一、需求
我們有三張表,我們需要分類統(tǒng)計一段時間內(nèi)抗生素的不同藥敏結(jié)果,即 report_item_drugs 表的 drugs_result, 在不同項目project_name 和不同抗生素 antibiotic_dict_name 下的占比,并將藥敏結(jié)果顯示在行上,效果如下:
![](/d/20211017/b3ff69511120a523525a255a4bbd9aa9.gif)
三張原始表(僅取需要的字段示例),分別是:
報告表
![](/d/20211017/9c1576246dbcf9591a39a996873a2c41.gif)
項目表
![](/d/20211017/e7de10a885f640cf237bf0c3a297bee7.gif)
抗生素表(藥敏結(jié)果drugs_result為一列值)
![](/d/20211017/94d9e2fcf8cadd1a2d082cc162f3cb81.gif)
二、實現(xiàn)
1、按照項目、抗生素分組求出檢出的總數(shù)
SELECT
A.project_name,A.antibiotic_dict_name,SUM(nums) AS 檢出總數(shù)
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
![](/d/20211017/9c0d984fd0aa1e2a34afb5507f2b2ebc.gif)
2、按照項目、抗生素、藥敏結(jié)果求出不同藥敏結(jié)果數(shù)量
SELECT i.project_name,d.antibiotic_dict_name,IF(d.drugs_result>'', d.drugs_result, '未填寫') AS drugs_result,COUNT(d.id) AS 數(shù)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
![](/d/20211017/330feb4b4dd60751090577e6eb0dc5bc.gif)
3、將兩個結(jié)果關(guān)聯(lián)到一起
SELECT
BB.project_name,BB.antibiotic_dict_name,BB.drugs_result,BB.`數(shù)量`,AA.`檢出總數(shù)`
FROM
(
SELECT
A.project_name,A.antibiotic_dict_name,SUM(nums) AS 檢出總數(shù)
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, '未填寫') AS drugs_result,COUNT(d.id) AS 數(shù)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.`檢出總數(shù)`>''
![](/d/20211017/580c09d021ee391b2a540aa3964950ec.gif)
4、一般來說,到上一步不同藥敏數(shù)量和總數(shù)都有了,可以直接求比例了
但是,我們需要的是將藥敏顯示到行上,直接求比不符合需求,所以我們需要將列轉(zhuǎn)換為行
我們借助于case when實現(xiàn)行列轉(zhuǎn)換,并將藥敏結(jié)果根據(jù)字典轉(zhuǎn)為方便閱讀的漢字
![](/d/20211017/f3f9956a39fb0c0445caa87b0c045abd.gif)
SELECT
C.project_name 項目名稱,C.antibiotic_dict_name 抗生素名稱,C.`檢出總數(shù)`,
SUM(CASE C.`drugs_result` WHEN 'D' THEN C.`數(shù)量` ELSE 0 END ) AS '劑量依賴性敏感',
CONCAT(SUM(CASE C.`drugs_result` WHEN 'D' THEN FORMAT(C.`數(shù)量`/C.`檢出總數(shù)`*100,2) ELSE 0 END),'%') AS '劑量依賴性敏感比率',
SUM(CASE C.`drugs_result` WHEN 'R' THEN C.`數(shù)量` ELSE 0 END ) AS '耐藥',
CONCAT(SUM(CASE C.`drugs_result` WHEN 'R' THEN FORMAT(C.`數(shù)量`/C.`檢出總數(shù)`*100,2) ELSE 0 END),'%') AS '耐藥比率',
SUM(CASE C.`drugs_result` WHEN 'S' THEN C.`數(shù)量` ELSE 0 END ) AS '敏感',
CONCAT(SUM(CASE C.`drugs_result` WHEN 'S' THEN FORMAT(C.`數(shù)量`/C.`檢出總數(shù)`*100,2) ELSE 0 END),'%') AS '敏感比率',
SUM(CASE C.`drugs_result` WHEN 'I' THEN C.`數(shù)量` ELSE 0 END ) AS '中介',
CONCAT(SUM(CASE C.`drugs_result` WHEN 'I' THEN FORMAT(C.`數(shù)量`/C.`檢出總數(shù)`*100,2) ELSE 0 END),'%') AS '中介比率',
SUM(CASE C.`drugs_result` WHEN 'n1' THEN C.`數(shù)量` ELSE 0 END ) AS '非敏感',
CONCAT(SUM(CASE C.`drugs_result` WHEN 'n1' THEN FORMAT(C.`數(shù)量`/C.`檢出總數(shù)`*100,2) ELSE 0 END),'%') AS '非敏感比率',
SUM(CASE C.`drugs_result` WHEN 'N' THEN C.`數(shù)量` ELSE 0 END ) AS '無',
CONCAT(SUM(CASE C.`drugs_result` WHEN 'N' THEN FORMAT(C.`數(shù)量`/C.`檢出總數(shù)`*100,2) ELSE 0 END),'%') AS '無比率',
SUM(CASE C.`drugs_result` WHEN '未填寫' THEN C.`數(shù)量` ELSE 0 END ) AS '未填寫',
CONCAT(SUM(CASE C.`drugs_result` WHEN '未填寫' THEN FORMAT(C.`數(shù)量`/C.`檢出總數(shù)`*100,2) ELSE 0 END),'%') AS '未填寫比率'
FROM
(
SELECT
BB.project_name,BB.antibiotic_dict_name,BB.drugs_result,BB.`數(shù)量`,AA.`檢出總數(shù)`
FROM
(
SELECT
A.project_name,A.antibiotic_dict_name,SUM(nums) AS 檢出總數(shù)
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, '未填寫') AS drugs_result,COUNT(d.id) AS 數(shù)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.`檢出總數(shù)`>''
) C
GROUP BY C.project_name,C.antibiotic_dict_name;
5、查看結(jié)果,成功轉(zhuǎn)換
![](/d/20211017/07f38bd153f01fe88951aa0c4eeba670.gif)
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