具有嵌套连接的PostgreSQL 9.2 row_to_json()

Question

我正在尝试使用row_to_json()PostgreSQL 9.2中添加的函数将查询结果映射到JSON .

我无法找出将连接行表示为嵌套对象的最佳方法(1:1关系)

这是我尝试过的(设置代码:表格,示例数据,然后是查询):

-- some test tables to start out with:
create table role_duties (
    id serial primary key,
    name varchar
);

create table user_roles (
    id serial primary key,
    name varchar,
    description varchar,
    duty_id int, foreign key (duty_id) references role_duties(id)
);

create table users (
    id serial primary key,
    name varchar,
    email varchar,
    user_role_id int, foreign key (user_role_id) references user_roles(id)
);

DO $$
DECLARE duty_id int;
DECLARE role_id int;
begin
insert into role_duties (name) values ('Script Execution') returning id into duty_id;
insert into user_roles (name, description, duty_id) values ('admin', 'Administrative duties in the system', duty_id) returning id into role_id;
insert into users (name, email, user_role_id) values ('Dan', 'someemail@gmail.com', role_id);
END$$;

查询本身:

select row_to_json(row)
from (
    select u.*, ROW(ur.*::user_roles, ROW(d.*::role_duties)) as user_role 
    from users u
    inner join user_roles ur on ur.id = u.user_role_id
    inner join role_duties d on d.id = ur.duty_id
) row;

我发现如果我使用了ROW(),我可以将结果字段分成一个子对象,但它似乎仅限于一个级别.我不能插入更多的AS XXX语句,因为我认为在这种情况下我应该需要.

我获得了列名,因为我转换为适当的记录类型,例如::user_roles,在该表的结果的情况下.

这是查询返回的内容:

{
   "id":1,
   "name":"Dan",
   "email":"someemail@gmail.com",
   "user_role_id":1,
   "user_role":{
      "f1":{
         "id":1,
         "name":"admin",
         "description":"Administrative duties in the system",
         "duty_id":1
      },
      "f2":{
         "f1":{
            "id":1,
            "name":"Script Execution"
         }
      }
   }
}

我想要做的是为连接生成JSON(再次1:1很好),我可以添加连接,并将它们表示为它们加入的父对象的子对象,即如下所示:

{
   "id":1,
   "name":"Dan",
   "email":"someemail@gmail.com",
   "user_role_id":1,
   "user_role":{
         "id":1,
         "name":"admin",
         "description":"Administrative duties in the system",
         "duty_id":1
         "duty":{
            "id":1,
            "name":"Script Execution"
         }
      }
   }
}

任何帮助表示赞赏.谢谢阅读.

Answer 1

更新:在PostgreSQL 9.4 中to_json,json_build_objectjson_objectjson_build_array由于需要明确命名所有字段,因此引入了很多,并且,虽然它很冗长:

select
        json_build_object(
                'id', u.id,
                'name', u.name,
                'email', u.email,
                'user_role_id', u.user_role_id,
                'user_role', json_build_object(
                        'id', ur.id,
                        'name', ur.name,
                        'description', ur.description,
                        'duty_id', ur.duty_id,
                        'duty', json_build_object(
                                'id', d.id,
                                'name', d.name
                        )
                )
    )
from users u
inner join user_roles ur on ur.id = u.user_role_id
inner join role_duties d on d.id = ur.duty_id;

对于旧版本,请继续阅读.

它不仅限于一排,它只是有点痛苦.您不能使用别名复合行类型AS,因此您需要使用别名子查询表达式或CTE来实现此效果:

select row_to_json(row)
from (
    select u.*, urd AS user_role
    from users u
    inner join (
        select ur.*, d
        from user_roles ur
        inner join role_duties d on d.id = ur.duty_id
    ) urd(id,name,description,duty_id,duty) on urd.id = u.user_role_id
) row;

通过http://jsonprettyprint.com/制作:

{
  "id": 1,
  "name": "Dan",
  "email": "someemail@gmail.com",
  "user_role_id": 1,
  "user_role": {
    "id": 1,
    "name": "admin",
    "description": "Administrative duties in the system",
    "duty_id": 1,
    "duty": {
      "id": 1,
      "name": "Script Execution"
    }
  }
}

array_to_json(array_agg(...))当你有1:很多关系时,你会想要使用顺便说一下.

理想情况下,上述查询应该写成:

select row_to_json(
    ROW(u.*, ROW(ur.*, d AS duty) AS user_role)
)
from users u
inner join user_roles ur on ur.id = u.user_role_id
inner join role_duties d on d.id = ur.duty_id;

...但PostgreSQL的ROW构造函数不接受AS列别名.可悲的是.

值得庆幸的是,他们优化了相同的.比较计划:

• 该嵌套子查询的版本 ; VS
• 后一个嵌套的ROW构造函数版本,删除了别名,以便执行

由于CTE是优化围栏,因此将嵌套子查询版本重新构造为使用链式CTE(WITH表达式)可能效果不佳,并且不会产生相同的计划.在这种情况下,你会遇到丑陋的嵌套子查询,直到我们得到一些改进row_to_json或ROW更直接地覆盖构造函数中的列名.

