GROUP BY Clause

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    The GROUP BY clause arranges aggregate values into groups, based on one or more fields.

    Purpose

    Use the GROUP BY clause to arrange aggregate values into groups of one or more fields. This GROUP BY clause follows the WHERE clause and can contain the optional GROUP AS, LETTING, and HAVING clauses.

    Syntax

    group-by-clause ::= 'GROUP' 'BY' group-term ( ',' group-term )*
                        group-as-clause? letting-clause? having-clause? 
                        | letting-clause
    Syntax diagram
    group-term

    Group Term

    group-as-clause

    GROUP AS Clause

    letting-clause

    LETTING Clause

    having-clause

    HAVING Clause

    Group Term

    group-term ::= expr ( ('AS')? alias )?
    Syntax diagram

    At least one group term is required.

    expr

    String or expression representing an aggregate function or field to group together.

    alias

    Assigns another name to the group term. For details, see AS Clause.

    Assigning an alias to the group term is optional. If you assign an alias, the AS keyword may be omitted.

    GROUP AS Clause

    group-as-clause ::= 'GROUP AS' alias
    Syntax diagram

    JSON is a hierarchical format, and a fully featured JSON query language needs to be able to produce hierarchies of its own, with computed data at every level of the hierarchy. The key feature of SQL++ that makes this possible is the GROUP AS clause.

    A query can include a GROUP AS clause only if it has a GROUP BY clause. The effect of a GROUP BY clause is to hide the original objects in each group, exposing only the grouping expressions and special aggregation functions on the non-grouping fields. The purpose of the GROUP AS clause is to make the original objects in the group visible to subsequent clauses. As a result, the query can generate output data both for the group as a whole and for the individual objects inside the group.

    For each group, GROUP AS preserves all of the objects in the group, as they were before grouping, in an array. The name of the array is the alias in the GROUP AS clause. You can then use the array name in the FROM clause of a subquery to process and return the individual objects in the group.

    In the GROUP AS array, each object is wrapped in an outer object that gives it the name of the alias specified in the FROM clause, or its implicit alias if no alias was specified. Identifying the objects with their aliases in this way helps avoid ambiguity.

    For example, a query has a FROM clause of FROM airline as a, route as r, and a GROUP AS clause of GROUP AS g. The result of the GROUP AS clause is an array of objects, each of which contains both an airline object and a route object. These objects might contain field names that are the same, such as "id". The resulting array g identifies each of the objects it contains separately with its alias, as follows:

    [
    	{
    		"a": { an original airline object },
    		"r": { an original route object }
    	},
    	{
    		"a": { an original airline object },
    		"r": { an original route object }
    	},
    	...
    ]

    LETTING Clause

    letting-clause ::= 'LETTING' alias '=' expr ( ',' alias '=' expr )*
    Syntax diagram

    [Optional] Stores the result of a sub-expression in order to use it in subsequent clauses.

    alias

    String or expression representing the name of the clause to be referred to.

    expr

    String or expression representing the value of the LETTING alias variable.

    HAVING Clause

    having-clause ::= 'HAVING' cond
    Syntax diagram

    [Optional] To return items where aggregate values meet the specified conditions.

    cond

    String or expression representing the clause of aggregate values.

    Limitations

    GROUP BY works only on a group key or aggregate function.

    A query needs a predicate on a leading index key to ensure that the optimizer can select a secondary index for the query. Without a matching predicate, the query will use the primary index. The simplest predicate is WHERE leading-index-key IS NOT MISSING. This is usually only necessary in queries which do not otherwise have a WHERE clause; for example, some GROUP BY and aggregate queries. For more details, refer to Index Selection.

