In Cloud Spanner SQL, an array is an ordered list consisting of zero or more
values of the same data type. You can construct arrays of simple data types,
such as INT64, and complex data types, such as STRUCTs. The current
exception to this is the ARRAY data type: arrays of arrays
are not supported. Arrays can include
NULL values.
With Cloud Spanner SQL, you can construct array literals,
build arrays from subqueries using the
ARRAY function,
and aggregate values into an array using the
ARRAY_AGG
function.
You can combine arrays using functions like
ARRAY_CONCAT(), and convert arrays to strings using ARRAY_TO_STRING().
Constructing arrays
Using array literals
You can build an array literal in Cloud Spanner SQL using brackets ([ and
]). Each element in an array is separated by a comma.
SELECT [1, 2, 3] as numbers;
SELECT ["apple", "pear", "orange"] as fruit;
SELECT [true, false, true] as booleans;
You can also create arrays from any expressions that have compatible types. For example:
SELECT [a, b, c]
FROM
(SELECT 5 AS a,
37 AS b,
406 AS c);
SELECT [a, b, c]
FROM
(SELECT CAST(5 AS INT64) AS a,
CAST(37 AS FLOAT64) AS b,
406 AS c);
Notice that the second example contains three expressions: one that returns an
INT64, one that returns a FLOAT64, and one that
declares a literal. This expression works because all three expressions share
FLOAT64 as a supertype.
To declare a specific data type for an array, use angle
brackets (< and >). For example:
SELECT ARRAY<FLOAT64>[1, 2, 3] as floats;
Arrays of most data types, such as INT64 or STRING, don't require
that you declare them first.
SELECT [1, 2, 3] as numbers;
You can write an empty array of a specific type using ARRAY<type>[]. You can
also write an untyped empty array using [], in which case Cloud Spanner SQL
attempts to infer the array type from the surrounding context. If
Cloud Spanner SQL cannot infer a type, the default type ARRAY<INT64> is used.
Using generated values
You can also construct an ARRAY with generated values.
Generating arrays of integers
GENERATE_ARRAY
generates an array of values from a starting and ending value and a step value.
For example, the following query generates an array that contains all of the odd
integers from 11 to 33, inclusive:
SELECT GENERATE_ARRAY(11, 33, 2) AS odds;
+--------------------------------------------------+
| odds |
+--------------------------------------------------+
| [11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33] |
+--------------------------------------------------+
You can also generate an array of values in descending order by giving a negative step value:
SELECT GENERATE_ARRAY(21, 14, -1) AS countdown;
+----------------------------------+
| countdown |
+----------------------------------+
| [21, 20, 19, 18, 17, 16, 15, 14] |
+----------------------------------+
Generating arrays of dates
GENERATE_DATE_ARRAY
generates an array of DATEs from a starting and ending DATE and a step
INTERVAL.
You can generate a set of DATE values using GENERATE_DATE_ARRAY. For
example, this query returns the current DATE and the following
DATEs at 1 WEEK intervals up to and including a later DATE:
SELECT
GENERATE_DATE_ARRAY('2017-11-21', '2017-12-31', INTERVAL 1 WEEK)
AS date_array;
+--------------------------------------------------------------------------+
| date_array |
+--------------------------------------------------------------------------+
| [2017-11-21, 2017-11-28, 2017-12-05, 2017-12-12, 2017-12-19, 2017-12-26] |
+--------------------------------------------------------------------------+
Accessing array elements
Consider the following table, sequences:
CREATE TABLE sequences (
id INT64 NOT NULL,
some_numbers ARRAY<INT64> NOT NULL
) PRIMARY KEY(id);
Assume the table is populated with the following rows:
+----+---------------------+
| id | some_numbers |
+----+---------------------+
| 1 | [0, 1, 1, 2, 3, 5] |
| 2 | [2, 4, 8, 16, 32] |
| 3 | [5, 10] |
+----+---------------------+
This table contains the column some_numbers of the ARRAY data type.
To access elements from the arrays in this column, you must specify which type
of indexing you want to use: either
OFFSET,
for zero-based indexes, or
ORDINAL,
for one-based indexes.
