php - 如何将多个值分配给php/mysql中另一个表的值

Question

this is my requirement_tbl

and my documents_tbl

和我的预期结果

-------------------------------------------------------------
| doc_id | document_name | requirement_id |requirement_name | 
-------------------------------------------------------------
|   1    |Certifications |       3        | requirementname |
|        |               |       1        | requirementname |
|        |               |       7        | requirementname |
|        |               |       9        | requirementname |
|    2   |CAV            |       1        | requirementname |
|        |               |       4        | requirementname |
|        |               |       3        | requirementname |
|        |               |       5        | requirementname |
-------------------------------------------------------------

此输出是否可以使用 PHP 在 HTML 中显示？有人可以告诉我怎么做吗？ 这与外键和连接表有关吗？ 提前致谢。

Answer 1

构建一个 sqlfiddle 会很有帮助对于诸如此类的问题，以便审阅者可以遵循通用的模式/数据集。给定以下架构:

CREATE TABLE `document` (
  `document_id` INT(11) PRIMARY KEY AUTO_INCREMENT,
  `document_name` VARCHAR(64) DEFAULT NULL,
  `document_amount` INT(4),
  `requirement1` INT(11) DEFAULT 0,
  `requirement2` INT(11) DEFAULT 0,
  `requirement3` INT(11) DEFAULT 0,
  `requirement4` INT(11) DEFAULT 0,
  `requirement5` INT(11) DEFAULT 0,
  `requirement6` INT(11) DEFAULT 0,
  `requirement7` INT(11) DEFAULT 0
);

CREATE TABLE `requirement` (
  `requirement_id` INT(11) PRIMARY KEY AUTO_INCREMENT,
  `document_id` INT(11) DEFAULT 0,
  `requirement_name` VARCHAR(64) DEFAULT NULL
);

主要障碍是每个文件的要求是非规范化的；可能有多达 7 个不同的需求选项分布在每个文档栏中。为此，您需要将它们组合在一起，使它们代表一个列，这可以通过 union 完成:

SELECT `document_id`, `requirement1` as 'requirement' FROM `document`
UNION
SELECT `document_id`, `requirement2` as 'requirement' FROM `document`
UNION
SELECT `document_id`, `requirement3` as 'requirement' FROM `document`
UNION
SELECT `document_id`, `requirement4` as 'requirement' FROM `document`
UNION
SELECT `document_id`, `requirement5` as 'requirement' FROM `document`
UNION
SELECT `document_id`, `requirement6` as 'requirement' FROM `document`
UNION
SELECT `document_id`, `requirement7` as 'requirement' FROM `document`

然后您就可以将您的查询连接到所需的结果集中:

SELECT d.document_id, d.document_name, r.requirement_id, r.requirement_name FROM
`document` as `d` JOIN
(
# correct for non-normalization via UNION on requirements
SELECT `document_id`, `requirement1` as 'requirement' FROM `document`
UNION
SELECT `document_id`, `requirement2` as 'requirement' FROM `document`
UNION
SELECT `document_id`, `requirement3` as 'requirement' FROM `document`
UNION
SELECT `document_id`, `requirement4` as 'requirement' FROM `document`
UNION
SELECT `document_id`, `requirement5` as 'requirement' FROM `document`
UNION
SELECT `document_id`, `requirement6` as 'requirement' FROM `document`
UNION
SELECT `document_id`, `requirement7` as 'requirement' FROM `document`
) `req`
ON d.document_id = req.document_id
JOIN `requirement` as `r`
ON req.requirement = r.requirement_id
WHERE req.requirement > 0
ORDER BY d.document_id, r.requirement_id;