Ada*_*amG 7 javascript node.js firebase google-cloud-firestore
我正在使用Firebase Cloud Firestore,但是,我认为这可能更像是JavaScript异步vs同步承诺返回问题.
我正在进行查询以从一个集合中获取ID,然后我循环查询该查询的结果,以根据该ID查找来自另一个集合的各个记录.
然后我想将每个找到的记录存储到一个数组中,然后返回整个数组.
results.length始终为0,因为return results在forEach完成之前触发.如果我results.length从forEach内部打印它有数据.
在从外部promise和外部函数本身返回之前,我怎么能等到forEach完成?
getFacultyFavoritesFirebase() {
var dbRef = db.collection("users").doc(global.user_id).collection("favorites");
var dbQuery = dbRef.where("type", "==", "faculty");
var dbPromise = dbQuery.get();
var results = [];
return dbPromise.then(function(querySnapshot) {
querySnapshot.forEach(function(doc) {
var docRef = db.collection("faculty").doc(doc.id);
docRef.get().then(function(doc) {
if (doc.exists) {
results.push(doc);
}
})
});
console.log(results.length);
return results;
})
.catch(function(error) {
console.log("Error getting documents: ", error);
});
}
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这里的诀窍是填充results承诺而不是结果.然后Promise.all(),您可以调用该数组的promise并获得所需的结果.当然,doc.exists在推动承诺之前你无法检查是否需要在解决之后处理Promise.all().例如:
function getFacultyFavoritesFirebase() {
var dbRef = db.collection("users").doc(global.user_id).collection("favorites");
var dbQuery = dbRef.where("type", "==", "faculty");
var dbPromise = dbQuery.get();
// return the main promise
return dbPromise.then(function(querySnapshot) {
var results = [];
querySnapshot.forEach(function(doc) {
var docRef = db.collection("faculty").doc(doc.id);
// push promise from get into results
results.push(docRef.get())
});
// dbPromise.then() resolves to a single promise that resolves
// once all results have resolved
return Promise.all(results)
})
.catch(function(error) {
console.log("Error getting documents: ", error);
});
}
getFacultyFavoritesFirebase
.then(results => {
// use results array here and check for .exists
}
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