Flutter:如果存在则获取提供者，否则返回 null 而不是异常

Question

当我们使用 BlocProvider.of<OrderBloc>(context)访问 Bloc 对象，如果没有 OrderBloc 则返回异常存在于当前上下文的祖先小部件上。返回异常如下:

No ancestor could be found starting from the context that was passed to BlocProvider.of<OrderBloc>().

但是我要退货null而不是异常时没有 OrderBloc存在于祖先小部件上。考虑以下场景:

var orderBloc = BlocProvider.of<OrderBloc>(context);

return Container(child: orderBloc == null
      ? Text('-')
      : BlocBuilder<OrderBloc, OrderState>(
          bloc: orderBloc,
          builder: (context, state) {
            // build something if orderBloc exists.
          },

        ),          
);

Answer 1

你可以像这样用 try/catch 包裹那一行:

var orderBloc;
try {
  orderBloc = BlocProvider.of<OrderBloc>(context);
} catch (e) {}

return Container(child: orderBloc == null
  ? Text('-')
  : BlocBuilder<OrderBloc, OrderState>(
      bloc: orderBloc,
      builder: (context, state) {
        // build something if orderBloc exists.
      },
    ),          
);

编辑:
如果你想减少样板文件:

extension ReadOrNull on BuildContext {
  T? readOrNull<T>() {
    try {
      return read<T>();
    } on ProviderNotFoundException catch (_) {
      return null;
    }
  }
}

那么你的代码将是:

var orderBloc = context.readOrNull<OrderBloc>();

return Container(child: orderBloc == null
  ? Text('-')
  : BlocBuilder<OrderBloc, OrderState>(
      bloc: orderBloc,
      builder: (context, state) {
        // build something if orderBloc exists.
      },
    ),          
);