mysql 判斷是否為子集的方法步驟
故事起源于一個查詢錯漏率的報表:有兩個查詢結(jié)果,分別是報告已經(jīng)添加的項目和報告應(yīng)該添加的項目,求報告無遺漏率
何為無遺漏?即,應(yīng)該添加的項目已經(jīng)被全部添加
報告無遺漏率也就是無遺漏報告數(shù)占報告總數(shù)的比率
這里以兩個報告示例(分別是已全部添加和有遺漏的報告)
首先,查出第一個結(jié)果——報告應(yīng)該添加的項目
SELECT r.id AS 報告ID,m.project_id 應(yīng)添加項目FROM report r INNER JOIN application a ON r.app_id=a.id INNER JOIN application_sample s ON a.id=s.app_id RIGHT JOIN application_sample_item si ON s.id=si.sample_id RIGHT JOIN set_project_mapping m ON si.set_id=m.set_idWHERE r.id IN (’44930’,’44927’)ORDER BY r.id,m.project_id;
然后,再查出第二個結(jié)果——報告已經(jīng)添加的項目
SELECT r.id AS 報告ID,i.project_id AS 已添加項目 FROM report r RIGHT JOIN report_item i ON r.id=i.report_idWHERE r.id IN (’44930’,’44927’);
以上就是我們要比較的結(jié)果集,不難看出報告44927是無遺漏的,而44930雖然項目數(shù)量一致,但實際是多添加了項目758,缺少了項目112,是有遺漏的報告
二、解決方案從問題看,顯然是一個判斷是否為子集的問題。可以分別遍歷已添加的項目和應(yīng)該添加的項目,如果應(yīng)該添加的項目在已添加的項目中都能匹配上,即代表應(yīng)該添加的項目是已添加的項目子集,也就是無遺漏。
通過循環(huán)遍歷比較確實可以解決這個問題,但是SQL中出現(xiàn)笛卡兒積的交叉連接往往意味著開銷巨大,查詢速度慢,那么有沒有辦法避免這一問題呢?
方案一:借助于函數(shù) FIND_IN_SET和GROUP_CONCAT, 首先認(rèn)識下兩個函數(shù)
FIND_IN_SET(str,strlist)
str: 需要查詢的字符串 strlist: 參數(shù)以英文”,”分隔,如 (1,2,6,8,10,22)FIND_IN_SET 函數(shù)返回了需要查詢的字符串在目標(biāo)字符串的位置
GROUP_CONCAT( [distinct] 要連接的字段 [order by 排序字段 asc/desc ] [separator ’分隔符’] )
GROUP_CONCAT()函數(shù)可以將多條記錄的同一字段的值,拼接成一條記錄返回。默認(rèn)以英文‘,’分割。
但是,GROUP_CONCAT()默認(rèn)長度為1024
所以,如果需要拼接的長度超過1024將會導(dǎo)致截取不全,需要修改長度
SET GLOBAL group_concat_max_len=102400;SET SESSION group_concat_max_len=102400;
從上述兩個函數(shù)介紹中,我們發(fā)現(xiàn)FIND_IN_SET和GROUP_CONCAT都以英文‘,’分割(加粗標(biāo)識)
所以,我們可以用GROUP_CONCAT將已添加項目的項目連接為一個字符串,然后再用FIND_IN_SET逐一查詢應(yīng)添加項目是否都存在于字符串
1、修改問題中描述中的SQL,用GROUP_CONCAT將已添加項目的項目連接為一個字符串
SELECT r.id,GROUP_CONCAT(i.project_id ORDER BY i.project_id,’’) AS 已添加項目列表 FROM report r LEFT JOIN report_item i ON r.id=i.report_idWHERE r.id IN (’44930’,’44927’)GROUP BY r.id;
2、用FIND_IN_SET逐一查詢應(yīng)添加項目是否都存在于字符串
