HOW TO RUN (more details in HW7):

STEP 1: download AsterixDB
https://www.apache.org/dyn/closer.lua/asterixdb/asterixdb-0.9.9/asterix-server-0.9.9-binary-assembly.zip

STEP 2: from the directory opt/local/bin run:

	start-sample-cluster.sh

or

	start-sample-cluster.bat

depending on whether you have a MacOS X or a Windows.

Your java installation must be up to date.  If you get errors, wait
for instructions for HW7.

STEP 3: in your browsers, type this URL: 127.0.0.1:19001

Notice that step 2 says incorrectly 19002, you must type 19001

STEP 4: run the queries below in the browser; make up your own.

When finished:

STEP 5: from the directory opt/local/bin run:

	stop-sample-cluster.sh

or

	stop-sample-cluster.bat

QUERIES

-- an array
select x
from [1,2,3] as x;

-- a multiset
select x
from {{1,2,3}} as x;

-- an array of records
select x.a
from [ {"a": 1, "b":10}, {"a": 2, "b":20} ] as x;

-- what about a where clause
select x.a
from [ {"a": 1, "b":10}, {"a": 2, "b":20} ] as x
where b=20;

-- don't confuse objects (i.e. records) with multisets
select x
from {{1,2,3}} as x;

-- but this is an error:
select x
from {1,2,3} as x;

-- can order the clauses in a logical way:
from [1,2,3] as x
select x;

-- a nested bag
from {{ {{1,2,3}}, {{2,4,6,8}} }} as x
select x;

-- flatten the nested bag
from {{ {{1,2,3}}, {{2,4,6,8}} }} as x,
     x as y
select y;

-- or compute the sums of the inner bags
from {{ {{1,2,3}}, {{2,4,6,8}} }} as x
select array_sum(x);

-- same with arrays: we can  unnest it
from [ [1,2,3], [2,4,6,8]] as x,
     x as y
select y;

-- or compute the inner sum
from [ [1,2,3], [2,4,6,8]] as x
select array_sum(x);


-- an array of objects with nested arrays
from [ {"a": 1, "b":[1,2,3]},
       {"a": 2, "b":[2,4,6,8]} ] as x
select x.b;

-- what does this return?
from [ {"a": 1, "b":[1,2,3]},
       {"a": 2, "b":[2,4,6,8]} ] as x,
     x.b as y
select x.a, y as b;

-- conversely, we can nest a flat collection
-- first, the easy way:
let c =
  [ {"a": 1, "b":10},
    {"a":2, "b":20},
    {"a":1, "b":30} ]
from c as x
select x.a, (from c as y where x.a=y.a select y.b);

-- see the small problem?  Try this:
let c =
  [ {"a": 1, "b":10},
    {"a":2, "b":20},
    {"a":1, "b":30} ]
from c as x
select distinct x.a, (from c as y where x.a=y.a select y.b);

-- there is also a hard way, with group by and group as
-- lets first examine what group by and group as do:
let c =
  [ {"a": 1, "b":10},
    {"a":2, "b":20},
    {"a":1, "b":30} ]
from c as x
group by x.a
group as g
select x.a, g;

-- so we still need a nested subquery:
let c =
  [ {"a": 1, "b":10},
    {"a":2, "b":20},
    {"a":1, "b":30} ]
from c as x
group by x.a
group as g
select x.a, (from g as y select y.x.b);