Select leadId count on two collection in Mongo DB
Collection 1 : leads
{
leadId:"abc123",
status:"OPENED",
stage:"start",
crossSell:
{
cc:
{
consent:true,
shown:[{first:true}]
}
}
}
Collection 2 : pdata
{
activeLeadId:"abc123",
status:"OPENED",
details:
[
{
rating:10
},
{
rating:9
}
]
}
Question : Find leadId count from leads collection join with pdata collection based on below conditions
leads.leadId = pdata.activeleadId and
leads.status = "OPENED" and
leads.crossSell.cc.consent = true and
leads.crossSell.cc.shown[0].first = true and
pdata.details.rating >= 5
You can try a aggregation query,
$match your conditions for leads collection
$lookup with pdata collection, pass leadId to match with pdata
match required conditions for pdata
$limit to return single document, because we don't need that data in response
$match condition to check is pdata is not empty
$count to get total number of records
db.leads.aggregate([
{
$match: {
status: "OPENED",
"crossSell.cc.consent": true,
"crossSell.cc.shown.first": true
}
},
{
"$lookup": {
"from": "pdata",
"let": { "leadId": "$leadId" },
"pipeline": [
{
$match: {
$expr: { $eq: ["$$leadId", "$activeLeadId"] },
"details.rating": { $gte: 5 }
}
},
{ $limit: 1 }
],
"as": "pdata"
}
},
{ $match: { pdata: { $ne: [] } } },
{ $count: "count" }
])
Playground
Related
the mongodb data like this:
{
"_id": "123dsadasfa454sdsaw",
"hashmap": {
"uuid-12sadsadw5": {
"name": "bob"
},
"uuid-12sadsadwew5": {
"name": "alice"
}
},
"age": 10
}
"hashmap" like java HashMap, the key is uuid like "uuid-12sadsadwew5" and the value is object.
I want to get the data which the name in "hashmap" value is not null. And I use sql :
db.tabl1.find({"hashmap.values.name":{$ne:null}})
but cannot get the right result
You can use this aggregation query:
First use $objectToArray to create an array with values k and v. As we don't know the key (k) we can search by value (v).
Then $unwind array
And $match values where name is not null.
And then regroup and recreate the object using $arrayToObject.
db.collection.aggregate([
{
"$set": {
"hashmap": {
"$objectToArray": "$hashmap"
}
}
},
{
"$unwind": "$hashmap"
},
{
"$match": {
"hashmap.v.name": {
"$ne": null
}
}
},
{
"$group": {
"_id": "$_id",
"hashmap": {
"$push": "$hashmap"
}
}
},
{
"$set": {
"hashmap": {
"$arrayToObject": "$hashmap"
}
}
}
])
Example here
How to find duplicate records count in mongodb
Here is how I get that in mysql
SELECT name, COUNT(*) c FROM table GROUP BY name HAVING c > 1;
Try this
db.table.group({
"key": {
"name": true
},
"initial": {
"c": 0
},
"reduce": function(obj, prev) {
if (true != null) if (true instanceof Array) prev.c += true.length;
else prev.c++;
}});
db.mycollection.aggregate(
// Pipeline
[
// Stage 1
{
$group: {
"_id": "$a",
count: {
$sum: 1
}
}
},
// Stage 2
{
$match: {
count: {
$gt: 1
}
}
},
]
);
Realise this topic has been asked many times - but the advice hasn't helped me solve this problem.
The following query is trying to determine the presence of sales on a given weekday using ISODay. Because the query will be run at the start of the month, I need to know how many occurrences of the specific ISOday occur in the month.
