1

我有一个查询，mysql想在PostgreSql.

这是查询：

-- mysql
SELECT cat.id, cat.category,
CONCAT('[',
      GROUP_CONCAT(CONCAT('{"rate":"',mod.rate,'", "model_name":"', mod.modelName, '", "capacity":"',mod.capacity,'"}')), ']') vehicles 
      FROM categories cat 
      LEFT JOIN models mod 
      ON cat.id = mod.category_id 
      GROUP BY cat.id

这将得到输出：

id  |  category  |  vehicles
__________________________________________________________________________
1   |  2-wheeler |  [ { "rate": "400", "model_name": "Pulsar 12", ... }, {...} ]
2   |  4-wheeler |  [ { "rate": "800", "model_name": "Honda 12", ... }, {...} ]

我正在尝试与PostgreSql.

这就是我卡住的地方：

-- postgres
SELECT cat.id, cat.category,
CONCAT('[',
      array_to_string(CONCAT('{"rate":"',mod.rate,'", "model_name":"', mod.modelName, '", "capacity":"',mod.capacity,'"}')), ']') vehicles 
      FROM categories cat 
      LEFT JOIN models mod 
      ON cat.id = mod.category_id 
      GROUP BY cat.id

我不知道如何继续前进。

此外，表格：

楷模：

id  |  modelName   |  rate  |  category_id
_____________________________________________
1   |  Pulsar - 12 |  400   |  1
2   |  Honda       |  800   |  2

类别：

id  | category
__________________
1   | 2-wheeler
2   | 4-wheeler
4

2 回答 2

3

您将不得不使用JSON_BUILD_OBJECT来构建单个 JSON 对象，然后JSON_AGG构建这些对象的数组：

SELECT cat.id, cat.category,
(SELECT json_agg(
  json_build_object(
    'rate',mod.rate,
    'model_name', mod.modelName, 
    'capacity',mod.capacity
  )) AS vehicles 
  FROM models AS mod WHERE category_id = cat.id
) AS models
FROM categories AS cat
于 2019-08-19T06:40:09.410 回答
1

您可以使用内置 PostgreSQL 的函数来构建 JSON 对象。

select cat.id, cat.category, json_agg(row_to_json(row(mod.rate, mod.modelName))) vehicles from categories cat
left join models mod on cat.id = mod.category_id
group by cat.id, cat.category;

结果将是

    id  category    vehicles
1   1   2-wheeler   [{"f1":400,"f2":"Pulsar - 12"}]
2   2   4-wheeler   [{"f1":800,"f2":"Honda"}]
于 2019-08-19T06:45:09.510 回答