    Examples

    Example 1. Group the unique landmarks by city and list the top 4 cities with the most landmarks in descending order
    SELECT city City, COUNT(DISTINCT name) LandmarkCount
    FROM landmark
    GROUP BY city
    ORDER BY LandmarkCount DESC
    LIMIT 4;
    Results
    [
      {
        "City": "San Francisco",
        "LandmarkCount": 797
      },
      {
        "City": "London",
        "LandmarkCount": 443
      },
      {
        "City": "Los Angeles",
        "LandmarkCount": 284
      },
      {
        "City": "San Diego",
        "LandmarkCount": 197
      }
    ]
    Example 2. Group airlines by country and also list all airlines for each country
    SELECT country, COUNT(*) as count, g as group_docs
    FROM airline a
    GROUP BY country
    GROUP AS g;
    Results
    [
        {
          "country": "United Kingdom",
          "count": 39,
          "group_docs": [
            {
              "a": {
                "id": 10642,
                "type": "airline",
                "name": "Jc royal.britannica",
                "iata": null,
                "icao": "JRB",
                "callsign": null,
                "country": "United Kingdom"
              }
            },
            {
              "a": {
                "id": 112,
                "type": "airline",
                "name": "Astraeus",
                "iata": "5W",
                "icao": "AEU",
                "callsign": "FLYSTAR",
                "country": "United Kingdom"
              }
            },
            ...
          ]
        },
        {
          "country": "United States",
          "count": 127,
          "group_docs": [
            {
              "a": {
                "id": 10,
                "type": "airline",
                "name": "40-Mile Air",
                "iata": "Q5",
                "icao": "MLA",
                "callsign": "MILE-AIR",
                "country": "United States"
              }
            },
            {
              "a": {
                "id": 10123,
                "type": "airline",
                "name": "Texas Wings",
                "iata": "TQ",
                "icao": "TXW",
                "callsign": "TXW",
                "country": "United States"
              }
            },
            ...
          ]
        },
        {
          "country": "France",
          "count": 17,
          "group_docs": [
            {
              "a": {
                "id": 1191,
                "type": "airline",
                "name": "Air Austral",
                "iata": "UU",
                "icao": "REU",
                "callsign": "REUNION",
                "country": "France"
              }
            },
            {
              "a": {
                "id": 1203,
                "type": "airline",
                "name": "Airlinair",
                "iata": "A5",
                "icao": "RLA",
                "callsign": "AIRLINAIR",
                "country": "France"
              }
            },
            ...
          ]
        }
      ]
    Example 3. Group airlines by country and also list the name and id of two airlines from each country
    SELECT country, COUNT(*) as count, 
        (SELECT g1.airline.name, g1.airline.id 
        FROM g g1 
        LIMIT 2) 
    AS airlines 
    FROM airline
    GROUP BY country
    GROUP AS g;
    Results
    [
            {
                "country": "United States",
                "count": 127,
                "airlines": [
                    {
                        "id": 10,
                        "name": "40-Mile Air"
                    },
                    {
                        "id": 10123,
                        "name": "Texas Wings"
                    }
                ]
            },
            {
                "country": "France",
                "count": 21,
                "airlines": [
                    {
                        "id": 1191,
                        "name": "Air Austral"
                    },
                    {
                        "id": 1203,
                        "name": "Airlinair"
                    }
                ]
            },
            {
                "country": "United Kingdom",
                "count": 39,
                "airlines": [
                    {
                        "id": 10642,
                        "name": "Jc royal.britannica"
                    },
                    {
                        "id": 112,
                        "name": "Astraeus"
                    }
                ]
            }
        ] 
    Example 4. Use LETTING to find cities that have a minimum number of things to see
    SELECT city City, COUNT(DISTINCT name) LandmarkCount
    FROM landmark
    GROUP BY city
    LETTING MinimumThingsToSee = 400
    HAVING COUNT(DISTINCT name) > MinimumThingsToSee;
    Results
    [
      {
        "City": "London",
        "LandmarkCount": 443
      },
      {
        "City": "San Francisco",
        "LandmarkCount": 797
      }
    ]
    Example 5. Use HAVING to specify cities that have more than 180 landmarks
    SELECT city City, COUNT(DISTINCT name) LandmarkCount
    FROM landmark
    GROUP BY city
    HAVING COUNT(DISTINCT name) > 180;
    Results
    [
      {
        "City": "London",
        "LandmarkCount": 443
      },
      {
        "City": "Los Angeles",
        "LandmarkCount": 284
      },
      {
        "City": "San Francisco",
        "LandmarkCount": 797
      },
      {
        "City": "San Diego",
        "LandmarkCount": 197
      }
    ]
    The above HAVING clause must use the aggregate function COUNT instead of its alias LandmarkCount.
    Example 6. Use HAVING to specify landmarks that begin with an "S" or higher
    SELECT city City, COUNT(DISTINCT name) LandmarkCount
    FROM landmark
    GROUP BY city
    HAVING city > "S"
    ORDER BY city;
    Results
    [
      {
        "City": "Sacramento",
        "LandmarkCount": 2
      },
      {
        "City": "Saint Albans",
        "LandmarkCount": 5
      },
      {
        "City": "Saint Andrews",
        "LandmarkCount": 13
      },
      {
        "City": "Saint Annes Head",
        "LandmarkCount": 1
      },
    // ...

    (execution: 1s docs: 138)

    Example 7. Using WHERE yields the same results as HAVING, however, WHERE is faster
    SELECT city City, COUNT(DISTINCT name) LandmarkCount
    FROM landmark
    WHERE city > "S"
    GROUP BY city
    ORDER BY city;
    Results
    [
      {
        "City": "Sacramento",
        "LandmarkCount": 2
      },
      {
        "City": "Saint Albans",
        "LandmarkCount": 5
      },
      {
        "City": "Saint Andrews",
        "LandmarkCount": 13
      },
      {
        "City": "Saint Annes Head",
        "LandmarkCount": 1
      },
    // ...

    (execution: 480.2ms docs: 138)

    The WHERE clause is faster because WHERE gets processed before any GROUP BY and doesn’t have access to aggregated values. HAVING gets processed after GROUP BY and is used to constrain the resultset to only those with aggregated values.
    Example 8. Using an alias for a group term
    SELECT Hemisphere, COUNT(DISTINCT name) AS LandmarkCount
    FROM landmark AS l
    GROUP BY CASE
      WHEN l.geo.lon <0 THEN "West"
      ELSE "East"
    END AS Hemisphere;
    Results
    [
      {
        "Hemisphere": "East",
        "LandmarkCount": 459
      },
      {
        "Hemisphere": "West",
        "LandmarkCount": 3885
      }
    ]
    The CASE expression categorizes each landmark into the Western hemisphere if its longitude is negative, or the Eastern hemisphere otherwise. The alias in the GROUP BY clause enables you to refer to the CASE expression in the SELECT clause.