如果你正在寻找一种快速的方法，而不关心让用户在一个按钮上点击多次。你也可以这样做:

// Constant whether button is clicked
bool isClicked = false;

然后在onPressed()函数中检查用户是否已经单击了按钮。

onPressed: () async {
    if (!isClicked) {
       isClicked = true;
       // await Your normal function
    } else {
       Toast.show(
          "You click already on this button", context,
          duration: Toast.LENGTH_LONG, gravity: Toast.BOTTOM);
    }
}

禁用点击:

onPressed: null

可以点击:

onPressed: () => fooFunction() 
// or
onPressed: fooFunction

组合:

onPressed: shouldEnable ? fooFunction : null

这是在Flutter中禁用按钮的最简单的方法是将空值分配给onPressed

ElevatedButton(
  style: ElevatedButton.styleFrom(
    primary: Colors.blue, // background
    onPrimary: Colors.white, // foreground
  ),
  onPressed: null,
  child: Text('ElevatedButton'),
),

我认为您可能需要引入一些帮助函数来构建按钮，以及一个有状态的小部件以及一些要关闭的属性。

Use a StatefulWidget/State and create a variable to hold your condition (e.g. isButtonDisabled) Set this to true initially (if that's what you desire) When rendering the button, don't directly set the onPressed value to either null or some function onPressed: () {} Instead, conditionally set it using a ternary or a helper function (example below) Check the isButtonDisabled as part of this conditional and return either null or some function. When the button is pressed (or whenever you want to disable the button) use setState(() => isButtonDisabled = true) to flip the conditional variable. Flutter will call the build() method again with the new state and the button will be rendered with a null press handler and be disabled.

这里是一些更多的上下文使用颤振计数器项目。

class MyHomePage extends StatefulWidget {
  @override
  _MyHomePageState createState() => new _MyHomePageState();
}

class _MyHomePageState extends State<MyHomePage> {
  int _counter = 0;
  bool _isButtonDisabled;

  @override
  void initState() {
    _isButtonDisabled = false;
  }

  void _incrementCounter() {
    setState(() {
      _isButtonDisabled = true;
      _counter++;
    });
  }

  @override
  Widget build(BuildContext context) {
    return new Scaffold(
      appBar: new AppBar(
        title: new Text("The App"),
      ),
      body: new Center(
        child: new Column(
          mainAxisAlignment: MainAxisAlignment.center,
          children: <Widget>[
            new Text(
              'You have pushed the button this many times:',
            ),
            new Text(
              '$_counter',
              style: Theme.of(context).textTheme.display1,
            ),
            _buildCounterButton(),
          ],
        ),
      ),
    );
  }

  Widget _buildCounterButton() {
    return new RaisedButton(
      child: new Text(
        _isButtonDisabled ? "Hold on..." : "Increment"
      ),
      onPressed: _isButtonDisabled ? null : _incrementCounter,
    );
  }
}

在这个例子中，我使用了一个内联三元来有条件地设置Text和onPressed，但它可能更适合你将其提取到一个函数中(你也可以使用相同的方法来更改按钮的文本):

Widget _buildCounterButton() {
    return new RaisedButton(
      child: new Text(
        _isButtonDisabled ? "Hold on..." : "Increment"
      ),
      onPressed: _counterButtonPress(),
    );
  }

  Function _counterButtonPress() {
    if (_isButtonDisabled) {
      return null;
    } else {
      return () {
        // do anything else you may want to here
        _incrementCounter();
      };
    }
  }

根据文件:

如果onPressed回调为空，则该按钮将被禁用 默认情况下类似于disabledColor中的平面按钮。

所以，你可以这样做:

RaisedButton(
  onPressed: calculateWhetherDisabledReturnsBool() ? null : () => whatToDoOnPressed,
  child: Text('Button text')
);

你也可以设置空白条件，在设置null的地方

         var isDisable=true;

   

          RaisedButton(
              padding: const EdgeInsets.all(20),
              textColor: Colors.white,
              color: Colors.green,
              onPressed:  isDisable
                  ? () => (){} : myClickingData(),
              child: Text('Button'),
            )