Panache with MongoDB find distinct

Question

I have documents with "tags" arrays as properties. Now I want to query all distinct tag-items.

{ 
  "name": "test1",
  "tags": ["tag1","tag2", "tag3"]
},
{
  "name": "test2",
  "tags": ["tag1"]
}

Solution in mongo shell:

db.ApiModel.distinct("tags")

which gives me:

["tag1", "tag2", "tag3"]

But how can I achieve the same result with Panache? The PanacheMongoEntity does not offer a specific distinct method. Nor do i know how to use the find method to achieve my goal or if it even is possible using this method.

All I could possibly think of is finding all tags with find("tags", "*") (is * the wildcard?) and then dealing with duplicates in Java, but I don't believe that's the intended use.

Answer 1

You can use either of the two methods to get distinct tags from the collection test.

public List<String> getDistinctTags() {
        return Tags
                .<Tags>mongoCollection()
                .distinct("tags", String.class)
                .into(new ArrayList<String>());
}

public List<String> getDistinctTags() {
        return Tags
                .<Tags>streamAll()
                .flatMap(e -> e.tags.stream())
                .distinct()
                .collect(toList());
}

Assuming the Tags class is defined as follows and represents the Panache entity:

@MongoEntity(collection="test")
public class Tags extends PanacheMongoEntity {
    public String name;
    public List<String> tags;
    // ...
}