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` rright join report_item i on r.id=i.report_idright join report_item_drugs d on d.report_item_id=i.idwhere 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`<>'' ) cgroup 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
