Reputation: 374
I have the following JSON which is stored in a jsonb field named "Data" in a PostgreSQL database:
{
"CompetitionData" :
{
"StartDate" : "12.06.2018",
"Name" : "TestCompetition",
"Competitors" :
[
{
"Id" : "100",
"Name" : "John",
"Age" : "24",
"Score" : "98",
"Shoes":
{
"Brand" : "Nike"
}
},
{
"Id" : "200",
"Name" : "Adam",
"Age" : "32",
"Score" : "78",
"Shoes":
{
"Brand" : "Adidas"
}
}
]
}
}
Im trying to get a specific entry in Competitors, like e.g.
SELECT * FROM Competitors WHERE Shoes = "Nike";
And the result must look like this:
{
"Id" : "100",
"Name" : "John",
"Age" : "24",
"Score" : "98",
"Shoes":
{
"Brand" : "Nike"
}
}
I tried the following query, but i keeps returning all competitors:
SELECT jsonb_array_elements(public."Competitions"."Data"->'CompetitionData'->'Competitors') as test
FROM public."Competitions" WHERE public."Competitions"."Data" @> '{"CompetitionData":{"Competitors":[{"Shoes":{"Brand":"Nike"}}]}}';
Is it possible to return just the competitor with Shoe Brand "Nike" ?
Upvotes: 1
Views: 172
Reputation: 31676
Use jsonb_array_elements
in the from clause
SELECT j.* FROM
t cross join lateral
jsonb_array_elements(data->'CompetitionData'->'Competitors') as j(comp)
where j.comp->'Shoes'->>'Brand' = 'Nike'
Upvotes: 2
Reputation: 781
Why do you want to save it in jsonb? Just normalize it into a database:
CREATE TABLE competitor_shoe (
name text PRIMARY KEY,
brand text NOT NULL
);
CREATE TABLE competitor (
id int PRIMARY KEY,
name text NOT NULL,
age int NOT NULL,
score int NOT NULL,
shoe text NOT NULL REFERENCES competitor_shoe(name)
);
CREATE TABLE competition (
name text PRIMARY KEY,
start_date date NOT NULL
);
CREATE TABLE competition_competitor (
competition text REFERENCES competition,
competitor int REFERENCES competitor,
PRIMARY KEY (competition,competitor)
);
INSERT INTO competitor_shoe
VALUES ('shoes1', 'Nike'),
('shoes2', 'Adidas');
INSERT INTO competitor
VALUES (100,'John',24,98,'shoes1'),
(200,'Adam',32,78,'shoes2');
INSERT INTO competition
VALUES (
'TestCompetition',
'12.06.2018'
);
INSERT INTO competition_competitor
VALUES ('TestCompetition', 100), ('TestCompetition', 200);
-- query the data
SELECT *
FROM competitor c
JOIN competitor_shoe cs
ON c.shoe = cs.name
WHERE brand = 'Nike';
-- query the data and return it as json object
SELECT to_jsonb(c) || jsonb_build_object('shoe', to_jsonb(cs)) as data
FROM competitor c
JOIN competitor_shoe cs
ON c.shoe = cs.name
WHERE brand = 'Nike';
Upvotes: 0