如何合并两个 JSON 数组中的记录？
How can I merge records inside two JSON arrays?
我有两个 Postgres SQL 查询返回 JSON 数组：
第一季度：
[
{"id": 1, "a": "text1a", "b": "text1b"},
{"id": 2, "a": "text2a", "b": "text2b"},
{"id": 2, "a": "text3a", "b": "text3b"},
...
]
q2:
[
{"id": 1, "percent": 12.50},
{"id": 2, "percent": 75.00},
{"id": 3, "percent": 12.50}
...
]
我希望结果是两个数组唯一元素的并集：
[
{"id": 1, "a": "text1a", "b": "text1b", "percent": 12.50},
{"id": 2, "a": "text2a", "b": "text2b", "percent": 75.00},
{"id": 3, "a": "text3a", "b": "text3b", "percent": 12.50},
...
]
如何使用 Postgres 9.4 中的 SQL 完成此操作？
假设数据类型jsonb并且您想要合并共享相同'id'值的每个JSON数组的记录。
Postgres 9.5
使用新的 concatenate operator || for jsonb values 使其更简单：
SELECT json_agg(elem1 || elem2) AS result
FROM (
SELECT elem1->>'id' AS id, elem1
FROM (
SELECT '[
{"id":1, "percent":12.50},
{"id":2, "percent":75.00},
{"id":3, "percent":12.50}
]'::jsonb AS js
) t, jsonb_array_elements(t.js) elem1
) t1
FULL JOIN (
SELECT elem2->>'id' AS id, elem2
FROM (
SELECT '[
{"id": 1, "a": "text1a", "b": "text1b", "percent":12.50},
{"id": 2, "a": "text2a", "b": "text2b", "percent":75.00},
{"id": 3, "a": "text3a", "b": "text3b", "percent":12.50}]'::jsonb AS js
) t, jsonb_array_elements(t.js) elem2
) t2 USING (id);
FULL [OUTER] JOIN 确保您不会丢失其他数组中不匹配的记录。
类型jsonb有方便的属性只保留记录中每个键的最新值。因此，结果中重复的 'id' 键会自动合并。
Postgres 9.5 手册还建议：
Note: The || operator concatenates the elements at the top level of
each of its operands. It does not operate recursively. For example, if
both operands are objects with a common key field name, the value of
the field in the result will just be the value from the right hand operand.
Postgres 9.4
有点不方便。我的想法是提取数组元素，然后提取所有 key/value 对，UNION 两个结果，每个 id 值聚合成一个新的 jsonb 值，最后聚合成一个数组。
SELECT json_agg(j) -- ::jsonb
FROM (
SELECT json_object_agg(key, value)::jsonb AS j
FROM (
SELECT elem->>'id' AS id, x.*
FROM (
SELECT '[
{"id":1, "percent":12.50},
{"id":2, "percent":75.00},
{"id":3, "percent":12.50}]'::jsonb AS js
) t, jsonb_array_elements(t.js) elem, jsonb_each(elem) x
UNION ALL -- or UNION, see below
SELECT elem->>'id' AS id, x.*
FROM (
SELECT '[
{"id": 1, "a": "text1a", "b": "text1b", "percent":12.50},
{"id": 2, "a": "text2a", "b": "text2b", "percent":75.00},
{"id": 3, "a": "text3a", "b": "text3b", "percent":12.50}]'::jsonb AS js
) t, jsonb_array_elements(t.js) elem, jsonb_each(elem) x
) t
GROUP BY id
) t;
对 jsonb 的强制转换删除了重复键。或者，您可以使用 UNION 折叠重复项（例如，如果您想要 json 作为结果）。测试哪个对你的情况来说更快。
相关：
对于任何单个 jsonb 元素，concat || 运算符的使用对我来说效果很好 strip_nulls 和另一个将结果转换回 jsonb（不是数组）的技巧。
select jsonb_array_elements(jsonb_strip_nulls(jsonb_agg(
'{
"a" : "unchanged value",
"b" : "old value",
"d" : "delete me"
}'::jsonb
|| -- The concat operator works as merge on jsonb, the right operand takes precedence
-- NOTE: it only works one JSON level deep
'{
"b" : "NEW value",
"c" : "NEW field",
"d" : null
}'::jsonb
)));
这给出了结果
{"a": "unchanged value", "b": "NEW value", "c": "NEW field"}
输入正确jsonb
我有两个 Postgres SQL 查询返回 JSON 数组：
第一季度：
[
{"id": 1, "a": "text1a", "b": "text1b"},
{"id": 2, "a": "text2a", "b": "text2b"},
{"id": 2, "a": "text3a", "b": "text3b"},
...
