PostgreSQL JSON数组查询

Question

我尝试使用以下示例查询json数组：如何使用新PostgreSQL JSON数据类型内的字段进行查询？

他们使用示例：

SELECT *
FROM   json_array_elements(
  '[{"name": "Toby", "occupation": "Software Engineer"},
    {"name": "Zaphod", "occupation": "Galactic President"} ]'
  ) AS elem
WHERE elem->>'name' = 'Toby';

但是我的Json数组看起来更像这样（如果使用示例）：

    {
    "people": [{
            "name": "Toby",
        "occupation": "Software Engineer"
    },
    {
        "name": "Zaphod",
        "occupation": "Galactic President"
    }
    ]
}

但我收到一个错误：错误：无法在非数组上调用json_array_elements

我的Json“数组”不是数组吗？我必须使用此Json字符串，因为它包含在数据库中，因此如果它不是数组，则必须告诉他们对其进行修复。或者，还有另一种查询方法吗？

我阅读了文档，但无济于事，并不断出错。

Answer 1

json数组具有一个键，people因此可以my_json->'people'在函数中使用：

with my_table(my_json) as (
values(
'{
    "people": [
        {
            "name": "Toby",
            "occupation": "Software Engineer"
        },
        {
            "name": "Zaphod",
            "occupation": "Galactic President"
        }
    ]
}'::json)
)
select t.*
from my_table t,
json_array_elements(my_json->'people') elem
where elem->>'name' = 'Toby';

该函数json_array_elements()取消嵌套json数组，并将其所有元素生成为行：

select elem->>'name' as name, elem->>'occupation' as occupation
from my_table t,
json_array_elements(my_json->'people') elem

  name  |     occupation     
--------+--------------------
 Toby   | Software Engineer
 Zaphod | Galactic President
(2 rows)    

如果您对Toby的职业感兴趣：

select elem->>'occupation' as occupation
from my_table t,
json_array_elements(my_json->'people') elem
where elem->>'name' = 'Toby'

    occupation     
-------------------
 Software Engineer
(1 row)