问题描述
我正在尝试为 firebase 创建一个云函数,该函数根据延迟值删除用户并在延迟数之后插入回.
I'm trying to create a cloud function for firebase that remove a user depending on the delay value and insert back the after the number of delay.
exports.delayqueue = functions.database.ref('/queues/{queueid}/members/{memberid}').onWrite(event => {
var members = event.data.ref.parent;
var user = event.data;
var queueid = members.parent;
var userid = event.params.memberid;
var delayfor = user.child('delay').val();
var name = user.child('name').val();
if(delayfor != 0){
members.child(event.params.memberid).remove();
join(userid,queueid,delayfor,name);
return;
}else{
return;
}
});
function join(userid,queueid,delayfor,name){
setTimeout(function(){
var ref = db.ref("queues/queueid/members/userid");
ref.set({
name: name,
timestamp: Date.now(),
delay : 0
});
}, delayfor*1000);
};
但它不起作用,有人可以帮忙吗?
But it's not working can someone help?
推荐答案
您需要将 setTimeout
包装在 Promise 中:
You'll need to wrap your setTimeout
in a Promise:
exports.delayqueue = functions.database.ref('/queues/{queueid}/members/{memberid}').onWrite(event => {
var members = event.data.ref.parent;
var user = event.data;
var queueid = members.parent;
var userid = event.params.memberid;
var delayfor = user.child('delay').val();
var name = user.child('name').val();
if (delayfor !== 0){
members.child(event.params.memberid).remove();
return join(userid,queueid,delayfor,name);
} else {
return;
}
});
function join(userid,queueid,delayfor,name){
return new Promise((resolve, reject) => {
setTimeout(function(){
var ref = db.ref("queues/queueid/members/userid");
ref.set({
name: name,
timestamp: Date.now(),
delay : 0
}).then(resolve, reject);
}, delayfor*1000);
});
};
请注意,等待 setTimeout 所花费的时间计费为函数执行时间,并且还受函数超时的影响.如果您只是延迟几秒钟,那可能没问题,但如果预计延迟是几分钟,那么这个解决方案将不可行或不划算.
Note that the time spent waiting for setTimeout is billed as function execution time and is also subject to the function timeout. If you're only delaying a few seconds, that might be okay, but if the delay is expected to be minutes this solution isn't going to be viable or cost-effective.
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