SQL（之一）-SQL经典题目

说明：以下SQL练习题（主要针对DBMS为ORACLE），题目来自网络，本人整理编辑完成，难度分为1-5星。

涉及的4个表的表间关系：

--0-查询每个学生、最高学科的成绩、及最高成绩的所在学科在班级中排名-★★★

--RANK() OVER(PARTITION BY XXX ORDER BY XXX ) 使用分析函数
SELECT A.SID,A.CID,A.SCORE,DENSE_RANK() OVER(PARTITION BY A.CID ORDER BY A.SCORE DESC NULLS LAST) AS 学生最高分数所在该科目排名,A.TOTAL_MAX_SCORE 当前科目最高分FROM (SELECT T.SID,T.CID,T.SCORE,MAX(T.SCORE) OVER(PARTITION BY T.SID) PRESON_MAX_SCORE, --个人最高分MAX(T.SCORE) OVER(PARTITION BY T.CID) TOTAL_MAX_SCORE --当前科目最高分FROM SC T) AWHERE A.SCORE = A.PRESON_MAX_SCOREORDER BY 2;

--1、查询"001"课程比"002"课程成绩高的所有学生的学号-★

SELECT T1.SID ,T1.SCORE ,T2.SCORE FROM SC T1
LEFT JOIN SC T2 ON T1.SID=T2.SID
WHERE T1.CID='001' AND T2.CID='002' AND T1.SCORE >T2.SCORE  ORDER BY 1;

--2、查询平均成绩大于60分的同学的学号和平均成绩-★

SELECT T1.SID ,ROUND(AVG(T1.SCORE),2) AVG_SCORE FROM SC T1
GROUP BY T1.SID  HAVING AVG(T1.SCORE)>60  ORDER BY 2 DESC,1 ;

--3、查询所有同学的学号、姓名、选课数、总成绩-★

SELECT T1.SID   ,T2.SNAME  ,COUNT(T1.CID) COUNT_CID  ,SUM(T1.SCORE)SUM_SCORE  FROM SC T1
LEFT JOIN STUDENT T2 ON T1.SID=T2.SID
GROUP BY T1.SID  ,T2.SNAME  ORDER BY 1,4 DESC,3 ;

--4、查询姓"李"的老师的个数-★

SELECT COUNT (1) FROM TEACHER T1 WHERE T1.TNAME LIKE '李%';

--5A、查询没学过"叶平"老师课的同学的学号、姓名-★★
--5B、查询没学过"叶平"老师讲授的任一门课程的学生姓名-★★

--思路：反其道而行之,查询学过"叶平"老师 任意课程 的学生,然后排除这些学生
--方式1：
SELECT DISTINCT T4.SID,T4.SNAME FROM STUDENT  T4 WHERE T4.SID NOT IN (
SELECT T3.SID FROM SC T3 WHERE T3.CID  IN (SELECT T2.CID FROM TEACHER T1 ,COURSE T2 WHERE T1.TID=T2.TID  AND  T1.TNAME ='叶平' )) ORDER BY 1 ;--方式2
SELECT  DISTINCT T4.SID,T4.SNAME  FROM STUDENT T4 WHERE T4.SID NOT IN (
SELECT DISTINCT  T1.SID FROM SC T1 WHERE    EXISTS
(SELECT * FROM COURSE T2, TEACHER T3 WHERE T3.TNAME='叶平'  AND T1.CID=T2.CID AND T2.TID=T3.TID)
)ORDER BY 1;

--6、查询学过"001"并且也学过编号"002"课程的同学的学号、姓名-★

SELECT DISTINCT T1.SID ,T3.SNAME   FROM SC T1LEFT JOIN SC T2 ON T1.SID=T2.SID LEFT JOIN STUDENT T3 ON T3.SID=T1.SID AND T3.SID=T2.SID
WHERE T1.CID='001' AND T2.CID='002'
ORDER BY 1 ;
--或者
SELECT T1.SID ,T3.SNAME  FROM SC T1,SC T2 ,STUDENT T3 WHERE  T1.CID='001'AND T2.CID='002' AND T1.SID=T2.SID
AND T1.SID=T3.SID
ORDER BY 1 ;

--7、查询学过"叶平"老师所教的所有课的同学的学号、姓名-★★

SELECT A.SID,B.SNAME FROM (SELECT  DISTINCT T1.SID,T1.CID  FROM SC T1 WHERE  T1.CID IN(SELECT T2.CID FROM TEACHER T1 ,COURSE T2 WHERE T1.TID=T2.TID  AND  T1.TNAME ='叶平')--查询出叶平老师所教课程的课程ID
)A --临时表
LEFT JOIN STUDENT B ON A.SID=B.SID GROUP BY A.SID,B.SNAME
HAVING COUNT(DISTINCT A.CID)>=
(SELECT COUNT(DISTINCT T2.CID ) FROM TEACHER T1 ,COURSE T2 WHERE T1.TID=T2.TID AND  T1.TNAME ='叶平')--查询出叶平老师所教课程的课程数量
ORDER BY 1  ;

--8、查询课程编号"002"的成绩比课程编号"001"课程低的所有同学的学号、姓名-★

SELECT T1.SID , T3.SNAME ,T1.SCORE AS SC_002 ,T2.SCORE  AS SC_001 FROM SC T1 LEFT JOIN SC T2 ON T1.SID=T2.SID LEFT JOIN STUDENT T3 ON T1.SID=T3.SID AND T3.SID=T2.SID
WHERE T1.CID='002' AND T2.CID='001' AND T1.SCORE<T2.SCORE
ORDER BY 3 DESC,1 ;

--9、查询所有课程成绩小于60分的同学的学号、姓名-★

--思路：若最大的成绩小于60分，则该同学所有课程成绩必定都小于60分
SELECT S1.SID 学号, S1.SNAME 姓名FROM STUDENT S1, SC S2WHERE S1.SID = S2.SIDGROUP BY S1.SID, S1.SNAME HAVING MAX(S2.SCORE) < 60;

--10、查询没有学全所有课的同学的学号、姓名-★★

SELECT T1.SID, T2.SNAME ,COUNT(DISTINCT T1.CID) AS 学生选课数量FROM SC T1,STUDENT T2WHERE T1.SID=T2.SID GROUP BY T1.SID,T2.SNAME
HAVING COUNT(DISTINCT T1.CID) <
(SELECT COUNT(DISTINCT T2.CID) AS SC_COUNT FROM COURSE T2)--总的课程数量。小于总的课程数量即为"没有学全所有课"ORDER BY 1;

--11A、查询至少有一门课与学号为"1001"的同学所学相同的同学的学号和姓名-★

SELECT SID 学号, SNAME 姓名 FROM STUDENT  WHERE SID IN(SELECT DISTINCT SID FROM SC WHERE CID IN(SELECT CID FROM SC  WHERE SID='1001' --查出'1001'同学所学选的所有课程ID) ) ORDER BY 1;

--11B、查询至少学过学号为"1001"同学所有一门课的其他同学学号和姓名-★
--与上题11A基本类似。区别：把1001同学排除掉，因为此题要求是"其他同学学号和姓名"

SELECT SID 学号, SNAME 姓名 FROM STUDENT  WHERE SID IN(SELECT DISTINCT SID FROM SC WHERE CID IN(SELECT CID FROM SC  WHERE SID='1001' --查出'1001'同学所学选的所有课程ID) ) AND SID<> '1001' ORDER BY 1;

--12 、查询各科成绩最高和最低的分：以如下形式显示：课程ID ，最高分，最低分-★

SELECT T.CID 课程ID ,MAX(T.SCORE)最高分 ,MIN(T.SCORE) 最低分 FROM SC T  GROUP BY T.CID ORDER BY 1 ;