无论如何,一般来说,原则是你想用列创建一个json对象a, b, c,你希望你可以写出非法的语法:

ROW(a, b, c) AS outername(name1, name2, name3)

您可以改为使用返回行类型值的标量子查询:

(SELECT x FROM (SELECT a AS name1, b AS name2, c AS name3) x) AS outername

要么:

(SELECT x FROM (SELECT a, b, c) AS x(name1, name2, name3)) AS outername

另外,请记住,您可以在json没有额外引用的情况下编写值,例如,如果您将a的输出json_agg放在a中row_to_json,则内部json_agg结果将不会作为字符串引用,它将直接合并为json.

例如在任意例子中:

SELECT row_to_json(
        (SELECT x FROM (SELECT
                1 AS k1,
                2 AS k2,
                (SELECT json_agg( (SELECT x FROM (SELECT 1 AS a, 2 AS b) x) )
                 FROM generate_series(1,2) ) AS k3
        ) x),
        true
);

输出是:

{"k1":1,
 "k2":2,
 "k3":[{"a":1,"b":2}, 
 {"a":1,"b":2}]}

请注意,json_agg产品[{"a":1,"b":2}, {"a":1,"b":2}]尚未再次转义text.

这意味着你可以编写 json操作来构造行,你不必总是创建非常复杂的PostgreSQL复合类型然后调用row_to_json输出.

Answer 2

我添加此解决方案是因为接受的响应未考虑 N:N 关系。又名：对象集合的集合

如果你有 N:N 关系，那么这个条款with就是你的朋友。在我的示例中，我想构建以下层次结构的树视图。

A Requirement - Has - TestSuites
A Test Suite - Contains - TestCases.

以下查询代表连接。

SELECT reqId ,r.description as reqDesc ,array_agg(s.id)
            s.id as suiteId , s."Name"  as suiteName,
            tc.id as tcId , tc."Title"  as testCaseTitle

from "Requirement" r 
inner join "Has"  h on r.id = h.requirementid 
inner join "TestSuite" s on s.id  = h.testsuiteid
inner join "Contains" c on c.testsuiteid  = s.id 
inner join "TestCase"  tc on tc.id = c.testcaseid
  GROUP BY r.id, s.id;

由于不能进行多次聚合，因此需要使用“WITH”。

with testcases as (
select  c.testsuiteid,ts."Name" , tc.id, tc."Title"  from "TestSuite" ts
inner join "Contains" c on c.testsuiteid  = ts.id 
inner join "TestCase"  tc on tc.id = c.testcaseid

),                
requirements as (
    select r.id as reqId ,r.description as reqDesc , s.id as suiteId
    from "Requirement" r 
    inner join "Has"  h on r.id = h.requirementid 
    inner join "TestSuite" s on s.id  = h.testsuiteid

    ) 
, suitesJson as (
 select  testcases.testsuiteid,  
       json_agg(
                json_build_object('tc_id', testcases.id,'tc_title', testcases."Title" )
            ) as suiteJson
    from testcases 
    group by testcases.testsuiteid,testcases."Name"
 ),
allSuites as (
    select has.requirementid,
           json_agg(
                json_build_object('ts_id', suitesJson.testsuiteid,'name',s."Name"  , 'test_cases', suitesJson.suiteJson )
            ) as suites
            from suitesJson inner join "TestSuite" s on s.id  = suitesJson.testsuiteid
            inner join "Has" has on has.testsuiteid  = s.id
            group by has.requirementid
),
allRequirements as (
    select json_agg(
            json_build_object('req_id', r.id ,'req_description',r.description , 'test_suites', allSuites.suites )
            ) as suites
            from allSuites inner join "Requirement" r on r.id  = allSuites.requirementid

)
 select * from allRequirements

它的作用是在小的项目集合中构建 JSON 对象，并将它们聚合到每个项目上with子句中。

结果：

[
  {
    "req_id": 1,
    "req_description": "<character varying>",
    "test_suites": [
      {
        "ts_id": 1,
        "name": "TestSuite",
        "test_cases": [
          {
            "tc_id": 1,
            "tc_title": "TestCase"
          },
          {
            "tc_id": 2,
            "tc_title": "TestCase2"
          }
        ]
      },
      {
        "ts_id": 2,
        "name": "TestSuite",
        "test_cases": [
          {
            "tc_id": 2,
            "tc_title": "TestCase2"
          }
        ]
      }
    ]
  },
  {
    "req_id": 2,
    "req_description": "<character varying> 2 ",
    "test_suites": [
      {
        "ts_id": 2,
        "name": "TestSuite",
        "test_cases": [
          {
            "tc_id": 2,
            "tc_title": "TestCase2"
          }
        ]
      }
    ]
  }
]