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onTap replace widget from another file flutter

0

Say my App() contains some widget and to make the code look pretty, i have created classes for child widgets. Now these child widgets contain onTap functions which are supposed to replace a widget on App(), so how do i approach this type of problem ?

flutter flutter-layout

share|improve this question

asked Mar 8 at 17:24

Femin DharamshiFemin Dharamshi

84

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  do you want to navigate to another screen? can you add some code and explain a little bit in detail?
  
  – Amol G
  Mar 8 at 17:51
  

  
  
  
  

add a comment | 

0

Say my App() contains some widget and to make the code look pretty, i have created classes for child widgets. Now these child widgets contain onTap functions which are supposed to replace a widget on App(), so how do i approach this type of problem ?

flutter flutter-layout

share|improve this question

asked Mar 8 at 17:24

Femin DharamshiFemin Dharamshi

84

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  do you want to navigate to another screen? can you add some code and explain a little bit in detail?
  
  – Amol G
  Mar 8 at 17:51
  

  
  
  
  

add a comment | 

Say my App() contains some widget and to make the code look pretty, i have created classes for child widgets. Now these child widgets contain onTap functions which are supposed to replace a widget on App(), so how do i approach this type of problem ?

flutter flutter-layout

share|improve this question

asked Mar 8 at 17:24

Femin DharamshiFemin Dharamshi

84

flutter flutter-layout

flutter flutter-layout

share|improve this question

asked Mar 8 at 17:24

Femin DharamshiFemin Dharamshi

84

share|improve this question

asked Mar 8 at 17:24

Femin DharamshiFemin Dharamshi

84

asked Mar 8 at 17:24

Femin DharamshiFemin Dharamshi

84

asked Mar 8 at 17:24

Femin DharamshiFemin Dharamshi

84

1 Answer
1

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0

Some code of what you're doing would be useful or where you're trying to "replace a widget".

If by replace you mean show a different widget in the place of another then you just use a boolean to decide which one to show. These are the steps I would follow to implement this.

1. Make the widget in your App() stateful and create a member variable boolean called showingOriginalWidget = true;


2. In your child widget classes take in a Function in the parameter called onSwapWidget.


3. In your onTap function in your child widget call onSwapWidget()


4. In your App() supply the widget that's performing this action with your Function to call back to

See below

childWidget(onSwapWidget: ()
 setState(()
 // toggle the original widget state
 showingOriginalWidget = !showOriginalWidget;
 );
);

5. Where you're showing your widgets add a condition so that you show either one depending on the value.

example

...
child: showingOriginalWidget ? originalWidget() : swappedOutWidget()
...

That should do the trick.

share|improve this answer

answered Mar 8 at 18:12

Filled StacksFilled Stacks

1,0121914

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0

Some code of what you're doing would be useful or where you're trying to "replace a widget".

If by replace you mean show a different widget in the place of another then you just use a boolean to decide which one to show. These are the steps I would follow to implement this.

1. Make the widget in your App() stateful and create a member variable boolean called showingOriginalWidget = true;


2. In your child widget classes take in a Function in the parameter called onSwapWidget.


3. In your onTap function in your child widget call onSwapWidget()


4. In your App() supply the widget that's performing this action with your Function to call back to

See below

childWidget(onSwapWidget: ()
 setState(()
 // toggle the original widget state
 showingOriginalWidget = !showOriginalWidget;
 );
);

5. Where you're showing your widgets add a condition so that you show either one depending on the value.

example

...
child: showingOriginalWidget ? originalWidget() : swappedOutWidget()
...

That should do the trick.

share|improve this answer

answered Mar 8 at 18:12

Filled StacksFilled Stacks

1,0121914

add a comment | 

0

Some code of what you're doing would be useful or where you're trying to "replace a widget".

If by replace you mean show a different widget in the place of another then you just use a boolean to decide which one to show. These are the steps I would follow to implement this.

1. Make the widget in your App() stateful and create a member variable boolean called showingOriginalWidget = true;


2. In your child widget classes take in a Function in the parameter called onSwapWidget.


3. In your onTap function in your child widget call onSwapWidget()


4. In your App() supply the widget that's performing this action with your Function to call back to

See below

childWidget(onSwapWidget: ()
 setState(()
 // toggle the original widget state
 showingOriginalWidget = !showOriginalWidget;
 );
);

5. Where you're showing your widgets add a condition so that you show either one depending on the value.

example

...
child: showingOriginalWidget ? originalWidget() : swappedOutWidget()
...

That should do the trick.

share|improve this answer

answered Mar 8 at 18:12

Filled StacksFilled Stacks

1,0121914

add a comment | 

Some code of what you're doing would be useful or where you're trying to "replace a widget".

If by replace you mean show a different widget in the place of another then you just use a boolean to decide which one to show. These are the steps I would follow to implement this.

1. Make the widget in your App() stateful and create a member variable boolean called showingOriginalWidget = true;


2. In your child widget classes take in a Function in the parameter called onSwapWidget.


3. In your onTap function in your child widget call onSwapWidget()


4. In your App() supply the widget that's performing this action with your Function to call back to

See below

childWidget(onSwapWidget: ()
 setState(()
 // toggle the original widget state
 showingOriginalWidget = !showOriginalWidget;
 );
);

5. Where you're showing your widgets add a condition so that you show either one depending on the value.

example

...
child: showingOriginalWidget ? originalWidget() : swappedOutWidget()
...

That should do the trick.

share|improve this answer

answered Mar 8 at 18:12

Filled StacksFilled Stacks

1,0121914

childWidget(onSwapWidget: ()
 setState(()
 // toggle the original widget state
 showingOriginalWidget = !showOriginalWidget;
 );
);

...
child: showingOriginalWidget ? originalWidget() : swappedOutWidget()
...

share|improve this answer

answered Mar 8 at 18:12

Filled StacksFilled Stacks

1,0121914

answered Mar 8 at 18:12

Filled StacksFilled Stacks

1,0121914

answered Mar 8 at 18:12

Filled StacksFilled Stacks

1,0121914