[英]How can I aggregate a JSON array with a variable number of keys in PostgreSQL?
我在PostgreSQL表中有一系列的行,如下所示:
-[ RECORD 1 ]---------------------------------------------------------------------
student | e04c0ae4709340cb8e03c52f444e723f
group | 1
subgroup | 1
variable | VAR1
status | { "track_A" : "Done", "track_B" : "Done", "track_C" : "To Do" }
-[ RECORD 2 ]---------------------------------------------------------------------
student | e04c0ae4709340cb8e03c52f444e723f
group | 1
subgroup | 1
variable | VAR2
status | { "track_A" : "To Do", "track_B" : "Done", "track_C" : "To Do" }
-[ RECORD 3 ]---------------------------------------------------------------------
student | 849d1e6a0c2b4530a2b550829df94556
group | 0
subgroup | 1
variable | VAR3
status | { "track_A" : "Done", "track_B" : "To Do", "track_C" : "To Do" }
我想按学生,组和子组对它们进行分组,并获取每个曲目的计数状态。 就像是:
-[ RECORD 1 ]---------------------------------------------------------------------
student | e04c0ae4709340cb8e03c52f444e723f
group | 1
subgroup | 1
totals | { "track_A" : {"done": 1, "to_do": 1}, {"track_B" : {"done": 0, "to_do": 2}, "track_C" : {"done": 0, "to_do": 2} }
问题在于轨道的数量可能会有所不同。 我确实知道它们的名称,但是它们不是静态的,因此无法进行简单的汇总。 有什么建议可以在PostgreSQL(9.5)中编写吗? 我不想遍历所有轨道并进行汇总,因为该操作将花费一些时间。
你可以使用json_each_text
为“unest”的价值观和json_object_agg
一遍结合起来。
数据:
DROP TABLE IF EXISTS tab;
CREATE TABLE tab(student VARCHAR(36), "group" INT, subgroup INT,
variable VARCHAR(20), status JSON);
INSERT INTO tab(student, "group", subgroup, variable, status)
VALUES
('e04c0ae4709340cb8e03c52f444e723f',1,1,'VAR1'
,'{ "track_A" : "Done", "track_B" : "Done", "track_C" : "To Do" }'),
('e04c0ae4709340cb8e03c52f444e723f',1,1,'VAR2'
, '{ "track_A" : "To Do", "track_B" : "Done", "track_C" : "To Do" }')
,('849d1e6a0c2b4530a2b550829df94556',0,1,'VAR3'
,'{ "track_A" : "Done", "track_B" : "To Do", "track_C" : "To Do" }');
查询:
WITH cte AS
(
SELECT student, "group", subgroup, k
,COUNT(CASE WHEN v='Done' THEN 1 END) AS Done
,COUNT(CASE WHEN v='To Do' THEN 1 END) AS To_do
FROM tab
,LATERAL json_each_text(status) s(k,v)
GROUP BY student, "group", subgroup, k
), cte2 AS
(
SELECT student, "group", subgroup, k, json_object_agg(s.status, s.cnt) AS j
FROM cte
,LATERAL (VALUES('Done', Done),('To Do', To_Do)) AS s(status, cnt)
GROUP BY student, "group", subgroup, k
)
SELECT student, "group", subgroup
,json_object_agg(k, j) AS totals
FROM cte2
GROUP BY student, "group", subgroup;
输出:
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