WITH sequences AS
(SELECT [0, 1, 1, 2, 3, 5] AS some_numbers
UNION ALL SELECT [2, 4, 8, 16, 32] AS some_numbers
UNION ALL SELECT [5, 10] AS some_numbers)
SELECT some_numbers,
some_numbers[OFFSET(1)] AS offset_1,
some_numbers[ORDINAL(1)] AS ordinal_1
FROM sequences;
+--------------------+----------+-----------+
| some_numbers | offset_1 | ordinal_1 |
+--------------------+----------+-----------+
| [0, 1, 1, 2, 3, 5] | 1 | 0 |
| [2, 4, 8, 16, 32] | 4 | 2 |
| [5, 10] | 10 | 5 |
+--------------------+----------+-----------+
This query shows how to use OFFSET() and ORDINAL():
SELECT some_numbers,
some_numbers[OFFSET(1)] AS offset_1,
some_numbers[ORDINAL(1)] AS ordinal_1
FROM sequences;
+---------------+----------+-----------+
| some_numbers | offset_1 | ordinal_1 |
+---------------+----------+-----------+
| [0,1,1,2,3,5] | 1 | 0 |
+---------------+----------+-----------+
| [2,4,8,16,32] | 4 | 2 |
+---------------+----------+-----------+
| [5,10] | 10 | 5 |
+---------------+----------+-----------+
Finding lengths
The ARRAY_LENGTH() function returns the length of an array.
WITH sequences AS
(SELECT [0, 1, 1, 2, 3, 5] AS some_numbers
UNION ALL SELECT [2, 4, 8, 16, 32] AS some_numbers
UNION ALL SELECT [5, 10] AS some_numbers)
SELECT some_numbers,
ARRAY_LENGTH(some_numbers) AS len
FROM sequences;
+--------------------+--------+
| some_numbers | len |
+--------------------+--------+
| [0, 1, 1, 2, 3, 5] | 6 |
| [2, 4, 8, 16, 32] | 5 |
| [5, 10] | 2 |
+--------------------+--------+
Here's an example query, assuming the same definition of the sequences table
as above, with the same sample rows:
SELECT some_numbers,
ARRAY_LENGTH(some_numbers) AS len
FROM sequences;
+---------------+------+
| some_numbers | len |
+---------------+------+
| [0,1,1,2,3,5] | 6 |
+---------------+------+
| [2,4,8,16,32] | 5 |
+---------------+------+
| [5,10] | 2 |
+---------------+------+
Convert elements in an array to rows in a table
To convert an ARRAY into a set of rows, also known as "flattening," use the
UNNEST
operator. UNNEST takes an ARRAY and returns a table with a single row for
each element in the ARRAY.
Because UNNEST destroys the order of the ARRAY elements, you may
wish to restore order to the table. To do so, use the optional WITH OFFSET
clause to return an additional column with the offset for each array element,
then use the ORDER BY clause to order the rows by their offset.
Example
SELECT *
FROM UNNEST(['foo', 'bar', 'baz', 'qux', 'corge', 'garply', 'waldo', 'fred'])
AS element
WITH OFFSET AS offset
ORDER BY offset;
+----------+--------+
| element | offset |
+----------+--------+
| foo | 0 |
| bar | 1 |
| baz | 2 |
| qux | 3 |
| corge | 4 |
| garply | 5 |
| waldo | 6 |
| fred | 7 |
+----------+--------+
To flatten an entire column of ARRAYs while preserving the values
of the other columns in each row, use a
CROSS JOIN
to join the table containing the ARRAY column to the UNNEST output of that
ARRAY column.
This is a correlated cross join: the UNNEST operator references the column of
ARRAYs from each row in the source table, which appears previously in the
FROM clause. For each row N in the source table, UNNEST flattens the
ARRAY from row N into a set of rows containing the ARRAY elements, and
then the CROSS JOIN joins this new set of rows with the single row N from
the source table.
Example
The following example uses UNNEST
to return a row for each element in the array column. Because of the
CROSS JOIN, the id column contains the id values for the row in
sequences that contains each number.
WITH sequences AS
(SELECT 1 AS id, [0, 1, 1, 2, 3, 5] AS some_numbers
UNION ALL SELECT 2 AS id, [2, 4, 8, 16, 32] AS some_numbers
UNION ALL SELECT 3 AS id, [5, 10] AS some_numbers)
SELECT id, flattened_numbers
FROM sequences
CROSS JOIN UNNEST(sequences.some_numbers) AS flattened_numbers;
+------+-------------------+
| id | flattened_numbers |
+------+-------------------+
| 1 | 0 |
| 1 | 1 |
| 1 | 1 |
| 1 | 2 |
| 1 | 3 |
| 1 | 5 |
| 2 | 2 |
| 2 | 4 |
| 2 | 8 |
| 2 | 16 |
| 2 | 32 |
| 3 | 5 |
| 3 | 10 |
+------+-------------------+
Note that for correlated cross joins the UNNEST operator is optional and the
CROSS JOIN can be expressed as a comma-join. Using this shorthand notation,
the above example becomes:
WITH sequences AS
(SELECT 1 AS id, [0, 1, 1, 2, 3, 5] AS some_numbers
UNION ALL SELECT 2 AS id, [2, 4, 8, 16, 32] AS some_numbers
UNION ALL SELECT 3 AS id, [5, 10] AS some_numbers)
SELECT id, flattened_numbers
FROM sequences, sequences.some_numbers AS flattened_numbers;
+------+-------------------+
| id | flattened_numbers |
+------+-------------------+
| 1 | 0 |
| 1 | 1 |
| 1 | 1 |
| 1 | 2 |
| 1 | 3 |
| 1 | 5 |
| 2 | 2 |
| 2 | 4 |
| 2 | 8 |
| 2 | 16 |
| 2 | 32 |
| 3 | 5 |
| 3 | 10 |
+------+-------------------+
Querying nested arrays
If a table contains an ARRAY of STRUCTs, you can
flatten the ARRAY to query the fields of the STRUCT.
