更新基于相同列值的行组合表

Question

我有一个名为table的表，这样：

| id | name  | city    |
|----|-------|---------|
| 0  | Rose  | Madrid  |
| 1  | Alex  | Lima    |
| 2  | Rose  | Sidney  |
| 3  | Mario | Glasgow |

我需要更新表，以便将共享相同name的两行组合成一个新行并删除。

| id | name  | city           |
|----|-------|----------------|
| 1  | Alex  | Lima           |
| 3  | Mario | Glasgow        |
| 4  | Rose  | Madrid, Sidney |

我不在乎它是否必须在多个 SQL 语句中完成。

到目前为止，我所做的只是列出受此影响的行。

SELECT *
FROM table
WHERE name IN (
  SELECT name
  FROM table
  GROUP BY name
  HAVING COUNT(*) > 1
);

Answer 1

假设id是自增主键，你需要一个INSERT和一个DELETE语句：

insert into tablename(name, city)
select name, group_concat(city, ',')
from tablename
group by name
having count(*) > 1;

delete from tablename
where instr(name, ',') = 0 
  and exists (
  select 1 from tablename t
  where t.id <> tablename.id and t.name = tablename.name 
  and ',' || t.city || ',' like '%,' || tablename.city || ',%'  
);

请参阅演示。
结果：

| id  | name  | city          |
| --- | ----- | ------------- |
| 1   | Alex  | Lima          |
| 3   | Mario | Glasgow       |
| 4   | Rose  | Madrid,Sidney |