var query = { eventType: 'Sale', site : 4, tank: 1, txnDate : { "$gt" : new Date('2018-08-01T00:00:00') } };
db.tankevent.aggregate([
{ $match: query },
{ $project : {
isoDay: { $isoDayOfWeek: "$txnDate" },
dayDate: { $dateToString: { format: "%d", date:"$txnDate" } }
}
},
{ $group:
{ _id : { isoday: "$isoDay", dday: "$dayDate" }, count: { "$sum" : 1 } }
},
{ $sort: { "_id.isoday": 1, "_id.dday": 1 } }
])
provides the following output
/* 1 */
{
"_id" : {
"isoday" : 1,
"dday" : "06"
},
"count" : 62.0
}
/* 2 */
{
"_id" : {
"isoday" : 1,
"dday" : "13"
},
"count" : 69.0
}
/* 3 */
{
"_id" : {
"isoday" : 1,
"dday" : "20"
},
"count" : 72.0
}
/* 4 */
{
"_id" : {
"isoday" : 2,
"dday" : "07"
},
"count" : 75.0
}
I am trying to have "count" represent the number of unique "dday" records - so using the output above, I want count to be "3" for isoDay = 1. At the moment count is reporting number of sales events that occurred for the group combination
All you need to do is have the grouping twice.
db.tankevent.aggregate([
{ $match: query },
{ $project : {
isoDay: { $isoDayOfWeek: "$txnDate" },
dayDate: { $dateToString: { format: "%d", date:"$txnDate" } }
}
},
{ $group:
{ _id : { isoday: "$isoDay", dday: "$dayDate" }, count: { "$sum" : 1 } }
},
{ $project : {
isoDay_Final: "$_id.isoday"
}
},
{ $group:
{ _id : "$isoDay_Final", count: { "$sum" : 1 } }
},
{ $sort: { "_id": 1 } }
])
I have this query that provides me the join I want to:
db.summoners.aggregate([
{ "$match": { "nick":"Luispfj" } },
{ "$unwind": "$matches" },
{
"$lookup": {
"from":"matches",
"localField":"matches.gameId",
"foreignField":"gameId",
"as":"fullMatches"
}
},
{ "$unwind": "$fullMatches" },
{
"$group": {
"_id": null,
"matches": { "$push":"$fullMatches" }
}
}
])
But when I run the unwind function the null entries are gone. How do I retrieve them (with their respective "gameId"s, if possible?
Also, is there a way to retrieve only the matches array, instead of it being a subproperty of the "null-id-object" it creates?
$unwind takes an optional field preserveNullAndEmptyArrays which by default is false. If you set it to true, unwind will output the documents that are null. Read more about $unwind
{
"$unwind": {
path: "$fullMatches",
preserveNullAndEmptyArrays: true
}
},
my friend is telling me that mongo is not worth learning since its very bad to do complex querying, something like this:
SELECT person, SUM(score), AVG(score), MIN(score), MAX(score), COUNT(*)
FROM demo
WHERE score > 0 AND person IN('bob','jake')
GROUP BY person;
he is telling me that if i want to do this query with mongo i have to write this
db.demo.group({
"key": {
"person": true
},
"initial": {
"sumscore": 0,
"sumforaverageaveragescore": 0,
"countforaverageaveragescore": 0,
"countstar": 0
},
"reduce": function(obj, prev) {
prev.sumscore = prev.sumscore + obj.score - 0;
prev.sumforaverageaveragescore += obj.score;
prev.countforaverageaveragescore++;
prev.minimumvaluescore = isNaN(prev.minimumvaluescore) ? obj.score : Math.min(prev.minimumvaluescore, obj.score);
prev.maximumvaluescore = isNaN(prev.maximumvaluescore) ? obj.score : Math.max(prev.maximumvaluescore, obj.score);
if (true != null) if (true instanceof Array) prev.countstar += true.length;
else prev.countstar++;
},
"finalize": function(prev) {
prev.averagescore = prev.sumforaverageaveragescore / prev.countforaverageaveragescore;
delete prev.sumforaverageaveragescore;
delete prev.countforaverageaveragescore;
},
"cond": {
"score": {
"$gt": 0
},
"person": {
"$in": ["bob", "jake"]
}
}
});
so having no mongodb background i dont know what to think and i've been searching arround and everyone says that mongo is better for a lot of stuff, still how do i do this query in mongo?
is it like my friend says? or is there a easier way to do this?
There is a much easier way to do that.
db.demo.aggregate([
{ $match: { score: { $gt: 0 }, person: { $in: ["bob", "jake"] } } },
{ $group: { _id: "$person", scoreSum: { $sum: "$score" }, scoreAvg: { $avg: "$score" }, scoreMin: { $min: "$score" }, scoreMax: { $max: "$score" }, count: { $sum: 1 } } }
])