SELECT Q.id,FIND_IN_SET(W.應(yīng)添加項目列表,Q.已添加項目列表) AS 是否遺漏 FROM ( -- 報告已經(jīng)添加的項目 SELECT r.id,GROUP_CONCAT(i.project_id ORDER BY i.project_id,’’) AS 已添加項目列表 FROM report r LEFT JOIN report_item i ON r.id=i.report_id WHERE r.id IN (’44930’,’44927’) GROUP BY r.id )Q, ( -- 報告應(yīng)該添加的項目 SELECT r.id,s.app_id,m.project_id 應(yīng)添加項目列表 FROM report r INNER JOIN application a ON r.app_id=a.id INNER JOIN application_sample s ON a.id=s.app_id INNER JOIN application_sample_item si ON s.id=si.sample_idINNER JOIN set_project_mapping m ON si.set_id=m.set_id WHERE r.id IN (’44930’,’44927’) ORDER BY r.id,m.project_id )W WHERE Q.id=W.id;
3、過濾掉有遺漏的報告
SELECT Q.id,CASE WHEN FIND_IN_SET(W.應(yīng)添加項目列表,Q.已添加項目列表)>0 THEN 1 ELSE 0 END AS 是否遺漏 FROM ( -- 報告已經(jīng)添加的項目 SELECT r.id,GROUP_CONCAT(i.project_id ORDER BY i.project_id,’’) AS 已添加項目列表 FROM report r LEFT JOIN report_item i ON r.id=i.report_id WHERE r.id IN (’44930’,’44927’) GROUP BY r.id )Q, ( -- 報告應(yīng)該添加的項目 SELECT r.id,s.app_id,m.project_id 應(yīng)添加項目列表 FROM report r INNER JOIN application a ON r.app_id=a.id INNER JOIN application_sample s ON a.id=s.app_id INNER JOIN application_sample_item si ON s.id=si.sample_idINNER JOIN set_project_mapping m ON si.set_id=m.set_id WHERE r.id IN (’44930’,’44927’) ORDER BY r.id,m.project_id )W WHERE Q.id=W.id GROUP BY Q.id HAVING COUNT(`是否遺漏`)=SUM(`是否遺漏`);
4、我們的最終目標(biāo)是求無遺漏率
SELECT COUNT(X.id) 無遺漏報告數(shù),Y.total 報告總數(shù), CONCAT(FORMAT(COUNT(X.id)/Y.total*100,2),’%’) AS 項目無遺漏率 FROM ( SELECT Q.id,CASE WHEN FIND_IN_SET(W.應(yīng)添加項目列表,Q.已添加項目列表)>0 THEN 1 ELSE 0 END AS 是否遺漏 FROM ( -- 報告已經(jīng)添加的項目 SELECT r.id,GROUP_CONCAT(i.project_id ORDER BY i.project_id,’’) AS 已添加項目列表 FROM report r LEFT JOIN report_item i ON r.id=i.report_id WHERE r.id IN (’44930’,’44927’) GROUP BY r.id )Q, ( -- 報告應(yīng)該添加的項目 SELECT r.id,s.app_id,m.project_id 應(yīng)添加項目列表 FROM report r INNER JOIN application a ON r.app_id=a.id INNER JOIN application_sample s ON a.id=s.app_id INNER JOIN application_sample_item si ON s.id=si.sample_idINNER JOIN set_project_mapping m ON si.set_id=m.set_id WHERE r.id IN (’44930’,’44927’) ORDER BY r.id,m.project_id )W WHERE Q.id=W.id GROUP BY Q.id HAVING COUNT(`是否遺漏`)=SUM(`是否遺漏`) )X, ( -- 總報告數(shù) SELECT COUNT(E.nums) AS total FROM ( SELECT COUNT(r.id) AS nums FROM report r WHERE r.id IN (’44930’,’44927’) GROUP BY r.id )E )Y ;
上述方案一雖然避免了逐行遍歷對比,但本質(zhì)上還是對項目的逐一對比,那么有沒有什么方式可以不用對比呢?
