8

我有一个订单表，我将订单摘要存储在jsonb列中

 {"users": [
   {"food": [{"name": "dinner", "price": "100"}], "room": "2", "user": "bob"}, 
   {"room": "3", "user": "foo"}
 ]}

现在我想users用他们的food->name.

我尝试了以下方法，但这也给了我没有食物的用户foo。

select 
  jsonb_array_elements(jsonb_array_elements(summary->'users')->'food')->>'name'  as food, 
  jsonb_array_elements(summary->'users')->>'user' as user_name 
from orders;

 food  | user_name 
 -------+-----------
 dinner | bob
 dinner | foo

我该怎么做这样的查询？

更新

我也有一个像这样的夏天有两种食物选择

{"users": [
  {"food": [{"name": "dinner", "price": "100"}, {"name": "breakfast", "price": "100"}], "room": "2", "user": "bob"}, 
  {"room": "3", "user": "foo"} 
]}

比我得到的：

   food    | user_name 
-----------+-----------
 dinner    | bob
 breakfast | foo

理想情况下我想得到

   food               | user_name 
----------------------+-----------
 dinner, breakfast    | bob
4

1 回答 1

8

好吧，如果你这样做

SELECT jsonb_array_elements(summary->'users') as users FROM orders;

你得到

┌──────────────────────────────────────────────────────────────────────────────────────────────────────────────────┐
│                                                      users                                                       │
├──────────────────────────────────────────────────────────────────────────────────────────────────────────────────┤
│ {"food": [{"name": "dinner", "price": "100"}, {"name": "breakfast", "price": "50"}], "room": "2", "user": "bob"} │
│ {"room": "3", "user": "foo"}                                                                                     │
└──────────────────────────────────────────────────────────────────────────────────────────────────────────────────┘

让我们把这个选择放在另一个里面，选择我们需要的：

SELECT users->'user' as user_name, users->'food'->0->'name' as food FROM (
    SELECT jsonb_array_elements(summary->'users') as users FROM orders
) as s;

┌───────────┬──────────┐
│ user_name │   food   │
├───────────┼──────────┤
│ "bob"     │ "dinner" │
│ "foo"     │ (null)   │
└───────────┴──────────┘

我们很接近。我们只需要添加一个WHERE.

SELECT users->'user' as user_name, users->'food'->0->'name' as food FROM (
    SELECT jsonb_array_elements(summary->'users') as users FROM orders
) as s WHERE (users->'food') is not null;

导致

┌───────────┬──────────┐
│ user_name │   food   │
├───────────┼──────────┤
│ "bob"     │ "dinner" │
└───────────┴──────────┘

如果您的食物阵列中有更多数据，例如

'{"users": [{"food": [{"name": "dinner", "price": "100"}, {"name" : "breakfast", "price" : "50"}], "room": "2", "user": "bob"}, {"room": "3", "user": "foo"}]}'

你可以做

SELECT users->'user' as user_name, jsonb_array_elements(users->'food')->>'name' as food FROM (
    SELECT jsonb_array_elements(summary->'users') as users FROM orders
) as s WHERE (users->'food') is not null;

和

┌───────────┬───────────┐
│ user_name │   food    │
├───────────┼───────────┤
│ "bob"     │ dinner    │
│ "bob"     │ breakfast │
└───────────┴───────────┘

重写上述查询以使用公用表表达式

WITH users_data AS (
    SELECT jsonb_array_elements(summary->'users') as users FROM orders
), user_food AS (
    SELECT users->'user' as user_name, jsonb_array_elements(users->'food')->>'name' as food 
    FROM users_data
    WHERE (users->'food') is not null  
) SELECT * FROM user_food;

现在我们只需要分组user_name

WITH users_data AS (
    SELECT jsonb_array_elements(summary->'users') as users FROM orders
), user_food AS (
    SELECT users->'user' as user_name, jsonb_array_elements(users->'food')->>'name' as food 
    FROM users_data
    WHERE (users->'food') is not null  
) SELECT user_name, array_agg(food) foods FROM user_food GROUP BY user_name;

最后结果

┌───────────┬────────────────────┐
│ user_name │       foods        │
├───────────┼────────────────────┤
│ "bob"     │ {dinner,breakfast} │
└───────────┴────────────────────┘

这是我能想到的最好的。如果您找到更好的方法，请告诉我。

于 2015-08-19T20:41:29.607 回答