Re: [GENERAL] json aggregation question - Mailing list pgsql-general

Hi Chris,

Maybe there is an another better solution;

1. sending values into jsonb_array_elements to getting elements (lateral join)

2. distinct to eliminate duplicates

3. regexp_replace to remove malformed Array literals

4. Casting into text array

SELECT 

 count(distinct tags ),

string_to_array(regexp_replace(string_agg(distinct elem::text , ','),'\[*\"*\s*\]*','','g'),',') AS list

from thing as t, jsonb_array_elements(t.tags->'tag1') elem

where tags->'tag2'?'t2val1'

count | tag1

2        | {val1,val2,val3}

28 Şub 2017 Sal, 19:22 tarihinde, Chris Withers <chris@simplistix.co.uk> şunu yazdı:

Hi All,

Given the following table:

# create table thing (id serial, tags jsonb);
# \d thing                        Table "public.thing"Column |  Type   |                     Modifiers                      
--------+---------+----------------------------------------------------id     | integer | not null default nextval('thing_id_seq'::regclass)tags   | jsonb   | 

...and the following data:

insert into thing (tags) values ('{"tag1": ["val1", "val2"], "tag2": ["t2val1"]}');
insert into thing (tags) values ('{"tag1": ["val3", "val1"], "tag2": ["t2val1"]}');
insert into thing (tags) values ('{"tag1": ["val2", "val1"], "tag2": ["t2val2"]}');

How can I aggregate the results of a query that equates to "show me the number of matching rows and the set of tag1 value that have a tag2 value of t2val1?

The closes I can get is:

# select count(*), json_agg(tags) from thing where tags->'tag2'?'t2val1';count |                                             json_agg                                             
-------+--------------------------------------------------------------------------------------------------    2 | [{"tag1": ["val1", "val2"], "tag2": ["t2val1"]}, {"tag1": ["val3", "val1"], "tag2": ["t2val1"]}]
(1 row)

...but I really want:

 count |     tag1                                             
-------+-------------------------    2 | ["val1", "val2", "val3"]
(1 row)

cheers,

Chris