Question

如何通过两个表中的字段交叉/存在来排序数据顺序？
序列存在于文件夹表和文件表中，如何对它们进行排序

查询

SELECT * FROM folder fo 
  LEFT JOIN file fi ON fi.parent_folder_id = fo.id
WHERE fo.parent_folder_id = $1 AND fi.parent_folder_id = $1
  ORDER BY  fo.sequence fi.sequence ??   ;    <<  my problem 

[1]

数据示例

folder
id | sequence | parent_folder_id | name
1  | 0        |                  | root
2  | 0        | 1                | 
3  | 2        | 1                |

file
id | sequence | parent_folder_id |
1  | 1        | 1                |

输出

folder(id:1, sequence:0 name:root)
  folder(id:2, sequence:0)
  file(id:1, sequence:1)
  folder(id:3 sequence:2)

Answer 1

如果要列出给定文件夹第一级下的所有文件和文件夹，我建议您使用UNION ALL并获取文件列表和文件夹列表：

(
    SELECT 'folder' AS type, fo.id, fo.sequence, fo.name
    FROM folder fo
    WHERE fo.parent_folder_id = 1
)
UNION ALL
(
    SELECT 'file' AS type, fi.id, fi.sequence, NULL
    FROM file fi
    WHERE fi.parent_folder_id = 1
)
ORDER BY sequence;

你在文件表中没有名字很奇怪，但我认为这可能只是对问题的过度简化。如果您还想在列表中包含父文件夹，我只需将其添加为另一个UNION ALL子查询，其中包含固定的较低顺序：

(
    SELECT 'folder' AS type, fo.id, -1 AS sequence, fo.name
    FROM folder fo
    WHERE fo.id = 1
)
UNION ALL
(
    SELECT 'folder' AS type, fo.id, fo.sequence, fo.name
    FROM folder fo
    WHERE fo.parent_folder_id = 1
)
UNION ALL
(
    SELECT 'file' AS type, fi.id, fi.sequence, NULL
    FROM file fi
    WHERE fi.parent_folder_id = 1
)
ORDER BY sequence;

如何通过两个表中的字段交叉/存在对数据顺序进行排序

1 个答案: