Reputation: 1996
Im looking into the Order class example and found that the Item class is not converted to Map.
class Order {
int count;
int itemNumber;
bool isRushed;
Item item;
Map<String, dynamic> toJson() => _$OrderToJson(this);
}
The generated .g file has this:
Map<String, dynamic> _$OrderToJson(Order instance) {
...
writeNotNull('item', instance.item);
...
return val;
}
The item in order map is still of Item type, but Im expecting it to be auto converted to Map as well. the generated .g file should has something like this
writeNotNull('item', instance.item.toJson());
I don't want to manually add this since it will be overwritten when .g file is regenerated. Why is the json_serializable lib not doing such a simple thing, or am I missing something? thanks.
Upvotes: 7
Views: 7029
Reputation: 3884
Please add the below line above your child-level data model class... So it will call in dept toJson method of each model...
@JsonSerializable(explicitToJson: true)
Upvotes: 2
Reputation: 14142
For firebase you'll want the any_map
option also since the maps from firebase are <dynamic, dynamic>
otherwise it will expect <String, dynamic>
Create this file as build.yaml
in the root of your flutter project, it doesn't exist by default.
targets:
$default:
builders:
json_serializable:
options:
any_map: true
explicit_to_json: true
Upvotes: 8
Reputation: 1996
Now I found the solution, just set this in build.yaml
explicit_to_json = true.
and regenerate the .g file. It should convert it to Map for you now.
Upvotes: 21
Reputation: 30103
json.encode(...)
will attempt to find a toJson()
method on the Item
class, and serialize the item correctly. There is no need for the generated serialization logic to call toJson()
.
You just have to make sure that the Item
class is also annotated with @JsonSerializable()
and implements toJson()
.
Also, you would never call toJson()
manually, instead you pass your order object to json.encode(...)
Upvotes: 2