【发布时间】:2020-09-13 17:51:29
【问题描述】:
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;
}
)
}
我想在将负载发送到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。
这里使用的正确语法是什么?
【问题讨论】:
标签: graphql apollo apollo-server