create on off toggle Icon for flutter torch

I tried to implement the possibility to use the flash of the phone as a torch in my flutter app. The on/ off button is located in the appbar. This runs fine except the light on and light off Button appear both at the same time. How can I make it, that either one or the other is shown. depending on whether the lamp is on or off?
Thank you very much for your help
I used the flutter torch_light: ^0.4.0

Class TorchController extends StatelessWidget {
const TorchController({super.key});
@override
Widget build(BuildContext context) {
return Scaffold(
body: FutureBuilder<bool>(
future: _isTorchAvailable(context),
builder: (BuildContext context, AsyncSnapshot<bool> snapshot) {
if (snapshot.hasData && snapshot.data!) {
return Column(
children: [
Expanded(
child: Center(
child: IconButton ( icon: const Icon(Icons.flashlight_on_outlined,size: 35,),
onPressed: () async {
_enableTorch(context);
},
),
),
),
Expanded(
child: Center(
child: IconButton (icon: const Icon(Icons.flashlight_off_outlined,size: 35,),
onPressed: () async {
_disableTorch(context);
},
),
),
),
],
);
} else if (snapshot.hasData) {
return const Center(
child: Text('No torch available.'),
);
} else {
return const Center(
child: CircularProgressIndicator(),
);
}
},
),
);
}
Future<bool> _isTorchAvailable(BuildContext context) async {
try {
return await TorchLight.isTorchAvailable();
} on Exception catch (_) {
_showMessage(
'Could not check if the device has an available torch',
context,
);
rethrow;
}
}
Future<void> _enableTorch(BuildContext context) async {
try {
await TorchLight.enableTorch();
} on Exception catch (_) {
_showMessage('Could not enable torch', context);
}
}
Future<void> _disableTorch(BuildContext context) async {
try {
await TorchLight.disableTorch();
} on Exception catch (_) {
_showMessage('Could not disable torch', context);
}
}
void _showMessage(String message, BuildContext context) {
ScaffoldMessenger.of(context)
.showSnackBar(SnackBar(content: Text(message)));
}
}
//Ende```
</details>
# 答案1
**得分**: 1
首先，将小部件从`stateless`更改为`stateful`小部件。然后
1. 定义一个变量来显示手电筒的状态
- isTorchOn = false;
2. 在`_enableTorch()`中将该值更新为`true`
**(不需要传递上下文，因为它现在是一个有状态的小部件)**
```dart
Future<void> _enableTorch(BuildContext context) async {
try {
await TorchLight.enableTorch();
setState(() => isTorchOn = true);
} on Exception catch (_) {
_showMessage('无法启用手电筒', context);
}
}

3. 对于_disableTorch()也要做同样的操作，将isTorchOn设置为false

Future<void> _disableTorch(BuildContext context) async {
    try {
        await TorchLight.disableTorch();
        setState(() => isTorchOn = false);
    } on Exception catch (_) {
        _showMessage('无法禁用手电筒', context);
    }
}

请注意，上述代码片段是在Dart编程语言中编写的，用于Flutter应用程序中的状态管理。

Future<void> _enableTorch(BuildContext context) async {
try {
await TorchLight.enableTorch();
setState(()=> isTorchOn = true);
} on Exception catch (_) {
_showMessage('Could not enable torch', context);
}
}

Future<void> _disableTorch(BuildContext context) async {
try {
await TorchLight.disableTorch();
setState(()=> isTorchOn = false);
} on Exception catch (_) {
_showMessage('Could not disable torch', context);
}
}