]
q2:
[
{"id": 1, "percent": 12.50},
{"id": 2, "percent": 75.00},
{"id": 3, "percent": 12.50}
...
]
我希望结果是两个数组唯一元素的并集：
[
{"id": 1, "a": "text1a", "b": "text1b", "percent": 12.50},
{"id": 2, "a": "text2a", "b": "text2b", "percent": 75.00},
{"id": 3, "a": "text3a", "b": "text3b", "percent": 12.50},
...
]
如何使用 Postgres 9.4 中的 SQL 完成此操作？
假设数据类型jsonb并且您想要合并共享相同'id'值的每个JSON数组的记录。
Postgres 9.5
使用新的 concatenate operator || for jsonb values 使其更简单：
SELECT json_agg(elem1 || elem2) AS result
FROM (
SELECT elem1->>'id' AS id, elem1
FROM (
SELECT '[
{"id":1, "percent":12.50},
{"id":2, "percent":75.00},
{"id":3, "percent":12.50}
]'::jsonb AS js
) t, jsonb_array_elements(t.js) elem1
) t1
FULL JOIN (
SELECT elem2->>'id' AS id, elem2
FROM (
SELECT '[
{"id": 1, "a": "text1a", "b": "text1b", "percent":12.50},
{"id": 2, "a": "text2a", "b": "text2b", "percent":75.00},
{"id": 3, "a": "text3a", "b": "text3b", "percent":12.50}]'::jsonb AS js
) t, jsonb_array_elements(t.js) elem2
) t2 USING (id);
FULL [OUTER] JOIN 确保您不会丢失其他数组中不匹配的记录。
类型jsonb有方便的属性只保留记录中每个键的最新值。因此，结果中重复的 'id' 键会自动合并。
Postgres 9.5 手册还建议：
Note: The
||operator concatenates the elements at the top level of each of its operands. It does not operate recursively. For example, if both operands are objects with a common key field name, the value of the field in the result will just be the value from the right hand operand.
Postgres 9.4
有点不方便。我的想法是提取数组元素，然后提取所有 key/value 对，UNION 两个结果，每个 id 值聚合成一个新的 jsonb 值，最后聚合成一个数组。
SELECT json_agg(j) -- ::jsonb
FROM (
SELECT json_object_agg(key, value)::jsonb AS j
FROM (
SELECT elem->>'id' AS id, x.*
FROM (
SELECT '[
{"id":1, "percent":12.50},
{"id":2, "percent":75.00},
{"id":3, "percent":12.50}]'::jsonb AS js
) t, jsonb_array_elements(t.js) elem, jsonb_each(elem) x
UNION ALL -- or UNION, see below
SELECT elem->>'id' AS id, x.*
FROM (
SELECT '[
{"id": 1, "a": "text1a", "b": "text1b", "percent":12.50},
{"id": 2, "a": "text2a", "b": "text2b", "percent":75.00},
{"id": 3, "a": "text3a", "b": "text3b", "percent":12.50}]'::jsonb AS js
) t, jsonb_array_elements(t.js) elem, jsonb_each(elem) x
) t
GROUP BY id
) t;
对 jsonb 的强制转换删除了重复键。或者，您可以使用 UNION 折叠重复项（例如，如果您想要 json 作为结果）。测试哪个对你的情况来说更快。
相关：
对于任何单个 jsonb 元素，concat || 运算符的使用对我来说效果很好 strip_nulls 和另一个将结果转换回 jsonb（不是数组）的技巧。
select jsonb_array_elements(jsonb_strip_nulls(jsonb_agg(
'{
"a" : "unchanged value",
"b" : "old value",
"d" : "delete me"
}'::jsonb
|| -- The concat operator works as merge on jsonb, the right operand takes precedence
-- NOTE: it only works one JSON level deep
'{
"b" : "NEW value",
"c" : "NEW field",
"d" : null
}'::jsonb
)));
这给出了结果
{"a": "unchanged value", "b": "NEW value", "c": "NEW field"}
输入正确jsonb