如何使用 sql 查询对列表列表进行排序？

Question

Abstract question

I have a sql-table that contains records in the following form:
(list_id, value) where the list_id is an Integer identifiing a specific list and the value is something that has an order.

I now struggle to write a sql query that returns all records of that table at first ordered by the rank the list has compared to the other lists and then ordered by the value .

The abstract problem is, that I want to sort a list of lists using sql.

Algorithm to compare two lists

The algorithm to compare two lists is the following:

data CompareRes = FirstSmaller | FirstGreater | Equal deriving Show

compareLists :: Ord a => [a] -> [a] -> CompareRes
compareLists [] [] = Equal
-- Longer lists are considered to be smaller
compareLists _ [] = FirstSmaller
compareLists [] _ = FirstGreater
compareLists (x:xs) (y:ys) 
  | x < y = FirstSmaller
  | x > y = FirstGreater
  | otherwise = compareLists xs ys

Details

In my specific case the values are all Date s. So my table looks like this:

CREATE TABLE `list_date` (
  `list_id` INT  NOT NULL,
  `date`    DATE NOT NULL,
  PRIMARY KEY (`list_id`, `date`)
);

I'm using a mysql:8.0 database so a solution using WINDOW -functions is acceptable.

Example

Data

INSERT INTO `list_date` VALUES
   (1, '2019-11-02'), (1, '2019-11-03'), (1, '2019-11-04'), (1, '2019-11-05'), (1, '2019-11-07'), (1, '2019-11-08'), (1, '2019-11-09'),
   (2, '2019-11-01'), (2, '2019-11-03'), (2, '2019-11-04'),
   (3, '2019-11-01'), (3, '2019-11-02'), (3, '2019-11-03'),
   (4, '2019-11-02'), (4, '2019-11-04'), (4, '2019-11-13'), (4, '2019-11-14'),
   (5, '2019-11-03'), (5, '2019-11-04'), (5, '2019-11-05'), (5, '2019-11-10'),
   (6, '2019-11-01'), (6, '2019-11-02'), (6, '2019-11-03'), (6, '2019-11-05');

Query

Where I really struggle is to create an expression that calculates the list_rank :

SELECT 
    `list_id`, 
    `date`,
    <PLEASE HELP> as `list_rank`
FROM 
    `list_date`
ORDER BY 
    `list_rank`, `date`;

Expected result

| list_id | date       | list_rank |
|---------|------------|-----------|
| 6       | 2019-11-01 | 1         |
| 6       | 2019-11-02 | 1         |
| 6       | 2019-11-03 | 1         |
| 6       | 2019-11-05 | 1         |
| 3       | 2019-11-01 | 2         |
| 3       | 2019-11-02 | 2         |
| 3       | 2019-11-03 | 2         |
| 2       | 2019-11-01 | 3         |
| 2       | 2019-11-03 | 3         |
| 2       | 2019-11-04 | 3         |
| 1       | 2019-11-02 | 4         |
| 1       | 2019-11-03 | 4         |
| 1       | 2019-11-04 | 4         |
| 1       | 2019-11-05 | 4         |
| 1       | 2019-11-07 | 4         |
| 1       | 2019-11-08 | 4         |
| 1       | 2019-11-09 | 4         |
| 4       | 2019-11-02 | 5         |
| 4       | 2019-11-04 | 5         |
| 4       | 2019-11-13 | 5         |
| 4       | 2019-11-14 | 5         |
| 5       | 2019-11-03 | 6         |
| 5       | 2019-11-04 | 6         |
| 5       | 2019-11-05 | 6         |
| 5       | 2019-11-10 | 6         |

or

That image is the current live result my application produces. Currently the sorting is implemented using Java.

Edit

After not receiving a better answer, I implemented a solution as suggested by @gordon-linoff:

SELECT 
    `list_id`, 
    `date`
FROM 
    `list_date`
        INNER JOIN (
            SELECT `sub`.`list_id`,
            GROUP_CONCAT(`sub`.`date` ORDER BY `sub`.`date` SEPARATOR '')  as `concat_dates`
            FROM `list_date` as `sub`
            GROUP BY `sub`.`list_id`
        ) `all_dates` ON (`all_dates`.`list_id` = `list_date`.`list_id`)
ORDER BY 
    `all_dates`.`concat_dates`, `date`;

I've also created an SQL Fiddle - So you can play around with your solution.

But this solution does not sort the lists as expected because longer lists are considered bigger than smaller lists.

So I am still hoping to receive a solution that solves 100% of my requirements:)

Answer 1

If I understand correctly, you can sort the lists by the dates concatenated together:

select ld.*
from list_date ld join
     (select list_id, group_concat(date) as dates
      from ld
      group by list_id
     ) ldc
     on ld.list_id = ldc.list_id
order by ldc.dates, ld.date;

Answer 2

Since it's for MySql 8 the window functions can be used for this (yay).

Here's a query that first calculates some metrics, to use in the calculation of the ranking:

SELECT 
 list_id, 
 `date`,
 DENSE_RANK() OVER (ORDER BY ListMinDate ASC, ListCount DESC, ListMaxDate, list_id) AS list_rank
FROM
(
  SELECT 
   list_id,
   `date`,
   COUNT(*) OVER (PARTITION BY list_id) AS ListCount,
   MIN(`date`) OVER (PARTITION BY list_id) AS ListMinDate,
   MAX(`date`) OVER (PARTITION BY list_id) AS ListMaxDate
  FROM list_date
) q
ORDER BY list_rank, `date`

A test on db<>fiddle here