如何禁用 ElevatedButton?
我希望按钮在字段中输入 10 个字符之前处于非活动状态。当输入 10 个字符时,该按钮处于活动状态。当它不活动时,它是灰色的,当它活动时,它是蓝色的。我怎样才能做到这一点? 这是带有按钮的输入代码:
Widget build(BuildContext context) {
return Scaffold(
body: Container(
width: MediaQuery.of(context).size.width,
height: MediaQuery.of(context).size.height,
child: Padding(
padding: EdgeInsets.fromLTRB(
20, MediaQuery.of(context).size.height * 0, 20, 0),
child: Column(
mainAxisAlignment: MainAxisAlignment.center,
children: <Widget>[
TextField(
onChanged: (String value) {
setState(() {
_showIcon = value.isNotEmpty;
});
},
controller: _inputController,
decoration: InputDecoration(
focusedBorder: UnderlineInputBorder(
borderSide: BorderSide(color: Colors.black, width: 2.0),
),
hintText: "(1201) 565-0123 ",
hintStyle: TextStyle(color: Colors.grey, fontSize: 15),
helperText: 'Enter your phone number',
helperStyle: TextStyle(color: Colors.grey, fontSize: 15),
suffixIcon: _showIcon
? IconButton(
onPressed: () {
setState(() {
_inputController.clear();
_showIcon = false;
});
},
icon: const Icon(Icons.close, color: Colors.grey),
) : null,
),
keyboardType: TextInputType.number,
inputFormatters: [maskFormatter],
),
Row(
mainAxisAlignment: MainAxisAlignment.end,
children: [
ElevatedButton(
onPressed: () {
},
child: const Icon(Icons.arrow_forward_rounded, size: 25),
style: ElevatedButton.styleFrom(
shape: CircleBorder(),
padding: EdgeInsets.all(15)
)
),
],
)
],
),
),
),
);
}
}
I want the button to be inactive until 10 characters are entered in the field. When 10 characters were entered, the button was active. And when it is inactive it is gray, and when it is active it is blue. How can I do that?
Here is the input code with the button:
Widget build(BuildContext context) {
return Scaffold(
body: Container(
width: MediaQuery.of(context).size.width,
height: MediaQuery.of(context).size.height,
child: Padding(
padding: EdgeInsets.fromLTRB(
20, MediaQuery.of(context).size.height * 0, 20, 0),
child: Column(
mainAxisAlignment: MainAxisAlignment.center,
children: <Widget>[
TextField(
onChanged: (String value) {
setState(() {
_showIcon = value.isNotEmpty;
});
},
controller: _inputController,
decoration: InputDecoration(
focusedBorder: UnderlineInputBorder(
borderSide: BorderSide(color: Colors.black, width: 2.0),
),
hintText: "(1201) 565-0123 ",
hintStyle: TextStyle(color: Colors.grey, fontSize: 15),
helperText: 'Enter your phone number',
helperStyle: TextStyle(color: Colors.grey, fontSize: 15),
suffixIcon: _showIcon
? IconButton(
onPressed: () {
setState(() {
_inputController.clear();
_showIcon = false;
});
},
icon: const Icon(Icons.close, color: Colors.grey),
) : null,
),
keyboardType: TextInputType.number,
inputFormatters: [maskFormatter],
),
Row(
mainAxisAlignment: MainAxisAlignment.end,
children: [
ElevatedButton(
onPressed: () {
},
child: const Icon(Icons.arrow_forward_rounded, size: 25),
style: ElevatedButton.styleFrom(
shape: CircleBorder(),
padding: EdgeInsets.all(15)
)
),
],
)
],
),
),
),
);
}
}
如果你对这篇内容有疑问,欢迎到本站社区发帖提问 参与讨论,获取更多帮助,或者扫码二维码加入 Web 技术交流群。

绑定邮箱获取回复消息
由于您还没有绑定你的真实邮箱,如果其他用户或者作者回复了您的评论,将不能在第一时间通知您!
发布评论
评论(2)
您可以调用
setState((){});
onon Changed
更新UI,或在_InputController
上添加侦听器。传递
开启:null
将提供禁用状态。更新UI可以完成
或
You can call
setState(() {});
ononChanged
to update the UI, or add listener on_inputController
.Passing
onPressed:null
will provide disable state.Updating UI can be done
Or
使用类似类似的变量,然后将null传递到OnPress函数将禁用该按钮。
PS请检查我刚刚为您提供概念的语法。
Use a variable like isEnabled and passing null to the onPress function will disable the button.
P.S Please check the syntax I have just provided you the concept.