mysql 行列转换的示例代码

一、需求

我们有三张表，我们需要分类统计一段时间内抗生素的不同药敏结果，即 report_item_drugs 表的 drugs_result， 在不同项目project_name 和不同抗生素 antibiotic_dict_name 下的占比，并将药敏结果显示在行上，效果如下：

三张原始表（仅取需要的字段示例），分别是：

报告表

项目表

抗生素表（药敏结果drugs_result为一列值）

二、实现

1、按照项目、抗生素分组求出检出的总数

SELECT
 A.project_name,A.antibiotic_dict_name,SUM(nums) AS 检出总数
FROM
(
      SELECT i.project_name,d.antibiotic_dict_name,d.drugs_result,COUNT(d.id) AS nums FROM `report` r
       RIGHT JOIN report_item i ON r.id=i.report_id
       RIGHT JOIN report_item_drugs d ON d.report_item_id=i.id
       WHERE r.report_status=2 AND r.add_date BETWEEN '2020-01-01' AND '2020-12-30'
       GROUP BY i.project_id,d.antibiotic_dict_id,d.drugs_result
 )  A
 GROUP BY A.project_name,A.antibiotic_dict_name

2、按照项目、抗生素、药敏结果求出不同药敏结果数量

SELECT i.project_name,d.antibiotic_dict_name,IF(d.drugs_result<>'', d.drugs_result, '未填写') AS drugs_result,COUNT(d.id) AS 数量
FROM `report` r
RIGHT JOIN report_item i ON r.id=i.report_id
RIGHT JOIN report_item_drugs d ON d.report_item_id=i.id
WHERE r.report_status=2 AND r.add_date BETWEEN '2020-01-01' AND '2020-12-30'
GROUP BY i.project_id,d.antibiotic_dict_id,d.drugs_result

3、将两个结果关联到一起

SELECT
      BB.project_name,BB.antibiotic_dict_name,BB.drugs_result,BB.`数量`,AA.`检出总数`
    FROM
        (
              SELECT
                A.project_name,A.antibiotic_dict_name,SUM(nums) AS 检出总数
              FROM
              (
                    SELECT i.project_name,d.antibiotic_dict_name,d.drugs_result,COUNT(d.id) AS nums FROM `report` r
                    RIGHT JOIN report_item i ON r.id=i.report_id
                    RIGHT JOIN report_item_drugs d ON d.report_item_id=i.id
                    WHERE r.report_status=2 AND r.add_date BETWEEN '2020-01-01' AND '2020-12-30'
                    GROUP BY i.project_id,d.antibiotic_dict_id,d.drugs_result
              )  A
              GROUP BY A.project_name,A.antibiotic_dict_name
        ) AA
        RIGHT JOIN
        (
              SELECT i.project_name,d.antibiotic_dict_name,IF(d.drugs_result<>'', d.drugs_result, '未填写') AS drugs_result,COUNT(d.id) AS 数量
              FROM `report` r
              RIGHT JOIN report_item i ON r.id=i.report_id
              RIGHT JOIN report_item_drugs d ON d.report_item_id=i.id
              WHERE r.report_status=2 AND r.add_date BETWEEN '2020-01-01' AND '2020-12-30'
              GROUP BY i.project_id,d.antibiotic_dict_id,d.drugs_result
        )BB ON AA.project_name=BB.project_name AND AA.antibiotic_dict_name=BB.antibiotic_dict_name
    WHERE AA.`检出总数`<>''

