Question

CREATE TABLE `locationcodes` (
  `id` int,
  `customer` varchar(100),
  `locationcode` varchar(50),
  `parentid` int
);

insert into locationcodes values (1, 'Test, Inc.', 'California', NULL);
insert into locationcodes values (2, 'Test, Inc.', 'Los Angeles', 1);
insert into locationcodes values (3, 'Test, Inc.', 'San Francisco', 1);
insert into locationcodes values (4, 'Test, Inc.', 'Sacramento', 1);

我想要一份父母地点及其子女的名单。如果没有孩子，则打印父母：

SQL：

SELECT DISTINCT parent.locationcode as 'Parent', parent.locationcode as 'Child', 1 AS `level`
FROM locationcodes parent
JOIN locationcodes child ON parent.id = child.parentid
WHERE parent.parentid IS NULL
AND parent.customer = 'Test, Inc.'
UNION
SELECT DISTINCT parent.locationcode as 'Parent', child.locationcode as 'Child', 2 AS `level`
FROM locationcodes parent 
JOIN locationcodes child ON parent.id = child.parentid
WHERE NOT child.parentid IS NULL
AND child.customer = 'Test, Inc.'
ORDER BY 1, 2

结果是正确的：

PARENT          CHILD           LEVEL
California      California          1
California      Los Angeles         2
California      Sacramento          2
California      San Francisco       2

我的问题是我是否尽可能高效地编写了SQL？

http://sqlfiddle.com/#!2/87a3d/3

Answer 1

如何只使用一个JOIN和一些内联IF？

SELECT IFNULL(parent.locationcode,child.locationcode) AS 'Parent',  child.locationcode AS 'Child', IF(child.parentid,2,1) AS `level`
FROM locationcodes child 
LEFT JOIN locationcodes parent ON parent.id = child.parentid
WHERE child.customer = 'Test, Inc.'
ORDER BY 1, 2

这将选择所有（子）位置，然后尝试尽可能加入父数据。应该比使用UNION和另一个JOIN更高效（而且简单！）。

演示：http://sqlfiddle.com/#!2/87a3d/21/0

在MySQL中，如何以最有效的方式编写此SQL查询？

1 个答案: