Min_doc_count on lower/lowest level nested aggregation

George_Lui · June 5, 2014, 9:51pm

I have this query with some nested aggregations

{

"aggs": {*
```
 "by_date": {*
```
```
    "date_histogram": {*
```
```
       "field": "timestamp",*
```
```
       "interval": "day"*
```
```
    },*
```
```
    "aggs": {*
```
```
       "new_users": {*
```
```
          "filter": {*
```
```
             "query": {*
```
```
                "match": {*
```

                   "action": "USER_ADD"*

```
                }*
```
```
             }*
```
```
          },*
```
```
          "aggs": {*
```
```
             "unique_users": {*
```
```
                "cardinality": {*
```
```
                   "field": "user"*
```
```
                }*
```
```
             }*
```
```
          }*
```
```
       }*
```
```
    }*
```
```
 }*
```
},*
"size": 0*
}

It yields results that look like this

"aggregations": {

```
 "by_date": {*
```
```
    "buckets": [*
```
```
       {*
```

          "key_as_string": "1970-01-07T00:00:00.000Z",*

```
          "key": 518400000,*
```
```
          "doc_count": 210,*
```
```
          "new_users": {*
```
```
             "doc_count": 0,*
```
```
             "unique_users": {*
```
```
                "value": 0*
```
```
             }*
```
```
          }*
```
```
       },*
```
```
       {*
```

          "key_as_string": "1970-01-09T00:00:00.000Z",*

```
          "key": 691200000,*
```
```
          "doc_count": 6,*
```
```
          "new_users": {*
```
```
             "doc_count": 0,*
```
```
             "unique_users": {*
```
```
                "value": 0*
```
```
             }*
```
```
          }*
```
```
       },*
```
```
  ......*
```

What I want to happen is apply min_doc_count on the most nested
sub-aggregation such that I don't see zero values for "unique_users" (in
this case) returned.

The issue is that min_doc_count can't be applied to my query other than the
date_histogram at the top level.

Does the ES query language support something like this? Any know
workarounds?

Thanks,
George

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