4、一般来说，到上一步不同药敏数量和总数都有了，可以直接求比例了

但是，我们需要的是将药敏显示到行上，直接求比不符合需求，所以我们需要将列转换为行

我们借助于case when实现行列转换，并将药敏结果根据字典转为方便阅读的汉字

SELECT
  C.project_name 项目名称,C.antibiotic_dict_name 抗生素名称,C.`检出总数`,
  SUM(CASE C.`drugs_result` WHEN 'D' THEN C.`数量` ELSE 0 END ) AS '剂量依赖性敏感',
  CONCAT(SUM(CASE C.`drugs_result` WHEN 'D' THEN FORMAT(C.`数量`/C.`检出总数`*100,2) ELSE 0 END),'%') AS '剂量依赖性敏感比率',
  SUM(CASE C.`drugs_result` WHEN 'R' THEN C.`数量` ELSE 0 END ) AS '耐药',
  CONCAT(SUM(CASE C.`drugs_result` WHEN 'R' THEN FORMAT(C.`数量`/C.`检出总数`*100,2) ELSE 0 END),'%') AS '耐药比率',
  SUM(CASE C.`drugs_result` WHEN 'S' THEN C.`数量` ELSE 0 END ) AS '敏感',
  CONCAT(SUM(CASE C.`drugs_result` WHEN 'S' THEN FORMAT(C.`数量`/C.`检出总数`*100,2) ELSE 0 END),'%') AS '敏感比率',
  SUM(CASE C.`drugs_result` WHEN 'I' THEN C.`数量` ELSE 0 END ) AS '中介',
  CONCAT(SUM(CASE C.`drugs_result` WHEN 'I' THEN FORMAT(C.`数量`/C.`检出总数`*100,2) ELSE 0 END),'%') AS '中介比率',
  SUM(CASE C.`drugs_result` WHEN 'n1' THEN C.`数量` ELSE 0 END ) AS '非敏感',
  CONCAT(SUM(CASE C.`drugs_result` WHEN 'n1' THEN FORMAT(C.`数量`/C.`检出总数`*100,2) ELSE 0 END),'%') AS '非敏感比率',
  SUM(CASE C.`drugs_result` WHEN 'N' THEN C.`数量` ELSE 0 END ) AS '无',
  CONCAT(SUM(CASE C.`drugs_result` WHEN 'N' THEN FORMAT(C.`数量`/C.`检出总数`*100,2) ELSE 0 END),'%') AS '无比率',
  SUM(CASE C.`drugs_result` WHEN '未填写' THEN C.`数量` ELSE 0 END ) AS '未填写',
  CONCAT(SUM(CASE C.`drugs_result` WHEN '未填写' THEN FORMAT(C.`数量`/C.`检出总数`*100,2) ELSE 0 END),'%') AS '未填写比率'
FROM
(
    SELECT
      BB.project_name,BB.antibiotic_dict_name,BB.drugs_result,BB.`数量`,AA.`检出总数`
    FROM
        (
              SELECT
                A.project_name,A.antibiotic_dict_name,SUM(nums) AS 检出总数
              FROM
              (
                    SELECT i.project_name,d.antibiotic_dict_name,d.drugs_result,COUNT(d.id) AS nums FROM `report` r
                    RIGHT JOIN report_item i ON r.id=i.report_id
                    RIGHT JOIN report_item_drugs d ON d.report_item_id=i.id
                    WHERE r.report_status=2 AND r.add_date BETWEEN '2020-01-01' AND '2020-12-30'
                    GROUP BY i.project_id,d.antibiotic_dict_id,d.drugs_result
              )  A
              GROUP BY A.project_name,A.antibiotic_dict_name
        ) AA
        RIGHT JOIN
        (
              SELECT i.project_name,d.antibiotic_dict_name,IF(d.drugs_result<>'', d.drugs_result, '未填写') AS drugs_result,COUNT(d.id) AS 数量
              FROM `report` r
              RIGHT JOIN report_item i ON r.id=i.report_id
              RIGHT JOIN report_item_drugs d ON d.report_item_id=i.id
              WHERE r.report_status=2 AND r.add_date BETWEEN '2020-01-01' AND '2020-12-30'
              GROUP BY i.project_id,d.antibiotic_dict_id,d.drugs_result
        )BB ON AA.project_name=BB.project_name AND AA.antibiotic_dict_name=BB.antibiotic_dict_name
    WHERE AA.`检出总数`<>''
) C
GROUP BY C.project_name,C.antibiotic_dict_name;

5、查看结果，成功转换

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