I have this database on Firebase:
{
"issues" : {
"-L04771_EjrLlv5u1-GU" : {
"issue" : "Test insert 1",
"last_edit" : "d8QICgTG5xR20RBzAXfzfu8gLgw2",
"owner" : "d8QICgTG5xR20RBzAXfzfu8gLgw2",
"owner_email" : "example1#gmail.com",
"status" : 1,
"url" : "http://www.example.com/example.html"
},
"-L047pIoqxkj4saaTYyQ" : {
"issue" : "Test insert 2",
"last_edit" : "d8QICgTG5xR20RBzAXfzfu8gLgw2",
"owner" : "d8QICgTG5xR20RBzAXfzfu8gLgw2",
"owner_email" : "example2#gmail.com",
"status" : 1,
"url" : "http://www.example.com/example.html"
}
}
}
I have to extract only those who have owner_email "example1#gmail.com".
Is possible?
You're probably looking for Firebase Queries and especially the equalTo-Query.
Your code would be something like this:
// Find all issues with owner_email = example1#gmail.com
var ref = firebase.database().ref("issues");
ref.orderByChild("owner_email").equalTo("example1#gmail.com").on("value", function(snapshot) {
// Loops through the matching issues
snapshot.forEach(function(child) {
console.log(child.key);
});
});
Related
Here is my database entry structure, it has a nested document called friends, I want to compare two different _id's friends list in dart using mongo_dart
{
"_id" : ObjectId("60ae06074e162995281b4666"),
"email" : "one#one.com",
"emailverified" : false,
"username" : "one#one.com",
"displayName" : "complete n00b",
"phonenumber" : "",
"dob" : "",
"points" : 0,
"friends" : [
{
"username" : "three#one.com",
"sent" : ISODate("2021-05-26T10:01:30.616Z")
},
{
"username" : "six#one.com",
"sent" : ISODate("2021-05-26T10:43:16.822Z")
}
]
}
Here is my code, but I am not getting any returns
Future<Map> commonFriends(store, myObjectId, theirObjectId) async {
var commonList = await store.aggregate([
{
'\$project': {
'friends': 1,
'commonToBoth': {
'\$setIntersection': [
{'_id': myObjectId, 'friends': '\$username'},
{'_id': theirObjectId, 'friends': '\$username'}
]
},
}
}
]);
return commonList;
}
I am getting an error from db.dart which is apart of mongo_dart package. The error is "Exception has occurred.
Map (4 items)"
How could I query this data in this way:
I would like to ignore the CurrentSubAdministrativeArea child and iterate every each sub child and find the right userKey
Actually I'm using this code, that isnt working:
self.ref.child("Ads").child("CurrentSubAdministrativeArea")
/*HERE I would like to ignore the childs*/
.queryOrdered(byChild: "userKey").queryEqual(toValue: uid).observeSingleEvent(of:.value, with: { (snapshot) in
--
{
"Ads" : {
"CurrentSubAdministrativeArea" : {
"Mantova" : {
"-L7ymBmmbHkNfhBRte9F" : {
"cost" : 200,
"date" : 1527256922000,
"info" : "Test",
"maxLimit" : 100,
"minLimit" : 10,
"personBadType" : [ "abitudinaria", "antipatica" ],
"personGoodType" : [ "simpatica", "felice" ],
"subAdministrativeArea" : "Mantova",
"title" : "Mantova Test",
"url" : "https://firebasestorage.googleapis.com/v0/b/team-34540.appspot.com/o/Mantova%20Test?alt=media&token=3a81ed1c-ecd6-4dc0-bd7c-45e093ce8188",
"userKey" : "OsJRc98sqxPx70iqxFtoqerMzHH2",
"via" : "viale dei test"
}
},
"Milano" : {
"-L6qywMC6nxi0fJNMHba" : {
"cost" : 454,
"date" : 1528298580000,
"info" : "Di pollo",
"maxLimit" : 100,
"minLimit" : 10,
"personBadType" : [ "abitudinaria", "antipatica" ],
"personGoodType" : [ "simpatica", "felice" ],
"subAdministrativeArea" : "Milano",
"title" : "Pollo 2",
"url" : "https://firebasestorage.googleapis.com/v0/b/team-34540.appspot.com/o/Pollo?alt=media&token=fc6a3ec8-5f9a-4347-bdad-2d9715af784d",
"userKey" : "OsJRc98sqxPx70iqxFtoqerMzHH2",
"via" : "viale test"
}
}
}
}
}
You could denormalize your data in such a way your query is easy to build and execute.
Together with the data structure you already have you would have another node (ie. another data structure) like
{
"AdsByUsers" : {
"OsJRc98sqxPx70iqxFtoqerMzHH2": {
"Mantova",
"Milano",
...
},
"abcde88qxPx70iqxFtoqerMzKh5": {
"Firenze",
...
}
With NoSQL database you should not hesitate to duplicate data in such a way your queries are easy and fast to execute.
Is it possible to create Salesreceipt without product/service value through QBO API? I have tried through API but it's not reflecting rate value and storing description value only.
If I remove ItemRef attribute(in request body) then it's reflecting rate and amount values and it's assigning some default and random product/service.
It is possible directly in QBO UI.
Request body where only description value storing:
{
"TxnDate" : "2016-05-27",
"Line" : [ {
"Amount" : 2222.00,
"Description" : "hi chk",
"DetailType" : "ItemReceiptLineDetail",
"ItemReceiptLineDetail" : {
"ItemRef" : { },
"Qty" : 1,
"UnitPrice" : 2222
} }
],
"CustomerRef" : {
"value" : "67"
},
"CustomerMemo" : {
"value" : "Thanks for your business! We appreciate referrals!"
},
"TotalAmt": 2222.00,
"PrivateNote" : "",
"CustomField" : [ {
"DefinitionId" : "1",
"Type" : "StringType",
"StringValue" : ""
} ]
}
Request body where default product/service assigning:
{
"TxnDate" : "2016-05-27",
"Line" : [ {
"Amount" : 2222.00,
"Description" : "hi chk",
"DetailType" : "ItemReceiptLineDetail",
"ItemReceiptLineDetail" : {
"Qty" : 1,
"UnitPrice" : 2222
} }
],
"CustomerRef" : {
"value" : "67"
},
"CustomerMemo" : {
"value" : "Thanks for your business! We appreciate referrals!"
},
"TotalAmt": 2222.00,
"PrivateNote" : "",
"CustomField" : [ {
"DefinitionId" : "1",
"Type" : "StringType",
"StringValue" : ""
} ]
}
No.
QuickBooks Online does not support this.
Consider the following example of mongo collection:
{"_id" : ObjectId("4f304818884672067f000001"), "hash" : {"call_id" : "1234"}, "something" : "AAA"}
{"_id" : ObjectId("4f304818884672067f000002"), "hash" : {"call_id" : "1234"}, "something" : "BBB"}
{"_id" : ObjectId("4f304818884672067f000003"), "hash" : {"call_id" : "1234"}, "something" : "CCC"}
{"_id" : ObjectId("4f304818884672067f000004"), "hash" : {"call_id" : "5555"}, "something" : "DDD"}
{"_id" : ObjectId("4f304818884672067f000005"), "hash" : {"call_id" : "5555"}, "something" : "CCC"}
I would like to query this collection and get only the first entry for each "call_id", in other words i'm trying to get unique results based on "call_id".
I tried to use .distinct method:
#result = Myobject.all.distinct('hash.call_id')
but the resulting array will contain only the unique call_id fields:
["1234", "5555"]
and I need all the other fields too.
Is it possible to make a query like this one?:
#result = Myobject.where('hash.call_id' => Myobject.all.distinct('hash.call_id'))
Thanks
You cannot simply return the document(or subset) by using the distinct. As per the documentation it only returns the distinct array of values based on the given key. But you can achieve this by using map-reduce
var _map = function () {
emit(this.hash.call_id, {doc:this});
}
var _reduce = function (key, values) {
var ret = {doc:[]};
var doc = {};
values.forEach(function (value) {
if (!doc[value.doc.hash.call_id]) {
ret.doc.push(value.doc);
doc[value.doc.hash.call_id] = true; //make the doc seen, so it will be picked only once
}
});
return ret;
}
The above code is self explanatory, on map function i am grouping it by key hash.call_id and returning the whole doc so it can be processed by reduce funcition.
On reduce function, just loop through the grouped result set and pick only one item from the grouped set (among the multiple duplicate key values - distinct simulation).
Finally create some test data
> db.disTest.insert({hash:{call_id:"1234"},something:"AAA"})
> db.disTest.insert({hash:{call_id:"1234"},something:"BBB"})
> db.disTest.insert({hash:{call_id:"1234"},something:"CCC"})
> db.disTest.insert({hash:{call_id:"5555"},something:"DDD"})
> db.disTest.insert({hash:{call_id:"5555"},something:"EEE"})
> db.disTest.find()
{ "_id" : ObjectId("4f30a27c4d203c27d8f4c584"), "hash" : { "call_id" : "1234" }, "something" : "AAA" }
{ "_id" : ObjectId("4f30a2844d203c27d8f4c585"), "hash" : { "call_id" : "1234" }, "something" : "BBB" }
{ "_id" : ObjectId("4f30a2894d203c27d8f4c586"), "hash" : { "call_id" : "1234" }, "something" : "CCC" }
{ "_id" : ObjectId("4f30a2944d203c27d8f4c587"), "hash" : { "call_id" : "5555" }, "something" : "DDD" }
{ "_id" : ObjectId("4f30a2994d203c27d8f4c588"), "hash" : { "call_id" : "5555" }, "something" : "EEE" }
and running this map reduce
> db.disTest.mapReduce(_map,_reduce, {out: { inline : 1}})
{
"results" : [
{
"_id" : "1234",
"value" : {
"doc" : [
{
"_id" : ObjectId("4f30a27c4d203c27d8f4c584"),
"hash" : {
"call_id" : "1234"
},
"something" : "AAA"
}
]
}
},
{
"_id" : "5555",
"value" : {
"doc" : [
{
"_id" : ObjectId("4f30a2944d203c27d8f4c587"),
"hash" : {
"call_id" : "5555"
},
"something" : "DDD"
}
]
}
}
],
"timeMillis" : 2,
"counts" : {
"input" : 5,
"emit" : 5,
"reduce" : 2,
"output" : 2
},
"ok" : 1,
}
You get the first document of the distinct set. You can do the same in mongoid by first stringify the map/reduce functions and call mapreduce like this
MyObject.collection.mapreduce(_map,_reduce,{:out => {:inline => 1},:raw=>true })
Hope it helps
I have user document collection like this:
User {
id:"001"
name:"John",
age:30,
friends:["userId1","userId2","userId3"....]
}
A user has many friends, I have the following query in SQL:
select * from user where in (select friends from user where id=?) order by age
I would like to have something similar in MongoDB.
To have everything with just one query using the $lookup feature of the aggregation framework, try this :
db.User.aggregate(
[
// First step is to extract the "friends" field to work with the values
{
$unwind: "$friends"
},
// Lookup all the linked friends from the User collection
{
$lookup:
{
from: "User",
localField: "friends",
foreignField: "_id",
as: "friendsData"
}
},
// Sort the results by age
{
$sort: { 'friendsData.age': 1 }
},
// Get the results into a single array
{
$unwind: "$friendsData"
},
// Group the friends by user id
{
$group:
{
_id: "$_id",
friends: { $push: "$friends" },
friendsData: { $push: "$friendsData" }
}
}
]
)
Let's say the content of your User collection is the following:
{
"_id" : ObjectId("573b09e6322304d5e7c6256e"),
"name" : "John",
"age" : 30,
"friends" : [
"userId1",
"userId2",
"userId3"
]
}
{ "_id" : "userId1", "name" : "Derek", "age" : 34 }
{ "_id" : "userId2", "name" : "Homer", "age" : 44 }
{ "_id" : "userId3", "name" : "Bobby", "age" : 12 }
The result of the query will be:
{
"_id" : ObjectId("573b09e6322304d5e7c6256e"),
"friends" : [
"userId3",
"userId1",
"userId2"
],
"friendsData" : [
{
"_id" : "userId3",
"name" : "Bobby",
"age" : 12
},
{
"_id" : "userId1",
"name" : "Derek",
"age" : 34
},
{
"_id" : "userId2",
"name" : "Homer",
"age" : 44
}
]
}
Edit: this answer only applies to versions of MongoDb prior to v3.2.
You can't do what you want in just one query. You would have to first retrieve the list of friend user ids, then pass those ids to the second query to retrieve the documents and sort them by age.
var user = db.user.findOne({"id" : "001"}, {"friends": 1})
db.user.find( {"id" : {$in : user.friends }}).sort("age" : 1);
https://docs.mongodb.org/manual/reference/operator/aggregation/lookup/
This is the doc for join query in mongodb , this is new feature from version 3.2.
So this will be helpful.
You can use in Moongoose JS .populate() and { populate : { path : 'field' } }.
Example:
Models:
mongoose.model('users', new Schema({
name:String,
status: true,
friends: [{type: Schema.Types.ObjectId, ref:'users'}],
posts: [{type: Schema.Types.ObjectId, ref:'posts'}],
}));
mongoose.model('posts', new Schema({
description: String,
comments: [{type: Schema.Types.ObjectId, ref:'comments'}],
}));
mongoose.model('comments', new Schema({
comment:String,
status: true
}));
If you want to see your friends' posts, you can use this.
Users.find(). //Collection 1
populate({path:'friends', //Collection 2
populate:{path:'posts' //Collection 3
}})
.exec();
If you want to see your friends' posts and also bring all the comments, you can use this and too, you can indentify the collection if this not find and the query is wrong.
Users.find(). //Collection 1
populate({path:'friends', //Collection 2
populate:{path:'posts', //Collection 3
populate:{path:'commets, model:Collection'//Collection 4 and more
}}})
.exec();
And to finish, if you want get only some fields of some Collection, you can use the propiertie select Example:
Users.find().
populate({path:'friends', select:'name status friends'
populate:{path:'comments'
}})
.exec();
MongoDB doesn't have joins, but in your case you can do:
db.coll.find({friends: userId}).sort({age: -1})
one kind of join a query in mongoDB, is ask at one collection for id that match , put ids in a list (idlist) , and do find using on other (or same) collection with $in : idlist
u = db.friends.find({"friends": ? }).toArray()
idlist= []
u.forEach(function(myDoc) { idlist.push(myDoc.id ); } )
db.friends.find({"id": {$in : idlist} } )
Only populate array friends.
User.findOne({ _id: "userId"})
.populate('friends')
.exec((err, user) => {
//do something
});
Result is same like this:
{
"_id" : "userId",
"name" : "John",
"age" : 30,
"friends" : [
{ "_id" : "userId1", "name" : "Derek", "age" : 34 }
{ "_id" : "userId2", "name" : "Homer", "age" : 44 }
{ "_id" : "userId3", "name" : "Bobby", "age" : 12 }
]
}
Same this: Mongoose - using Populate on an array of ObjectId
You can use playOrm to do what you want in one Query(with S-SQL Scalable SQL).
var p = db.sample1.find().limit(2) ,
h = [];
for (var i = 0; i < p.length(); i++)
{
h.push(p[i]['name']);
}
db.sample2.find( { 'doc_name': { $in : h } } );
it works for me.
You can do it in one go using mongo-join-query. Here is how it would look like:
const joinQuery = require("mongo-join-query");
joinQuery(
mongoose.models.User,
{
find: {},
populate: ["friends"],
sort: { age: 1 },
},
(err, res) => (err ? console.log("Error:", err) : console.log("Success:", res.results))
);
The result will have your users ordered by age and all of the friends objects embedded.
How does it work?
Behind the scenes mongo-join-query will use your Mongoose schema to determine which models to join and will create an aggregation pipeline that will perform the join and the query.