spring data mongodb: sum aggregation by domain name - spring

Question

I have collection which is contain id, domain . In the collection the same domain saved multiple times. I want to aggregate and get result like
google.com 4
times.com 5
My code
public List<DomainDTO> domainAggregation() {
Aggregation pipeline = newAggregation(
group(fields("id","domain")),
group("domain").count().as("count"),
sort(Sort.Direction.DESC, previousOperation(), "domain")
);
AggregationResults groupResults = mongoTemplate.aggregate(
pipeline, Domains.class, DomainDTO.class);
List<DomainDTO> domainReport = groupResults.getMappedResults();
return domainReport;
}
DomainDTO consist
private String domain;
private Integer count;
Domain entity consist
private String id;
private String searchId;
private String domain;
private Date searchDate;
private String searchName;
private Integer count;
Result json is
{
"domain": null,
"count": 2
},
{
"domain": null,
"count": 1
},
{
"domain": null,
"count": 2
},
{
"domain": null,
"count": 48
},
Domain name is not passed, and not sorted. Could not find bug. Any suggestions?

Answer 1

Your current query outputs something like
{ "$group" : {
"_id" : { "id" : "$id" , "domain" : "$domain"}
} } ,
{ "$group" : { "_id" : "$_id.domain" , "count" : { "$sum" : 1}}} ,
{ "$sort" : { "_id" : -1 , "_id.domain" : -1}}
I believe you're intending something like
{ "$group" : { "_id" : "$domain" , "count" : { "$sum" : 1}}} ,
{ "$sort" : { "_id" : -1}}
Aggregation Java Code:
Aggregation pipeline = newAggregation(
group("domain").count().as("count"),
sort(Sort.Direction.DESC, previousOperation())
);
You will need a $project stage to map _id back to domain in your DomainDTO class.
Aggregation pipeline = newAggregation(
group("domain").count().as("count"),
sort(Sort.Direction.DESC, previousOperation()),
project(bind("domain", "_id")).andExclude("_id").andInclude("count")
);
Mongo Shell
{ "$group" : { "_id" : "$domain" , "count" : { "$sum" : 1}}} ,
{ "$sort" : { "_id" : -1}},
{ "$project" : { "domain" : "$_id" , "_id" : 0 , "count" : 1}