You can also flatten ARRAY type fields of STRUCT values.
Querying STRUCT elements in an ARRAY
The following example uses UNNEST with CROSS JOIN to flatten an ARRAY of
STRUCTs.
SELECT race,
participant.name,
participant.splits
FROM
(SELECT "800M" AS race,
[STRUCT("Rudisha" as name, [23.4, 26.3, 26.4, 26.1] AS splits),
STRUCT("Makhloufi" AS name, [24.5, 25.4, 26.6, 26.1] AS splits),
STRUCT("Murphy" AS name, [23.9, 26.0, 27.0, 26.0] AS splits),
STRUCT("Bosse" AS name, [23.6, 26.2, 26.5, 27.1] AS splits),
STRUCT("Rotich" AS name, [24.7, 25.6, 26.9, 26.4] AS splits),
STRUCT("Lewandowski" AS name, [25.0, 25.7, 26.3, 27.2] AS splits),
STRUCT("Kipketer" AS name, [23.2, 26.1, 27.3, 29.4] AS splits),
STRUCT("Berian" AS name, [23.7, 26.1, 27.0, 29.3] as splits)]
AS participants
) AS r
CROSS JOIN UNNEST(r.participants) AS participant;
+------+-------------+-----------------------+
| race | name | splits |
+------+-------------+-----------------------+
| 800M | Rudisha | [23.4,26.3,26.4,26.1] |
+------+-------------+-----------------------+
| 800M | Makhloufi | [24.5,25.4,26.6,26.1] |
+------+-------------+-----------------------+
| 800M | Murphy | [23.9,26,27,26] |
+------+-------------+-----------------------+
| 800M | Bosse | [23.6,26.2,26.5,27.1] |
+------+-------------+-----------------------+
| 800M | Rotich | [24.7,25.6,26.9,26.4] |
+------+-------------+-----------------------+
| 800M | Lewandowski | [25,25.7,26.3,27.2] |
+------+-------------+-----------------------+
| 800M | Kipketer | [23.2,26.1,27.3,29.4] |
+------+-------------+-----------------------+
| 800M | Berian | [23.7,26.1,27,29.3] |
+------+-------------+-----------------------+
You can find specific information from repeated fields. For example, the following query returns the fastest racer in an 800M race.
This example does not involve flattening an array, but does represent a common way to get information from a repeated field.
Example
WITH races AS (
SELECT "800M" AS race,
[STRUCT("Rudisha" as name, [23.4, 26.3, 26.4, 26.1] as splits),
STRUCT("Makhloufi" as name, [24.5, 25.4, 26.6, 26.1] as splits),
STRUCT("Murphy" as name, [23.9, 26.0, 27.0, 26.0] as splits),
STRUCT("Bosse" as name, [23.6, 26.2, 26.5, 27.1] as splits),
STRUCT("Rotich" as name, [24.7, 25.6, 26.9, 26.4] as splits),
STRUCT("Lewandowski" as name, [25.0, 25.7, 26.3, 27.2] as splits),
STRUCT("Kipketer" as name, [23.2, 26.1, 27.3, 29.4] as splits),
STRUCT("Berian" as name, [23.7, 26.1, 27.0, 29.3] as splits)]
AS participants)
SELECT
race,
(SELECT name
FROM UNNEST(participants)
ORDER BY (
SELECT SUM(duration)
FROM UNNEST(splits) AS duration) ASC
LIMIT 1) AS fastest_racer
FROM races;
+------+---------------+
| race | fastest_racer |
+------+---------------+
| 800M | Rudisha |
+------+---------------+
SELECT race,
(SELECT name
FROM UNNEST(participants)
ORDER BY (
SELECT SUM(duration)
FROM UNNEST(splits) AS duration) ASC
LIMIT 1) AS fastest_racer
FROM
(SELECT "800M" AS race,
[STRUCT("Rudisha" as name, [23.4, 26.3, 26.4, 26.1] AS splits),
STRUCT("Makhloufi" AS name, [24.5, 25.4, 26.6, 26.1] AS splits),
STRUCT("Murphy" AS name, [23.9, 26.0, 27.0, 26.0] AS splits),
STRUCT("Bosse" AS name, [23.6, 26.2, 26.5, 27.1] AS splits),
STRUCT("Rotich" AS name, [24.7, 25.6, 26.9, 26.4] AS splits),
STRUCT("Lewandowski" AS name, [25.0, 25.7, 26.3, 27.2] AS splits),
STRUCT("Kipketer" AS name, [23.2, 26.1, 27.3, 29.4] AS splits),
STRUCT("Berian" AS name, [23.7, 26.1, 27.0, 29.3] as splits)]
AS participants
) AS r;
+------+---------------+
| race | fastest_racer |
+------+---------------+
| 800M | Rudisha |
+------+---------------+
Querying ARRAY-type fields in a STRUCT
You can also get information from nested repeated fields. For example, the following statement returns the runner who had the fastest lap in an 800M race.
