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

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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

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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

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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

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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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原文链接：https://blog.csdn.net/kk_gods/article/details/111933336