Displaying widget conditionally in flutter

Question

I am just writing a simple application in flutter and I came across an situation where I need to display widget conditionally. When I use the ternary operator it is working perfectly fine.

Widget build(BuildContext context) { return MaterialApp( home: Scaffold( appBar: AppBar( title: Text("I quiz you, OK?"), ), body: (_questionIndex < _questionList.length) ? Quiz( questionList: _questionList, questionIndex: _questionIndex, answerSelected: _answerSelected) : Result(_finalScore, _reset), ), ); } //Woks fine

but when I replace it with if-else block I am getting an error as

Expected an identifier.dart(missing_identifier)

Expected to find ')'.dart(expected_token)

Widget build(BuildContext context) { return MaterialApp( home: Scaffold( appBar: AppBar( title: Text("I quiz you, OK?"), ), body: if(_questionIndex < _questionList.length) { Quiz( questionList: _questionList, questionIndex: _questionIndex, answerSelected: _answerSelected) } else { Result(_finalScore, _reset) }, ), ); } //Getting error

What could be the issue as all brackets are balanced perfectly. I know I could just use ternary operator but there might arise a situation where if-else ladder would be required.

Note: Quiz() and Result() are custom widgets.

Valentin Vignal · Accepted Answer · 2022-02-14 13:46:48Z

You cannot use a if/else statement as you did, here are several ways to do it:

body: () { if (_questionIndex < _questionList.length) { return Quiz(/* ... */); } else { return Result(/* ... */); } }(),

or

body: Builder( builder: (context) { if (_questionIndex < _questionList.length) { return Quiz(/* ... */); } else { return Result(/* ... */); } }, ),

thanks bro i was searching for something like this for a long time 👏🏻

Osama Kashif · Accepted Answer · 2022-02-14 14:10:49Z

you can create a method returning a widget and easily return what ever you want.

 Widget func(){ if(_questionIndex < _questionList.length){ return Quiz( questionList: _questionList, questionIndex: _questionIndex, answerSelected: _answerSelected);} else if(_questionIndex >= _questionList.length){ return Result(_finalScore, _reset); } else{ //widget is must return Container(); }} Widget build(BuildContext context) { return MaterialApp( home: Scaffold( appBar: AppBar( title: Text("I quiz you, OK?"), ), body: func(); }

Md. Yeasin Sheikh · Accepted Answer · 2022-02-16 19:01:00Z

if..else is statement and ? ternary is expression on dart(flutter).

We can't pass something to the body that is not an expression.

I asked this question on Flutter community channel, you can check it from here.

You can use the ternary operator as you did,

body: condition ? whileTrue() : whenFalse();

method as @Osama Kashif answered Or inline function

body: () { if (_questionIndex < _questionList.length) return Quiz(..); else return Result(..); }(),

Maurice Raguse · Accepted Answer · 2022-02-14 13:51:57Z

Use Visibility

 Visibility( visible:_questionIndex < _questionList.length replacement:Result(/* ... */) , child:Quiz(/* ... */) )

Or offstage:

 Offstage( offstage: _questionIndex < _questionList.length child:Result(/* ... */) ), Offstage( offstage: _questionIndex >= _questionList.length child:Quiz(/* ... */) )

Dharman · Accepted Answer · 2022-02-14 13:59:18Z

You can simply use the ternary operator here, like

body: _questionIndex < _questionList.length ? Quiz( questionList: _questionList, questionIndex: _questionIndex, answerSelected: _answerSelected) : Result(_finalScore, _reset)

In case you're not familiar with ternary operators

statement ? true : false

if the statement is true the first statement/block after ? will run, otherwise the latter one

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