Apollo GraphQL：使用 withFilter 时修改有效负载？

Question

我有一个使用 withFilter 的有效订阅：

   User_Presence_Subscription: {
        subscribe: withFilter(
            () => pubsub.asyncIterator(USER_PRESENCE_UPDATED_CHANNEL),
            (payload, args, context) => {
                if (typeof (payload) === 'undefined') {
                    return false;
                }
                const localUserId = (typeof(context) == 'undefined' || typeof(context.userId) == 'undefined') ? null : context.userId;
                const ids_to_watch = args.ids_to_watch;
                const usersWithUpdatedPresence = payload.User_Presence_Subscription;

                let result = false;
                console.log("User_Presence_Subscription - args == ", args, result);
                return result;
            }
        )
    }

我想在将有效负载发送给客户端之前对其进行修改。我尝试添加一个 resolve 函数 as shown in the docs:

   User_Presence_Subscription: {
        resolve: (payload, args, context) => {
            debugger; <== NEVER ACTIVATES
            return {
                User_Presence_Subscription: payload,
            };
        },
        subscribe: withFilter(
            () => pubsub.asyncIterator(USER_PRESENCE_UPDATED_CHANNEL),
            (payload, args, context) => {
                if (typeof (payload) === 'undefined') {
                    return false;
                }
                const localUserId = (typeof(context) == 'undefined' || typeof(context.userId) == 'undefined') ? null : context.userId;
                const ids_to_watch = args.ids_to_watch;
                const usersWithUpdatedPresence = payload.User_Presence_Subscription;

                let result = false;
                console.log("User_Presence_Subscription - args == ", args, result);
                return result;
            }
        )
    }

...但是 resolve 函数中的 debugger 行永远不会被命中。

此处使用的正确语法是什么？

Answer 1

已解决。解析器没有被命中的唯一原因是在我的测试代码中，我从 withFilter 函数返回 false。当它 returns true 解析器按预期命中时。

Apollo GraphQL：使用 withFilter 时修改有效负载？

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