答案當(dāng)然是有的。我們可以根據(jù)統(tǒng)計數(shù)量判斷是否完全包含。
1、使用union all 將已添加項目與應(yīng)添加項目聯(lián)表,不去重
( -- 應(yīng)該添加的項目SELECT r.id,m.project_idFROM report r INNER JOIN application a ON r.app_id=a.idINNER JOIN application_sample s ON a.id=s.app_idINNER JOIN application_sample_item si ON s.id=si.sample_id INNER JOIN set_project_mapping m ON si.set_id=m.set_idWHERE r.id IN (’44930’,’44927’)ORDER BY r.id,m.project_id)UNION ALL( -- 已經(jīng)添加的項目select r.id,i.project_id from report r,report_item i where r.id = i.report_id and r.id IN (’44930’,’44927’)group by r.app_id,i.project_id )
從結(jié)果可以看出,項目同一個報告下有重復(fù)的項目,分別代表了應(yīng)該添加和已經(jīng)添加的項目
2、根據(jù)聯(lián)表結(jié)果,統(tǒng)計報告重合的項目數(shù)量
# 應(yīng)該添加與已經(jīng)添加的項目重疊數(shù)量select tt.id,count(*) count from ( select t.id,t.project_id,count(*) from ( (-- 應(yīng)該添加的項目SELECT r.id,m.project_idFROM report r INNER JOIN application a ON r.app_id=a.id INNER JOIN application_sample s ON a.id=s.app_id INNER JOIN application_sample_item si ON s.id=si.sample_id INNER JOIN set_project_mapping m ON si.set_id=m.set_idWHERE r.id IN (’44930’,’44927’)ORDER BY r.id,m.project_id ) UNION ALL (-- 已經(jīng)添加的項目select r.id,i.project_id from report r,report_item i where r.id = i.report_id and r.id IN (’44930’,’44927’)group by r.app_id,i.project_id ) ) t GROUP BY t.id,t.project_id HAVING count(*) >1 ) tt group by tt.id
3、將第二步的數(shù)量與應(yīng)該添加的數(shù)量作比較,如果相等,則代表無遺漏
select bb.id,aa.count 已添加,bb.count 需添加, CASE WHEN aa.count/bb.count=1 THEN 1 ELSE 0 END AS ’是否遺漏’ from (# 應(yīng)該添加與已經(jīng)添加的項目重疊數(shù)量select tt.id,count(*) count from ( select t.id,t.project_id,count(*) from ( (-- 應(yīng)該添加的項目SELECT r.id,m.project_idFROM report r INNER JOIN application a ON r.app_id=a.id INNER JOIN application_sample s ON a.id=s.app_id INNER JOIN application_sample_item si ON s.id=si.sample_id INNER JOIN set_project_mapping m ON si.set_id=m.set_idWHERE r.id IN (’44930’,’44927’)ORDER BY r.id,m.project_id ) UNION ALL (-- 已經(jīng)添加的項目select r.id,i.project_id from report r,report_item i where r.id = i.report_id and r.id IN (’44930’,’44927’)group by r.app_id,i.project_id ) ) t GROUP BY t.id,t.project_id HAVING count(*) >1 ) tt group by tt.id ) aa RIGHT JOIN( -- 應(yīng)該添加的項目數(shù)量 SELECT r.id,s.app_id,COUNT(m.project_id) count FROM report r INNER JOIN application a ON r.app_id=a.id INNER JOIN application_sample s ON a.id=s.app_id INNER JOIN application_sample_item si ON s.id=si.sample_id INNER JOIN set_project_mapping m ON si.set_id=m.set_id WHERE r.id IN (’44930’,’44927’) GROUP BY r.id ORDER BY r.id,m.project_id) bb ON aa.id = bb.id ORDER BY aa.id
4、求出無遺漏率
select SUM(asr.`是否遺漏`) AS 無遺漏數(shù),COUNT(asr.id) AS 總數(shù),CONCAT(FORMAT(SUM(asr.`是否遺漏`)/COUNT(asr.id)*100,5),’%’) AS 報告無遺漏率from ( select bb.id,aa.count 已添加,bb.count 需添加, CASE WHEN aa.count/bb.count=1 THEN 1 ELSE 0 END AS ’是否遺漏’ from ( # 應(yīng)該添加與已經(jīng)添加的項目重疊數(shù)量 select tt.id,count(*) count from ( select t.id,t.project_id,count(*) from (( -- 應(yīng)該添加的項目 SELECT r.id,m.project_id FROM report r INNER JOIN application a ON r.app_id=a.id INNER JOIN application_sample s ON a.id=s.app_id INNER JOIN application_sample_item si ON s.id=si.sample_id INNER JOIN set_project_mapping m ON si.set_id=m.set_id WHERE r.id IN (’44930’,’44927’) ORDER BY r.id,m.project_id)UNION ALL( -- 已經(jīng)添加的項目 select r.id,i.project_id from report r,report_item i where r.id = i.report_id and r.id IN (’44930’,’44927’) group by r.app_id,i.project_id) ) t GROUP BY t.id,t.project_id HAVING count(*) >1 ) tt group by tt.id ) aa RIGHT JOIN ( -- 應(yīng)該添加的項目數(shù)量 SELECT r.id,s.app_id,COUNT(m.project_id) count FROM report r INNER JOIN application a ON r.app_id=a.id INNER JOIN application_sample s ON a.id=s.app_id INNER JOIN application_sample_item si ON s.id=si.sample_id INNER JOIN set_project_mapping m ON si.set_id=m.set_id WHERE r.id IN (’44930’,’44927’) GROUP BY r.id ORDER BY r.id,m.project_id ) bb ON aa.id = bb.id ORDER BY aa.id) asr;
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