Question

我有一张桌子：

CREATE TABLE IF NOT EXISTS `columns` (
`id` int(11) NOT NULL AUTO_INCREMENT,
`section_id` int(11) NOT NULL,
`columntype` int(11) NOT NULL,
`name` varchar(128) COLLATE utf8_bin NOT NULL,
`refer_type` int(11) NOT NULL,
PRIMARY KEY (`id`));

具有以下内容：

INSERT INTO `columns` (`id`, `section_id`, `columntype`, `name`, `refer_type`) VALUES
(9, 35, 1, 'Headline1', 0),
(10, 36, 1, 'Headline2', 0),
(11, 36, 5, 'Headline3', 10),
(12, 36, 3, 'Headline4', 0),
(13, 36, 4, 'Headline5', 10);

列refer_from包含有关必须从中提取和计算包含数据的其他列的信息。在我的webapp中更改Headline2时，应更改Headline3和Headline5的内容。我正在webapp中执行最终计算，但是我需要一个能给我一个结果的SQL，比如

    id    name      columntype refer_type
    10    Headline2 1          4, 5
    11    Headline3 5          null
    12    Headline4 3          null
    13    Headline5 4          null

我尝试使用类似SELECT id, name, columntype, (SELECT GROUP_CONCAT(refer_type SEPARATOR "," )...的内容，但在所有四个结果行上都给了4,5。

提前致谢！

Answer 1

如果我做对了，你可能需要这样的选择：

SELECT c.id, c.name, c.columntype, ref.reftypes
FROM columns c LEFT JOIN
      (SELECT GROUP_CONCAT(columntype SEPARATOR  "," ) reftypes, refer_type
       FROM columns
       GROUP BY refer_type) ref
ON c.id = ref.refer_type;

Answer 2

这样做：

SELECT c1.id, c1.name, c1.columntype, GROUP_CONCAT(c2.columntype ORDER BY c2.columntype) AS refer_type
FROM columns AS c1
LEFT JOIN columns AS c2 ON c1.id = c2.refer_type
GROUP BY c1.id
ORDER BY c1.id

DEMO

Answer 3

谢谢！这两个SQL都很棒！由于简单，我选择了Bamar的答案。

自引用MySQL表

3 个答案: