Question

我正在使用以下Postgres表构建智能储物柜预订系统：

CREATE TABLE lockers (
  id serial primary key
)

CREATE TABLE doors (
  id serial primary key,
  locker_id integer NOT NULL,
  size integer
);

CREATE TABLE packages (
  id serial primary key,
  locker_id integer NOT NULL,
  size integer
);

可以通过设置包裹的locker_id列将包裹保留给储物柜。这些软件包在交付给储物柜时最终将获得door_id，但这不在此问题的范围内。现在，我只是对预订感兴趣，此设置为我提供了很大的灵活性（例如，如果在预订时将小包装预订给只有一个大门的储物柜，但该储物柜有较小的门可用在交付时，我不需要重写door_id，只需将其分配给可能的最小门即可。

一切正常，但是现在我想编写一个查询，该查询可以选择所有具有给定大小的新包装有空间的储物柜。我的问题是我不能只用JOIN选择空门，因为包裹只知道储物柜。对于每个储物柜，我基本上都需要做类似Find the smallest possible door for each package, and see if the new package fits in any of the remaining doors的事情。我可以像这样用JavaScript轻松编写：

const canFitPackage = (reservations, doors, newPackage) => {
  const sortedReservations = reservations
    .slice()
    .sort((a, b) => a.size - b.size)
    .reverse();
  const sortedDoors = doors.slice().sort((a, b) => a.size - b.size);

  for (let i = 0; i < sortedReservations.length; i++) {
    const res = sortedReservations[i];
    for (let j = 0; j < sortedDoors.length; j++) {
      const door = sortedDoors[j];
      if (door.size >= res.size) {
        sortedDoors.splice(j, 1);
        break;
      }
    }
  }
  return sortedDoors.some(door => door.size >= newPackage.size);
};

我花了几天的时间试图弄清楚如何在SQL中做到这一点，但到目前为止我还很走运。我想知道这是否真的可以在SQL调用中完成，或者我是否需要编写Postgres函数。任何帮助表示赞赏。

更新

经过数小时的尝试，我提出了这个SQL查询，它可以为我做这件事。我确信这是缓慢且无效的，因此感谢您的任何投入和技巧。

WITH 
locker_doors AS (
  SELECT * FROM doors 
  --ADD INNER JOIN ON LOCKER 
  --ADD WHERE CLAUSE ON LOCKERS
),
all_combinations AS (
  SELECT locker_doors.id AS door_id, locker_doors.size AS door_size, locker_doors.locker_id, packages.id AS package_id, packages.size AS package_size
  FROM locker_doors 
  JOIN packages ON locker_doors.locker_id = packages.locker_id AND locker_doors.size >= packages.size
  ORDER BY packages.size DESC, locker_doors.size ASC
), 
distinct_doors AS (
  SELECT DISTINCT ON (door_id) * FROM all_combinations
), 
package_placements AS (
  SELECT DISTINCT ON (package_id) * FROM distinct_doors
)
SELECT DISTINCT ON (locker_id) locker_id 
FROM locker_doors 
WHERE id NOT IN (SELECT door_id FROM package_placements)

Answer 1

您可以使用计算列对doors条记录的剩余空间进行排序，并获得最小的记录。

http://sqlfiddle.com/#!17/a8840/2

 WITH newPackage (packsize) as (
   values (1)--I used with clause to declare the new package. You can change the size of the new package from here to test different cases
)
select  d.*, 
(d.size - coalesce((select p.size from packages p where  p.locker_id = d.locker_id), 0)) remaining_size 
from 
doors d, 
newPackage
where 
d.size - coalesce((select p.size from packages p where  p.locker_id = d.locker_id), 0) >= newPackage.packsize
order by 
(d.size - coalesce((select p.size from packages p where  p.locker_id = d.locker_id), 0)) 
limit 1

Answer 2

locker_doors AS (
  SELECT * FROM doors 
  --ADD INNER JOIN ON LOCKER 
  --ADD WHERE CLAUSE ON LOCKERS
),
all_combinations AS (
  SELECT locker_doors.id AS door_id, locker_doors.size AS door_size, locker_doors.locker_id, packages.id AS package_id, packages.size AS package_size
  FROM locker_doors 
  JOIN packages ON locker_doors.locker_id = packages.locker_id AND locker_doors.size >= packages.size
  ORDER BY packages.size DESC, locker_doors.size ASC
), 
distinct_doors AS (
  SELECT DISTINCT ON (door_id) * FROM all_combinations
), 
package_placements AS (
  SELECT DISTINCT ON (package_id) * FROM distinct_doors
)
SELECT DISTINCT ON (locker_id) locker_id 
FROM locker_doors 
WHERE id NOT IN (SELECT door_id FROM package_placements)

SQL查询以选择有储物空间的储物柜

2 个答案: