Postgres JSON密钥计数 [英] Postgres json key count

问题描述

给出N条包含json列的记录

Given N records that contain a json column

|ID    |Name   |JSON                             
|01    |TEST1  |{"key1" : "value1", "key2": "value2", "key4": "value4"}
|02    |TEST1  |{"key1" : "value1"}
|03    |TEST2  |{"key1" : "value1", "key2": "value2", "key3":"value3"}
...

为一组键计算每个json值的出现的最佳策略是什么，因此对于上面的示例，我将限制为key1，key2，key3并得到:-

What would be the best strategy to count the occurrences of each json value for a set of keys, such that for the example above I would restrict to key1, key2, key3 and get:-

|value1|value2|value3|
|3     |2     |1     |

值将改变，所以我真的不想明确地寻找它们.

The values will change so I don't really want to look for them explicitly.

推荐答案

CREATE TABLE test (id INT4 PRIMARY KEY, some_name TEXT, j json);
copy test FROM stdin;
01  TEST1   {"key1" : "value1", "key2": "value2", "key4": "value4"}
02  TEST1   {"key1" : "value1"}
03  TEST2   {"key1" : "value1", "key2": "value2", "key3":"value3"}
\.
with unpacked as (
    SELECT (json_each_text(j)).* FROM test
)
SELECT value, count(*) FROM unpacked WHERE key in ('key1', 'key2', 'key3') group by value;

返回:

 value  | count 
--------+-------
 value1 |     3
 value3 |     1
 value2 |     2
(3 rows)

像显示的那样返回它并不是一个好主意(如果存在40亿个不同的值，您想做什么?)，但是您始终可以在应用程序中进行透视，或修改查询以进行透视

Returning it like you showed doesn't strike me as great idea (what would you want to do if there are 4 billion different values?), but you can always pivot in your app, or modify the query to do the pivoting.

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