WITH races AS (
SELECT "800M" AS race,
[STRUCT("Rudisha" as name, [23.4, 26.3, 26.4, 26.1] as splits),
STRUCT("Makhloufi" as name, [24.5, 25.4, 26.6, 26.1] as splits),
STRUCT("Murphy" as name, [23.9, 26.0, 27.0, 26.0] as splits),
STRUCT("Bosse" as name, [23.6, 26.2, 26.5, 27.1] as splits),
STRUCT("Rotich" as name, [24.7, 25.6, 26.9, 26.4] as splits),
STRUCT("Lewandowski" as name, [25.0, 25.7, 26.3, 27.2] as splits),
STRUCT("Kipketer" as name, [23.2, 26.1, 27.3, 29.4] as splits),
STRUCT("Berian" as name, [23.7, 26.1, 27.0, 29.3] as splits)]
AS participants)
SELECT
race,
(SELECT name
FROM UNNEST(participants),
UNNEST(splits) AS duration
ORDER BY duration ASC LIMIT 1) AS runner_with_fastest_lap
FROM races;
+------+-------------------------+
| race | runner_with_fastest_lap |
+------+-------------------------+
| 800M | Kipketer |
+------+-------------------------+
SELECT race,
(SELECT name
FROM UNNEST(participants),
UNNEST(splits) AS duration
ORDER BY duration ASC LIMIT 1) AS runner_with_fastest_lap
FROM
(SELECT "800M" AS race,
[STRUCT("Rudisha" as name, [23.4, 26.3, 26.4, 26.1] AS splits),
STRUCT("Makhloufi" AS name, [24.5, 25.4, 26.6, 26.1] AS splits),
STRUCT("Murphy" AS name, [23.9, 26.0, 27.0, 26.0] AS splits),
STRUCT("Bosse" AS name, [23.6, 26.2, 26.5, 27.1] AS splits),
STRUCT("Rotich" AS name, [24.7, 25.6, 26.9, 26.4] AS splits),
STRUCT("Lewandowski" AS name, [25.0, 25.7, 26.3, 27.2] AS splits),
STRUCT("Kipketer" AS name, [23.2, 26.1, 27.3, 29.4] AS splits),
STRUCT("Berian" AS name, [23.7, 26.1, 27.0, 29.3] as splits)]
AS participants
) AS r;
+------+-------------------------+
| race | runner_with_fastest_lap |
+------+-------------------------+
| 800M | Kipketer |
+------+-------------------------+
Notice that the preceding query uses the comma operator (,) to perform an
implicit CROSS JOIN. It is equivalent to the following example, which uses
an explicit CROSS JOIN.
