状态的更改。如果您发现其中一个按钮发生了变化，请将另外 2 个更改为关闭状态。
回答by ThePikeman
A simple onChangeListener will do:
一个简单的 onChangeListener 将执行以下操作：
public class TestProjectActivity extends Activity {
ToggleButton one;
ToggleButton two;
ToggleButton three;
ToggleButton four;
/** Called when the activity is first created. */
@Override
public void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.main);
one = (ToggleButton) findViewById(R.id.toggleButton1);
two = (ToggleButton) findViewById(R.id.toggleButton2);
three = (ToggleButton) findViewById(R.id.toggleButton3);
four = (ToggleButton) findViewById(R.id.toggleButton4);
one.setOnCheckedChangeListener(changeChecker);
two.setOnCheckedChangeListener(changeChecker);
three.setOnCheckedChangeListener(changeChecker);
four.setOnCheckedChangeListener(changeChecker);
}
OnCheckedChangeListener changeChecker = new OnCheckedChangeListener() {
@Override
public void onCheckedChanged(CompoundButton buttonView, boolean isChecked) {
if (isChecked){
if (buttonView == one) {
two.setChecked(false);
three.setChecked(false);
four.setChecked(false);
}
if (buttonView == two) {
one.setChecked(false);
three.setChecked(false);
four.setChecked(false);
}
if (buttonView == three) {
two.setChecked(false);
one.setChecked(false);
four.setChecked(false);
}
if (buttonView == four) {
two.setChecked(false);
three.setChecked(false);
one.setChecked(false);
}
}
}
};
}
}
回答by xtempore
ThePikeman's solution is okay, but depending on how many buttons you have you might want to consider an array that you can iterate over.
ThePikeman 的解决方案是好的，但根据您拥有的按钮数量，您可能需要考虑一个可以迭代的数组。
For a small number of buttons, Pikeman's code can be simplified to save some typing...
对于少量按钮，可以简化 Pikeman 的代码以节省一些输入...
public void onCheckedChanged(CompoundButton buttonView, boolean isChecked) {
if (isChecked){
if (buttonView != one) {
one.setChecked(false);
}
if (buttonView != two) {
two.setChecked(false);
}
if (buttonView != three) {
three.setChecked(false);
}
if (buttonView != four) {
four.setChecked(false);
}
}
}
回答by aya salama
I use the following function;
我使用以下功能；
private void setIsChecked(CheckBox checkBox){
buttonOne.setChecked(false);
buttonTwo.setChecked(false);
buttonThree.setChecked(false);
checkBox.setChecked(true)
}
now all of them are deemed except for the one you choose
现在所有的都被认为除了你选择的那个
ex
前任
setIsChecked(buttonOne);
Now only button one is checked and etc..
现在只有按钮一被选中等等。
