[英]Getting biggest document from a collection in the last 24 hours in Firestore
I have a collection of trades in my Firestore database.我的 Firestore 数据库中有一组交易。 The trades have the following relevant data that I need in the query:
这些交易具有我在查询中需要的以下相关数据:
time: 1589535935410 (milliseconds)
tradeValue: 12343.017
isBuyerMaker: true || false
I need to run a query that gets the biggest TradeValue (one in isBuyerMaker true and also in false) in the last 24 hours.我需要运行一个查询,以获取过去 24 小时内最大的 TradeValue(一个在 isBuyerMaker 中为 true,也为 false)。 How can I achieve this through Firestore queries?
如何通过 Firestore 查询实现此目的?
I tried running this query:我尝试运行此查询:
const biggestBuyQuery = firebase
.firestore()
.collection("cex-trades")
.orderBy("time", "desc")
.where("isBuyerMaker", "==", true)
.where("time", ">", seconds24h())
.orderBy("tradeValue", "desc")
.limit(1);
This does get the documents from the last 24 hours but since it is being ordered by time, I do not get the biggest tradeValue这确实获得了过去 24 小时的文件,但由于它是按时间排序的,所以我没有获得最大的 tradeValue
From the tags I'm going to assume you're working with the JS firestore API.根据标签,我假设您正在使用 JS firestore API。
Edit : I've added workaround that should work for this:编辑:我添加了适用于此的解决方法:
let query = collection
.where("isBuyerMaker", "==", /* true / false */)
.where("time", ">", /* now.milliseconds - (24*60*60*1000) */)
.where("tradeValue", ">=", 0) // New workaround
.orderBy("tradeValue", "desc")
.limit(1)
.get()
I'd run a compound query with a few where clauses:我会运行一个带有几个 where 子句的复合查询:
let query = collection
.where("isBuyerMaker", "==", /* true / false */)
.where("time", ">", /* now.milliseconds - (24*60*60*1000) */)
.orderBy("tradeValue", "desc")
.limit(1)
.get()
and collect the 0th document from the promise when it resolved并在解决时从 promise 中收集第 0 个文档
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