WITH races AS (
SELECT "800M" AS race,
[STRUCT("Rudisha" as name, [23.4, 26.3, 26.4, 26.1] as splits),
STRUCT("Makhloufi" as name, [24.5, 25.4, 26.6, 26.1] as splits),
STRUCT("Murphy" as name, [23.9, 26.0, 27.0, 26.0] as splits),
STRUCT("Bosse" as name, [23.6, 26.2, 26.5, 27.1] as splits),
STRUCT("Rotich" as name, [24.7, 25.6, 26.9, 26.4] as splits),
STRUCT("Lewandowski" as name, [25.0, 25.7, 26.3, 27.2] as splits),
STRUCT("Kipketer" as name, [23.2, 26.1, 27.3, 29.4] as splits),
STRUCT("Berian" as name, [23.7, 26.1, 27.0, 29.3] as splits)]
AS participants)
SELECT
race,
(SELECT name
FROM UNNEST(participants)
CROSS JOIN UNNEST(splits) AS duration
ORDER BY duration ASC LIMIT 1) AS runner_with_fastest_lap
FROM races;
+------+-------------------------+
| race | runner_with_fastest_lap |
+------+-------------------------+
| 800M | Kipketer |
+------+-------------------------+
SELECT race,
(SELECT name
FROM UNNEST(participants)
CROSS JOIN UNNEST(splits) AS duration
ORDER BY duration ASC LIMIT 1) AS runner_with_fastest_lap
FROM
(SELECT "800M" AS race,
[STRUCT("Rudisha" as name, [23.4, 26.3, 26.4, 26.1] AS splits),
STRUCT("Makhloufi" AS name, [24.5, 25.4, 26.6, 26.1] AS splits),
STRUCT("Murphy" AS name, [23.9, 26.0, 27.0, 26.0] AS splits),
STRUCT("Bosse" AS name, [23.6, 26.2, 26.5, 27.1] AS splits),
STRUCT("Rotich" AS name, [24.7, 25.6, 26.9, 26.4] AS splits),
STRUCT("Lewandowski" AS name, [25.0, 25.7, 26.3, 27.2] AS splits),
STRUCT("Kipketer" AS name, [23.2, 26.1, 27.3, 29.4] AS splits),
STRUCT("Berian" AS name, [23.7, 26.1, 27.0, 29.3] as splits)]
AS participants
) AS r;
+------+-------------------------+
| race | runner_with_fastest_lap |
+------+-------------------------+
| 800M | Kipketer |
+------+-------------------------+
Flattening arrays with a CROSS JOIN excludes rows that have empty
or NULL arrays. If you want to include these rows, use a LEFT JOIN.
WITH races AS (
SELECT "800M" AS race,
[STRUCT("Rudisha" as name, [23.4, 26.3, 26.4, 26.1] as splits),
STRUCT("Makhloufi" as name, [24.5, 25.4, 26.6, 26.1] as splits),
STRUCT("Murphy" as name, [23.9, 26.0, 27.0, 26.0] as splits),
STRUCT("Bosse" as name, [23.6, 26.2, 26.5, 27.1] as splits),
STRUCT("Rotich" as name, [24.7, 25.6, 26.9, 26.4] as splits),
STRUCT("Lewandowski" as name, [25.0, 25.7, 26.3, 27.2] as splits),
STRUCT("Kipketer" as name, [23.2, 26.1, 27.3, 29.4] as splits),
STRUCT("Berian" as name, [23.7, 26.1, 27.0, 29.3] as splits),
STRUCT("Nathan" as name, ARRAY<FLOAT64>[] as splits),
STRUCT("David" as name, NULL as splits)]
AS participants)
SELECT
name, sum(duration) AS finish_time
FROM races, races.participants LEFT JOIN participants.splits duration
GROUP BY name;
+-------------+--------------------+
| name | finish_time |
+-------------+--------------------+
| Murphy | 102.9 |
| Rudisha | 102.19999999999999 |
| David | NULL |
| Rotich | 103.6 |
| Makhloufi | 102.6 |
| Berian | 106.1 |
| Bosse | 103.4 |
| Kipketer | 106 |
| Nathan | NULL |
| Lewandowski | 104.2 |
+-------------+--------------------+
SELECT
name, sum(duration) as duration
FROM
(SELECT "800M" AS race,
[STRUCT("Rudisha" AS name, [23.4, 26.3, 26.4, 26.1] AS splits),
STRUCT("Makhloufi" AS name, [24.5, 25.4, 26.6, 26.1] AS splits),
STRUCT("Murphy" AS name, [23.9, 26.0, 27.0, 26.0] AS splits),
STRUCT("Bosse" AS name, [23.6, 26.2, 26.5, 27.1] AS splits),
STRUCT("Rotich" AS name, [24.7, 25.6, 26.9, 26.4] AS splits),
STRUCT("Lewandowski" AS name, [25.0, 25.7, 26.3, 27.2] AS splits),
STRUCT("Kipketer" AS name, [23.2, 26.1, 27.3, 29.4] AS splits),
STRUCT("Nathan" as name, ARRAY<FLOAT64>[] as splits),
STRUCT("David" as name, NULL as splits)]
AS participants) AS races,
races.participants LEFT JOIN participants.splits duration
GROUP BY name;
+-------------+--------------------+
| name | duration |
+-------------+--------------------+
| Murphy | 102.9 |
| Rudisha | 102.19999999999999 |
| David | NULL |
| Rotich | 103.6 |
| Makhloufi | 102.6 |
| Bosse | 103.4 |
| Kipketer | 106 |
| Nathan | NULL |
| Lewandowski | 104.2 |
+-------------+--------------------+
Creating arrays from subqueries
A common task when working with arrays is turning a subquery result into an
array. In Cloud Spanner SQL, you can accomplish this using the
ARRAY() function.
