到目前为止,当我需要在小部件中使用条件语句时,我已经做了以下工作(使用中心和容器作为简化的虚拟示例):

new Center(
  child: condition == true ? new Container() : new Container()
)

虽然当我尝试使用if/else语句时,它会导致一个死亡代码警告:

new Center(
  child: 
    if(condition == true){
      new Container();
    }else{
      new Container();
    }
)

有趣的是,我尝试了一个switch case语句,它给了我同样的警告,因此我不能运行代码。我做错了什么,或者它是这样的,不能使用if/else或开关语句而不颤振认为有死代码?


当前回答

像这样做

Widget showIf(bool shouldShow, Widget widget) {
if (shouldShow) {
  return widget;
} else {
  return Container();
}}

所以当你想用条件来展示某物的时候你会说

Column(children: [showIf(myConditionIsTrue, myComplexWidget)])

其他回答

另一种选择:对于“switch’s”这样的语句,有很多条件,我喜欢使用map:

return Card(
        elevation: 0,
        margin: EdgeInsets.all(1),
        child: conditions(widget.coupon)[widget.coupon.status] ??
            (throw ArgumentError('invalid status')));


conditions(Coupon coupon) => {
      Status.added_new: CheckableCouponTile(coupon.code),
      Status.redeemed: SimpleCouponTile(coupon.code),
      Status.invalid: SimpleCouponTile(coupon.code),
      Status.valid_not_redeemed: SimpleCouponTile(coupon.code),
    };

在不使用条件语句的情况下,更容易向条件列表中添加/删除元素。

另一个例子:

var condts = {
  0: Container(),
  1: Center(),
  2: Row(),
  3: Column(),
  4: Stack(),
};

class WidgetByCondition extends StatelessWidget {
  final int index;
  WidgetByCondition(this.index);
  @override
  Widget build(BuildContext context) {
    return condts[index];
  }
}

如果你使用小部件列表,你可以使用这个:

class HomePage extends StatelessWidget {
  bool notNull(Object o) => o != null;
  @override
  Widget build(BuildContext context) {
    var condition = true;
    return Scaffold(
      appBar: AppBar(
        title: Text("Provider Demo"),
      ),
      body: Center(
          child: Column(
        children: <Widget>[
          condition? Text("True"): null,
          Container(
            height: 300,
            width: MediaQuery.of(context).size.width,
            child: Text("Test")
          )
        ].where(notNull).toList(),
      )),
    );
  }
}

更好的方法

 Column(
        children: [
            if (firstCondition == true) ...[
              DayScreen(),
            ] else if(secondCondition == true)...[
              StatsScreen(),
            ], else...[
              StatsScreen(),
            ],
        ],
     ),

像这样做

Widget showIf(bool shouldShow, Widget widget) {
if (shouldShow) {
  return widget;
} else {
  return Container();
}}

所以当你想用条件来展示某物的时候你会说

Column(children: [showIf(myConditionIsTrue, myComplexWidget)])

Flutter Widget可以在不破坏代码树的情况下有条件地用父元素包装子树

import 'package:flutter/widgets.dart';

/// Conditionally wrap a subtree with a parent widget without breaking the code tree.
///
/// [condition]: the condition depending on which the subtree [child] is wrapped with the parent.
/// [child]: The subtree that should always be build.
/// [conditionalBuilder]: builds the parent with the subtree [child].
///
/// ___________
/// Usage:
/// ```dart
/// return ConditionalParentWidget(
///   condition: shouldIncludeParent,
///   child: Widget1(
///     child: Widget2(
///       child: Widget3(),
///     ),
///   ),
///   conditionalBuilder: (Widget child) => SomeParentWidget(child: child),
///);
/// ```
///
/// ___________
/// Instead of:
/// ```dart
/// Widget child = Widget1(
///   child: Widget2(
///     child: Widget3(),
///   ),
/// );
///
/// return shouldIncludeParent ? SomeParentWidget(child: child) : child;
/// ```
///
class ConditionalParentWidget extends StatelessWidget {
  const ConditionalParentWidget({
    Key key,
    @required this.condition,
    @required this.child,
    @required this.conditionalBuilder,
  }) : super(key: key);

  final Widget child;
  final bool condition;
  final Widget Function(Widget child) conditionalBuilder;

  @override
  Widget build(BuildContext context) {
    return condition ? this.conditionalBuilder(this.child) : this.child;
  }
}