在我看来，最好和最干净的方式，我更喜欢的是创建一个helper类型定义函数类coditional_widget.dart。

typedef IfWidget = List<Widget> Function(bool, Widget);
typedef IfElseWidget = Widget Function(bool, Widget, Widget);
typedef ElseEmptyWidget = Widget Function(bool, Widget);

IfWidget ifTrueWidget =
    (bool condition, Widget child) => [if (condition) child];

IfElseWidget ifElseWidget =
    (bool condition, Widget isTrueChild, Widget isFalseChild) =>
        condition ? isTrueChild : isFalseChild;

ElseEmptyWidget elseEmptyWidget = (bool condition, Widget isTrueChild) =>
    condition ? isTrueChild : const SizedBox.shrink();

如何使用

// IfWidget 
  ** Row/ Column / Wrap child etc.
   children: <Widget>[
      ...ifWidget(title != null, Text('Only Display for True Conditon')),
      ]

// elseEmptyWidget
  ** Row/ Column / Wrap child etc.
   children: <Widget>[
          
      elseEmptyWidget(title!=null,Text('Only Display for True Conditon')),
      ]



// ifElseWidget
      ** Row/ Column / Wrap child etc.
       children: <Widget>[
            ifElseWidget(true,Text('Only Display for True Conditon'),Text('Only Display for false Conditon')),
          ]

这只是几个你可以添加更多的

最简单的方法:

// the ternary operator:
<conditon>
  ? Widget1(...)
  : Widget2(...)

// Or:
if (condition)
    Widget1(...)

// With else/ if else
if (condition1)
    Widget1(...)
else if (condition2)
    Widget2(...)
else
    Widget3(...),

如果你想在一个条件下呈现多个小部件，你可以使用扩展操作符(为此，你必须在行，列或堆栈小部件中):

if (condition) ...[
    Widget1(...),
    Widget2(...),
  ],

// with else / else if:
if (condition1) ...[
    Widget1(...),
    Widget2(...),
  ]
else if(condition2)...[
    Widget3(...),
    Widget4(...),
]
else ...[
    Widget3(...),
    Widget4(...),
],

这是一篇很棒的文章和对话。我尝试使用如上所述的三元运算符。但是代码没有工作，导致了前面提到的错误。

Column(children: [ condition? Text("True"): null,],);

上面的三元例子是miss leading。Dart将响应一个错误，返回的是null而不是小部件。你不能返回null。正确的方法是返回一个小部件:

Column(children: [ condition? Text("True"): Text("false"),],); 

为了让三元函数正常工作，您需要返回一个Widget。如果你不想返回任何东西，你可以返回一个空容器。

Column(children: [ condition? Text("True"): Container(),],); 

祝你好运。

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;
  }
}

有两种可能:

如果您只使用一个小部件

解决方案= >

     Visibility(
       visible: condition == true, 
       child: Text(""),
      ),
    OR

     Offstage(
       offstage: condition == false, 
       child: Text(""),
     ),

如果您正在使用两个或更多小部件

解决方案= >

      bool _visibility = false;
     
      isVisible?
          Widget1 
           :
          WIdget2

在Dart中，if/else和switch是语句而不是表达式。它们不返回值，所以不能将它们传递给构造函数参数。如果您的构建方法中有很多条件逻辑，那么尝试简化它是一个很好的实践。例如，您可以将自包含逻辑移动到方法，并使用if/else语句来初始化稍后使用的局部变量。

使用方法和if/else

Widget _buildChild() {
  if (condition) {
    return ...
  }
  return ...
}

Widget build(BuildContext context) {
  return new Container(child: _buildChild());
}

使用if/else

Widget build(BuildContext context) {
  Widget child;
  if (condition) {
    child = ...
  } else {
    child = ...
  }
  return new Container(child: child);
}