flutter - 如何修复 flutter 中的 "A dismissed Dismissible widget is still part of the tree."错误

Question

我正在构建一个带有可关闭小部件、firebase 和 StreamBuilder 的 flutter 应用程序，并收到以下错误“已关闭的可关闭小部件仍然是树的一部分。”

请找到以下相同的代码。

Expanded(
                  child: StreamBuilder(
                    stream: Firestore.instance
                        .document('/users/User1/Trips/${widget.tripId}')
                        .collection('TropDocs')
                        .snapshots(),
                    builder: (context, snapshot) {
                      if (!snapshot.hasData) return const Text("Loading....");
                      return ListView.builder(
                        itemExtent: 150.0,
                        itemCount: snapshot.data.documents.length,
                        itemBuilder: (context, index) {

final item = snapshot.data.documents[index];
                          final itemID =
                              snapshot.data.documents[index].documentID;
                          final list = snapshot.data.documents;      
return Dismissible(
   key: Key(itemID),
              // We also need to provide a function that tells our app
              // what to do after an item has been swiped away.
              onDismissed: (direction) {
                // Remove the item from our data source.

                //fBtDoc.deleteTraveldoc(item);
                //Firestore.instance.collection('/users/User1/Trips/${widget.tripId}/TropDocs/').document('$itemID').delete();
                setState(() {
                  list.removeAt(index);
                });

                // Then show a snackbar!
                Scaffold.of(context)
                    .showSnackBar(SnackBar(content: Text("$item dismissed")));
              },
              // Show a red background as the item is swiped away
              background: Container(color: Colors.red),
              child: _buildlistitem(
                            context, snapshot.data.documents[index])
);

                        }
                      );
                    },
                  ),
                )

Answer 1

可能，我迟到了，但我认为这对其他人有帮助。有同样的问题，甚至设置

key: Key(itemId[index]),

没用。然而，

key: UniqueKey(),

非常适合我