For example, consider the following operation on the sequences table:
WITH sequences AS
(SELECT [0, 1, 1, 2, 3, 5] AS some_numbers
UNION ALL SELECT [2, 4, 8, 16, 32] AS some_numbers
UNION ALL SELECT [5, 10] AS some_numbers)
SELECT some_numbers,
ARRAY(SELECT x * 2
FROM UNNEST(some_numbers) AS x) AS doubled
FROM sequences;
+--------------------+---------------------+
| some_numbers | doubled |
+--------------------+---------------------+
| [0, 1, 1, 2, 3, 5] | [0, 2, 2, 4, 6, 10] |
| [2, 4, 8, 16, 32] | [4, 8, 16, 32, 64] |
| [5, 10] | [10, 20] |
+--------------------+---------------------+
SELECT some_numbers,
ARRAY(SELECT x * 2
FROM UNNEST(some_numbers) AS x) AS doubled
FROM sequences;
+---------------+----------------+
| some_numbers | doubled |
+---------------+----------------+
| [0,1,1,2,3,5] | [0,2,2,4,6,10] |
+---------------+----------------+
| [2,4,8,16,32] | [4,8,16,32,64] |
+---------------+----------------+
| [5,10] | [10,20] |
+---------------+----------------+
This example starts with a table named sequences. This table contains a column,
some_numbers, of type ARRAY<INT64>.
The query itself contains a subquery. This subquery selects each row in the
some_numbers column and uses
UNNEST to return the
array as a set of rows. Next, it multiplies each value by two, and then
recombines the rows back into an array using the ARRAY() operator.
Filtering arrays
The following example uses a WHERE clause in the ARRAY() operator's subquery
to filter the returned rows.
Note: In the following examples, the resulting rows are not ordered.
WITH sequences AS
(SELECT [0, 1, 1, 2, 3, 5] AS some_numbers
UNION ALL SELECT [2, 4, 8, 16, 32] AS some_numbers
UNION ALL SELECT [5, 10] AS some_numbers)
SELECT
ARRAY(SELECT x * 2
FROM UNNEST(some_numbers) AS x
WHERE x < 5) AS doubled_less_than_five
FROM sequences;
+------------------------+
| doubled_less_than_five |
+------------------------+
| [0, 2, 2, 4, 6] |
| [4, 8] |
| [] |
+------------------------+
SELECT
ARRAY(SELECT x * 2
FROM UNNEST(some_numbers) AS x
WHERE x < 5) AS doubled_less_than_five
FROM sequences;
+------------------------+
| doubled_less_than_five |
+------------------------+
| [0,2,2,4,6] |
+------------------------+
| [4,8] |
+------------------------+
| [] |
+------------------------+
Notice that the third row contains an empty array, because the elements in the
corresponding original row ([5, 10]) did not meet the filter requirement of
x < 5.
You can also filter arrays by using SELECT DISTINCT to return only
unique elements within an array.
WITH sequences AS
(SELECT [0, 1, 1, 2, 3, 5] AS some_numbers)
SELECT ARRAY(SELECT DISTINCT x
FROM UNNEST(some_numbers) AS x) AS unique_numbers
FROM sequences;
+-----------------+
| unique_numbers |
+-----------------+
| [0, 1, 2, 3, 5] |
+-----------------+
SELECT ARRAY(SELECT DISTINCT x
FROM UNNEST(some_numbers) AS x) AS unique_numbers
FROM sequences
WHERE id = 1;
+----------------+
| unique_numbers |
+----------------+
| [0,1,2,3,5] |
+----------------+
You can also filter rows of arrays by using the
IN keyword. This
keyword filters rows containing arrays by determining if a specific
value matches an element in the array.
WITH sequences AS
(SELECT [0, 1, 1, 2, 3, 5] AS some_numbers
UNION ALL SELECT [2, 4, 8, 16, 32] AS some_numbers
UNION ALL SELECT [5, 10] AS some_numbers)
SELECT
ARRAY(SELECT x
FROM UNNEST(some_numbers) AS x
WHERE 2 IN UNNEST(some_numbers)) AS contains_two
FROM sequences;
+--------------------+
| contains_two |
+--------------------+
| [0, 1, 1, 2, 3, 5] |
| [2, 4, 8, 16, 32] |
| [] |
+--------------------+
SELECT
ARRAY(SELECT x
FROM UNNEST(some_numbers) AS x
WHERE 2 IN UNNEST(some_numbers)) AS contains_two
FROM sequences;
+---------------+
| contains_two |
+---------------+
| [0,1,1,2,3,5] |
+---------------+
| [2,4,8,16,32] |
+---------------+
| [] |
+---------------+
Notice again that the third row contains an empty array, because the array in
the corresponding original row ([5, 10]) did not contain 2.
