Max
Not available in this language yet.
Max( value_1, value_2, ... )
Max( value_1, value_2, ... )
Max( value_1, value_2, ... )
Not available in this language yet.
Max( value_1, value_2, ... )
Description
The Max
function returns the largest value in a list of values.
The run time of To work around this, you may specify a larger query timeout via the driver that you are using. |
Parameters
Argument | Type | Definition and Requirements |
---|---|---|
|
List of values. |
A single value or a list of values. |
Examples
The query below executes an array of independent max operations and returns the results in an array. The result array position matches the execution array position. The top operation in the execution array, max of the values 1, 5, and 22, returns a long value of 22 in the top position of the result array.
Not available in this language yet.
result, err := client.Query(
f.Arr{
f.Max(1, 5, 22),
f.Max(1, 0, 3, -1),
f.Max(-1, 12, 3, -1),
f.Max(10)})
[22 3 12 10]
System.out.println(
client.query(
Arr(
Max(Value(1), Value(5), Value(22)),
Max(Value(1), Value(0), Value(3), Value(-1)),
Max(Value(-1), Value(12), Value(3), Value(-1)),
Max(Value(10))
)
).get()
);
[
22,
3,
12,
10
]
client.query([
q.Max(1, 5, 22),
q.Max(1, 0, 3, -1),
q.Max(-1, 12, 3, -1),
q.Max(10),
]).then((ret) => console.log(ret))
[ 22, 3, 12, 10 ]
Not available in this language yet.
println(Await.result(
client.query(
Arr(
Max(1, 5, 22),
Max(1, 0, 3, -1),
Max(-1, 12, 3, -1),
Max(10)
)
),
5.seconds
))
[22, 3, 12, 10]
The following query uses the same approach as the previous query to
demonstrate using Max
with various types of values:
Not available in this language yet.
result, err := client.Query(
f.Arr{
f.Max("A", "B", "C", "D"),
f.Max(f.Time("1970-01-01T00:00:00Z"), f.Time("1980-01-01T00:00:00Z")),
f.Max(f.Date("1970-01-01"), f.Date("1930-01-01")),
f.Max("A", 1),
f.Max(true, false),
f.Max(f.Obj{"x": 10}, f.Obj{"x": 11}),
f.Max(f.Arr{"A"}, f.Arr{"B"}, f.Arr{"C"}),
f.Max(f.Arr{"X"}, f.Arr{"A", "B"})})
[D {0 62451129600 <nil>} {0 62135596800 <nil>} A true map[x:11] [C] [X]]
System.out.println(
client.query(
Arr(
Max(Value("A"), Value("B"), Value("C"), Value("D")),
Max(Time("1970-01-01T00:00:00Z"), Time("1980-01-01T00:00:00Z")),
Max(Date("1970-01-01"), Date("1930-01-01")),
Max(Value("A"), Value(1)),
Max(Value(true), Value(false)),
Max(Obj("x", Value(10)), Obj("x", Value(11))),
Max(Arr(Value("A")), Arr(Value("B")), Arr(Value("C"))),
Max(Arr(Value("X")), Arr(Value("A"), Value("B")))
)
).get()
);
["D", 1980-01-01T00:00:00Z, 1970-01-01, "A", true, {x: 11}, ["C"], ["X"]]
client.query([
q.Max('A', 'B', 'C', 'D'),
q.Max(q.Time('1970-01-01T00:00:00Z'), q.Time('1980-01-01T00:00:00Z')),
q.Max(q.Date('1970-01-01'), q.Date('1930-01-01')),
q.Max('A', 1),
q.Max(true, false),
q.Max({ x: 10 }, { x: 11 }),
q.Max(['A'], ['B'], ['C']),
q.Max(['X'], ['A', 'B']),
])
.then((ret) => console.log(ret))
[ 'D',
Time("1980-01-01T00:00:00Z"),
Date("1970-01-01"),
'A',
true,
{ x: 11 },
[ 'C' ],
[ 'X' ] ]
Not available in this language yet.
println(Await.result(client.query(
Arr(
Max("A", "B", "C", "D"),
Max(Time("1970-01-01T00:00:00Z"), Time("1980-01-01T00:00:00Z")),
Max(Date("1970-01-01"), Date("1930-01-01")),
Max("A", 1),
Max(true, false),
Max(Obj("x" -> 10), Obj("x" -> 11)),
Max(Arr("A"), Arr("B"), Arr("C")),
Max(Arr("X"), Arr("A", "B"))
)
),
5.seconds))
["D", 1980-01-01T00:00:00Z, 1970-01-01, "A", true, {x: 11}, ["C"], ["X"]]
Is this article helpful?
Tell Fauna how the article can be improved:
Visit Fauna's forums
or email docs@fauna.com
Thank you for your feedback!