在 Postgresql 中将统计信息聚合到 JSON 中

Question

所以我正在尝试将概览统计信息计算为 JSON，但无法将它们整理到查询中。

有2张桌子：

appointments
- time timestamp
- patients int


assignments
- user_id int
- appointment_id int

我想按用户、按小时计算当天的患者人数。理想情况下，它应该是这样的：

[ 
  {hour: "2015-07-01T08:00:00.000Z", assignments: [
    {user_id: 123, patients: 3}, 
    {user_id: 456, patients: 10}, 
    {user_id: 789, patients: 4},
  ]},
  {hour: "2015-07-01T09:00:00.000Z", assignments: [
    {user_id: 456, patients: 1},
    {user_id: 789, patients: 6}
  ]},
  {hour: "2015-07-01T10:00:00.000Z", assignments: []}
  ...
]

我有点接近了：

with assignments_totals as (
    select user_id,sum(patients),date_trunc('hour',appointments.time) as hour
    from assignments
    inner join appointments on appointments.id = assignments.appointment_id
    group by date_trunc('hour',sales.time),user_id
  ), hours as (
    select to_char(date_trunc('hour',time),'YYYY-MM-DD"T"HH24:00:00.000Z') as hour, array_to_json(array_agg(DISTINCT assignment_totals)) as patients
    from appointments 
    left join assignment_totals on date_trunc('hour',sales.time) = assignment_totals.hour
    where time >= '2015-07-01T07:00:00.000Z' and time < '2015-07-02T07:00:00.000Z' 
    group by date_trunc('hour',time)
    order by date_trunc('hour',time) 
  )
  select array_to_json(array_agg(hours)) as hours from hours;

哪些输出：

[ 
  {hour: "2015-07-01T08:00:00.000Z", assignments: [
    {user_id: 123, patients: 3, hour: "2015-07-01T08:00:00.000Z" }, 
    {user_id: 456, patients: 10, hour: "2015-07-01T08:00:00.000Z"}, 
    {user_id: 789, patients: 4, hour: "2015-07-01T08:00:00.000Z"},
  ]},
  {hour: "2015-07-01T09:00:00.000Z", assignments: [
    {user_id: 456, patients: 1, hour: "2015-07-01T09:00:00.000Z"},
    {user_id: 789, patients: 6, hour: "2015-07-01T09:00:00.000Z"}
  ]},
  {hour: "2015-07-01T10:00:00.000Z", assignments: [null]}
  ...
]

虽然这可行，但有 2 个问题可能相互独立，也可能不相互独立：

1. 如果该小时没有约会，我仍然希望将该小时包含在数组中（如示例中的上午 10 点），但要有一个空的“分配”数组。现在它在那里放了一个空值，我不知道如何在仍然保留时间的同时摆脱它。
2. 我必须将小时与 user_id 和约会一起包含在分配条目中，因为我需要它将 assignments_totals 查询加入小时查询。但这是不必要的，因为它已经在父级中。
3. 我觉得应该可以在 1 个 cte 和 1 个查询中完成，现在我使用的是 2 个 cte...但不知道如何压缩它并使其工作。

我想做类似的事情

  hours as (
    select to_char(date_trunc('hour',time),'YYYY-MM-DD"T"HH24:00:00.000Z') as hour, sum(appointments.patients) OVER(partition by assignments.user_id) as appointments
    from appointments 
    left join assignments on appointments.id = assignments.appointment_id
    where time >= '2015-07-01T07:00:00.000Z' and time < '2015-07-02T07:00:00.000Z'  
    group by date_trunc('hour',time)
    order by date_trunc('hour',time) 
  )
  select array_to_json(array_agg(hours)) as hours from hours

但如果不给我一个“属性必须在分组依据或聚合函数错误中”，我就无法让它工作。

有人知道如何解决这些问题吗？提前致谢！

Answer 1

您上次查询的主要问题似乎在于将window functions 与aggregate functions 混为一谈。窗口函数使用OVER 语法，当SELECT 子句中有其他字段时，它们本身不需要GROUP BY。另一方面，当SELECT 子句中有其他（非聚合函数）字段时，聚合函数使用GROUP BY。这种差异的一个实际后果是窗口函数不会自动DISTINCT。

窗口函数导致的NULL 值问题可以通过简单的COALESCE 解决，这样使用零而不是null。

因此，要使用窗口函数编写查询，请使用以下内容：

WITH hours AS
(
    SELECT DISTINCT to_char(date_trunc('hour', ap.time), 'YYYY-MM-DD"T"HH:00:00.000Z') AS hour,
           COALESCE(SUM(ap.patients) OVER (PARTITION BY asgn.user_id), 0) AS appointment_count
    FROM   appointments ap
    LEFT JOIN assignments asgn ON ap.id = asgn.appointment_id
    WHERE  ap.time >= '2015-07-01T07:00:00.000Z'
    AND    ap.time < '2015-07-02T07:00:00.000Z'
)
SELECT array_to_json(array_agg(hours)) AS hours
FROM   hours
ORDER BY hour

使用聚合函数：

WITH hours AS
(
    SELECT to_char(date_trunc('hour', ap.time), 'YYYY-MM-DD"T"HH:00:00.000Z') AS hour,
           SUM(COALESCE(ap.patients, 0)) AS appointment_count,
           asgn.user_id
    FROM   appointments ap
    LEFT JOIN assignments asgn ON ap.id = asgn.appointment_id
    WHERE  ap.time >= '2015-07-01T07:00:00.000Z'
    AND    ap.time < '2015-07-02T07:00:00.000Z'
    GROUP BY asgn.user_id, to_char(date_trunc('hour', ap.time), 'YYYY-MM-DD"T"HH:00:00.000Z')
)
SELECT array_to_json(array_agg(hours)) AS hours
FROM   hours
ORDER BY hour

我的语法可能不太正确，因此在使用此解决方案或类似解决方案之前请仔细检查（并随时编辑以更正任何错误）。

Answer 2

我对此感到沮丧主要是因为我没有查看 Postgres 9.4 文档，该文档具有处理 json 的新功能。

我找到的解决方案建立在原始查询的基础上，但随后使用 json_array_elements 分解分配数组，使用 where 过滤，然后再次构建它。本质上似乎毫无意义：

json_agg(json_array_elements(json_agg(*)))

但它对性能的影响很小，可以让我到达我需要去的地方。如果您找到更好的解决方案，请随时发表评论！在

  with assignment_totals as (
    select 
      date_trunc('hour',appointments.time) as hour, 
      user_id, 
      coalesce(sum(patients),0) as patients
    from appointments
    left outer join assignments on appointment.id = assignments.appointment_id
    where time >= '2015-07-01T07:00:00.000Z' and time < '2015-07-02T07:00:00.000Z' 
    group by date_trunc('hour',appointments.time),user_id
  ), hours as (
    select 
      to_char(assignment_totals.hour,'YYYY-MM-DD"T"HH24:00:00.000Z') as hour,
      (
        select coalesce(json_agg(json_build_object('user_id',(t->'user_id'),'patients',(t->'patients')) order by (t->>'user_id')),'[]'::json) 
        from json_array_elements(json_agg(assignment_totals)) t 
        where (t->>'patients') != '0'
      ) as patients
    from assignment_totals 
    group by assignment_totals.hour
    order by assignment_totals.hour
  )
  select array_to_json(array_agg(hours)) as hours from hours

感谢 Andrew 指出我可以将空值合并为 0。但我仍然想过滤掉患者 = 0 的条目。这解决了我的所有问题，让我能够使用 where 过滤它们，然后让我能够通过使用 json_build_object 构建新的 json 对象来节省时间。