Scanning arrays
To check if an array contains a specific value, use the IN
operator with UNNEST. To
check if an array contains a value matching a condition, use the EXISTS
function with UNNEST.
Scanning for specific values
To scan an array for a specific value, use the IN operator with UNNEST.
Example
The following example returns true if the array contains the number 2.
SELECT 2 IN UNNEST([0, 1, 1, 2, 3, 5]) AS contains_value;
+----------------+
| contains_value |
+----------------+
| true |
+----------------+
To return the rows of a table where the array column contains a specific value,
filter the results of IN UNNEST using the WHERE clause.
Example
The following example returns the id value for the rows where the array
column contains the value 2.
WITH sequences AS
(SELECT 1 AS id, [0, 1, 1, 2, 3, 5] AS some_numbers
UNION ALL SELECT 2 AS id, [2, 4, 8, 16, 32] AS some_numbers
UNION ALL SELECT 3 AS id, [5, 10] AS some_numbers)
SELECT id AS matching_rows
FROM sequences
WHERE 2 IN UNNEST(sequences.some_numbers)
ORDER BY matching_rows;
+---------------+
| matching_rows |
+---------------+
| 1 |
| 2 |
+---------------+
Scanning for values that satisfy a condition
To scan an array for values that match a condition, use UNNEST to return a
table of the elements in the array, use WHERE to filter the resulting table in
a subquery, and use EXISTS to check if the filtered table contains any rows.
Example
The following example returns the id value for the rows where the array
column contains values greater than 5.
WITH sequences AS
(SELECT 1 AS id, [0, 1, 1, 2, 3, 5] AS some_numbers
UNION ALL SELECT 2 AS id, [2, 4, 8, 16, 32] AS some_numbers
UNION ALL SELECT 3 AS id, [5, 10] AS some_numbers)
SELECT id AS matching_rows FROM sequences
WHERE EXISTS (SELECT *
FROM UNNEST(some_numbers) AS x
WHERE x > 5);
+---------------+
| matching_rows |
+---------------+
| 2 |
| 3 |
+---------------+
Scanning for STRUCT field values that satisfy a condition
To search an array of STRUCTs for a field whose value matches a condition, use
UNNEST to return a table with a column for each STRUCT field, then filter
non-matching rows from the table using WHERE EXISTS.
Example
The following example returns the rows where the array column contains a
STRUCT whose field b has a value greater than 3.
WITH sequences AS
(SELECT 1 AS id, [STRUCT(0 AS a, 1 AS b)] AS some_numbers
UNION ALL SELECT 2 AS id, [STRUCT(2 AS a, 4 AS b)] AS some_numbers
UNION ALL SELECT 3 AS id, [STRUCT(5 AS a, 3 AS b), STRUCT (7 AS a, 4 AS b)]
AS some_numbers)
SELECT id AS matching_rows
FROM sequences
WHERE EXISTS (SELECT 1
FROM UNNEST(some_numbers)
WHERE b > 3);
+---------------+
| matching_rows |
+---------------+
| 2 |
| 3 |
+---------------+
Arrays and aggregation
With Cloud Spanner SQL, you can aggregate values into an array using
ARRAY_AGG().
WITH fruits AS
(SELECT "apple" AS fruit
UNION ALL SELECT "pear" AS fruit
UNION ALL SELECT "banana" AS fruit)
SELECT ARRAY_AGG(fruit) AS fruit_basket
FROM fruits;
+-----------------------+
| fruit_basket |
+-----------------------+
| [apple, pear, banana] |
+-----------------------+
Consider the following table, fruits:
CREATE TABLE fruits (
fruit STRING(MAX),
id INT64 NOT NULL
) PRIMARY KEY(id);
Assume the table is populated with the following data:
+----+--------------+
| id | fruit |
+----+--------------+
| 1 | "apple" |
| 2 | "pear" |
| 3 | "banana" |
+----+--------------+
This query shows how to use ARRAY_AGG():
SELECT ARRAY_AGG(fruit) AS fruit_basket
FROM fruits;
+---------------------+
| fruit_basket |
+---------------------+
| [apple,pear,banana] |
+---------------------+
The array returned by ARRAY_AGG() is in an arbitrary order, since the order in
which the function concatenates values is not guaranteed.
You can also apply aggregate functions such as SUM() to the elements in an
array. For example, the following query returns the sum of array elements for
each row of the sequences table.
