How to remove a rendered widget?

[rupinderjeet]

In onLayout(), I have some code like this:

final source = getChild(#source);
final sourceSize = source.layout(constraints);
source.position(const Offset(-10000, 0));

I am using sourceSize to determine source's size and apply this size to another widget. This is working for me. But, I want to remove source completely instead of hiding it at (-10000, 0). How can I do this?

[pingbird]

Call source.ignore(): https://pub.dev/documentation/boxy/latest/boxy/BaseBoxyChild/ignore.html

[rupinderjeet]

Thank you. This has helped me remove weird offset I was using. However, I have two more related questions.

1. Is the order of using ignore correct in below code?

final source = getChild(#source);
final sourceSize = source.layout(constraints);
source.position(Offset.zero);
source.ignore();

2. On a page, If a Hero widget, with tag photo_hero, is nested as a child of this dummy source widget, Boxy will layout it to get the size. But, this also means that I can no longer use the same Hero tag on that page again. Is this right? Hero widget tagged with photo_hero will be used with source widget and I will somehow need to tell actual widget and its Hero widget to use a different tag. Is this right?

[pingbird]

Yes, that is correct, the child still exists in the Element tree preventing you from having an identical Hero. I'm not sure if the framework allows you to create and destroy a child in the same layout pass, that requires some more testing.

[rupinderjeet]

Thank you for your help and this project. There is a lot to learn for me in this codebase. You can close this issue, my query is answered.