WITH sequences AS
(SELECT [0, 1, 1, 2, 3, 5] AS some_numbers
UNION ALL SELECT [2, 4, 8, 16, 32] AS some_numbers
UNION ALL SELECT [5, 10] AS some_numbers)
SELECT some_numbers,
(SELECT SUM(x)
FROM UNNEST(s.some_numbers) x) AS sums
FROM sequences s;
+--------------------+------+
| some_numbers | sums |
+--------------------+------+
| [0, 1, 1, 2, 3, 5] | 12 |
| [2, 4, 8, 16, 32] | 62 |
| [5, 10] | 15 |
+--------------------+------+
SELECT some_numbers,
(SELECT SUM(x)
FROM UNNEST(s.some_numbers) x) AS sums
FROM sequences s;
+---------------+------+
| some_numbers | sums |
+---------------+------+
| [0,1,1,2,3,5] | 12 |
+---------------+------+
| [2,4,8,16,32] | 62 |
+---------------+------+
| [5,10] | 15 |
+---------------+------+
Cloud Spanner SQL also supports an aggregate function, ARRAY_CONCAT_AGG(),
which concatenates the elements of an array column across rows.
WITH aggregate_example AS
(SELECT [1,2] AS numbers
UNION ALL SELECT [3,4] AS numbers
UNION ALL SELECT [5, 6] AS numbers)
SELECT ARRAY_CONCAT_AGG(numbers) AS count_to_six_agg
FROM aggregate_example;
+--------------------------------------------------+
| count_to_six_agg |
+--------------------------------------------------+
| [1, 2, 3, 4, 5, 6] |
+--------------------------------------------------+
Note: The array returned by ARRAY_CONCAT_AGG() is
non-deterministic, since the order in which the function concatenates values is
not guaranteed.
Converting arrays to strings
The ARRAY_TO_STRING() function allows you to convert an ARRAY<STRING> to a
single STRING value or an ARRAY<BYTES> to a single BYTES value where the
resulting value is the ordered concatenation of the array elements.
The second argument is the separator that the function will insert between inputs to produce the output; this second argument must be of the same type as the elements of the first argument.
Example:
WITH greetings AS
(SELECT ["Hello", "World"] AS greeting)
SELECT ARRAY_TO_STRING(greeting, " ") AS greetings
FROM greetings;
+-------------+
| greetings |
+-------------+
| Hello World |
+-------------+
The optional third argument takes the place of NULL values in the input
array.
If you omit this argument, then the function ignores
NULLarray elements.If you provide an empty string, the function inserts a separator for
NULLarray elements.
Example:
SELECT
ARRAY_TO_STRING(arr, ".", "N") AS non_empty_string,
ARRAY_TO_STRING(arr, ".", "") AS empty_string,
ARRAY_TO_STRING(arr, ".") AS omitted
FROM (SELECT ["a", NULL, "b", NULL, "c", NULL] AS arr);
+------------------+--------------+---------+
| non_empty_string | empty_string | omitted |
+------------------+--------------+---------+
| a.N.b.N.c.N | a..b..c. | a.b.c |
+------------------+--------------+---------+
Combining arrays
In some cases, you might want to combine multiple arrays into a single array.
You can accomplish this using the ARRAY_CONCAT() function.
SELECT ARRAY_CONCAT([1, 2], [3, 4], [5, 6]) as count_to_six;
+--------------------------------------------------+
| count_to_six |
+--------------------------------------------------+
| [1, 2, 3, 4, 5, 6] |
+--------------------------------------------------+
Building arrays of arrays
Cloud Spanner SQL does not support building
arrays of arrays
directly. Instead, you must create an array of structs, with each struct
containing a field of type ARRAY. To illustrate this, consider the following
points table:
CREATE TABLE points (
point ARRAY<INT64>,
id INT64 NOT NULL
) PRIMARY KEY(id);
Assume the table is populated with the following rows:
+----+----------+
| id | point |
+----+----------+
| 1 | [1, 5] |
| 2 | [2, 8] |
| 3 | [3, 7] |
| 4 | [4, 1] |
| 5 | [5, 7] |
+----+----------+
Now, let's say you wanted to create an array consisting of each point in the
points table. To accomplish this, wrap the array returned from each row in a
STRUCT, as shown below.
WITH points AS
(SELECT [1, 5] as point
UNION ALL SELECT [2, 8] as point
UNION ALL SELECT [3, 7] as point
UNION ALL SELECT [4, 1] as point
UNION ALL SELECT [5, 7] as point)
SELECT ARRAY(
SELECT STRUCT(point)
FROM points)
AS coordinates;
+-------------------+
| coordinates |
+-------------------+
| [{point: [1,5]}, |
| {point: [2,8]}, |
| {point: [5,7]}, |
| {point: [3,7]}, |
| {point: [4,1]}] |
+--------------------+
SELECT ARRAY(
SELECT STRUCT(point)
FROM points)
AS coordinates;
+--------------+
| coordinates |
+--------------+
| point: [1,5] |
| point: [2,8] |
| point: [3,7] |
| point: [4,1] |
| point: [5,7] |
+--------------+