--13、把"SC"表中"叶平"老师教的课的成绩都更改为此课程的平均成绩-★★★

--*推荐方法1：使用 MERGE INTO函数
MERGE INTO SC T1
USING (SELECT T3.CID ,AVG(T3.SCORE) AVG_SCOREFROM SC T3 WHERE  T3.CID IN (SELECT T2.CID FROM COURSE T2 WHERE T2.TID IN(SELECT T1.TID FROM TEACHER  T1 WHERE T1.TNAME='叶平'))GROUP BY T3.CID
) A --当成一个临时表A
ON (T1.CID=A.CID) --ON 条件要用括号
WHEN MATCHED THENUPDATE SET T1.SCORE=A.AVG_SCORE ;--方法2 较为繁琐，但是思路清晰，容易理解
UPDATE SC T4
SET T4.SCORE=(SELECT A.AVG_SCORE FROM (SELECT T3.CID ,AVG(T3.SCORE) AVG_SCOREFROM SC T3 WHERE  T3.CID IN (SELECT T2.CID FROM COURSE T2 WHERE T2.TID IN(SELECT T1.TID FROM TEACHER  T1 WHERE T1.TNAME='叶平'))GROUP BY T3.CID --当成一个临时表A
)   A WHERE A.CID=T4.CID  )WHERE T4.CID IN (SELECT  B.CID FROM  --要加限制条件，否则不匹配的学科 成绩会更新为NULL
(SELECT T3.CID ,AVG(T3.SCORE) AVG_SCOREFROM SC T3 WHERE  T3.CID IN (SELECT T2.CID FROM COURSE T2 WHERE T2.TID IN(SELECT T1.TID FROM TEACHER  T1 WHERE T1.TNAME='叶平'))GROUP BY T3.CID --当成一个临时表B   临时表A和临时表B完全一样 A=B
)   B) ;

方法1

方法2

--14、查询和"1002"号的同学学习的课程完全相同的其他同学学号和姓名-★

SELECT T3.SID,T3.SNAME FROM STUDENT T3 WHERE T3.SID IN(SELECT T2.SID FROM SC T2 WHERE T2.CID IN(SELECT CID FROM SC T1 WHERE T1.SID='1002') AND T2.SID<>'1002'GROUP BY T2.SID  HAVING COUNT(DISTINCT T2.CID)=(SELECT COUNT(DISTINCT CID )FROM SC T1 WHERE T1.SID='1002')
) ORDER BY 1;

--15、删除学习"叶平"老师课的SC表记录-★

DELETE FROM SC T3 WHERE T3.CID IN(SELECT T2.CID FROM COURSE T2 WHERE T2.TID IN(SELECT T1.TID FROM TEACHER T1 WHERE T1.TNAME='叶平')
);

--16、向SC表中插入一些记录，这些记录要求符合以下条件:①、没有上过编号"002"课程的同学学号 ②、插入"002"号课程的平均成绩-★★

--思路：首先查询没有上过002 课程的学生编码 使用MINUS 或者使用NOT IN
--方式1、使用 NOT IN
INSERT INTO  SC
SELECT SID,'002',
(SELECT ROUND(AVG(SCORE),2) FROM SC WHERE CID='002')
FROM STUDENT WHERE SID NOT IN (SELECT SID FROM SC WHERE CID='002');--方式2、使用MINUS
INSERT INTO SC
SELECT A.SID,B.CID,B.AVG_SOCRE FROM
(SELECT DISTINCT T2.SID FROM STUDENT T2
MINUS
SELECT  DISTINCT T1.SID FROM SC T1 WHERE T1.CID='002' ) A ,
(SELECT T3.CID,ROUND(AVG(T3.SCORE),2) AVG_SOCRE FROM SC T3 WHERE CID='002' GROUP BY T3.CID)B  ;

--17、按平均成绩从高到低显示所有学生的"数据库"、"企业管理"、"英语"三门的课程成绩，按如下形式显示：学生ID,数据库,企业管理,英语,有效课程数,有效平均分--★★

SELECT T1.SID 学生ID,SUM(DECODE(T1.CID,NULL,0,1)) 企业管理,SUM(DECODE(T2.CID,NULL,0,1)) 数据库,SUM(DECODE(T3.CID,NULL,0,1)) 英语,SUM(DECODE(T1.CID,NULL,0,1)+DECODE(T2.CID,NULL,0,1)+DECODE(T3.CID,NULL,0,1))有效课程数,ROUND(SUM(NVL(T1.SCORE,0)+NVL(T2.SCORE,0)+NVL(T3.SCORE,0))/(SUM(DECODE(T1.CID,NULL,0,1))+SUM(DECODE(T2.CID,NULL,0,1))+SUM(DECODE(T3.CID,NULL,0,1))),2) 有效平均分
FROM  SC T1
LEFT JOIN (SELECT * FROM  SC T WHERE T.CID='004' )T2  ON T1.SID=T2.SID
LEFT JOIN (SELECT * FROM  SC T WHERE T.CID='006' )T3  ON T1.SID=T3.SID
WHERE T1.CID='001'
GROUP BY T1.SID
ORDER BY 1 ;

--18、查询各科成绩最高和最低的分：以如下形式显示：课程ID，课程名称，最高分，最低分-★

SELECT  T.CID 课程ID ,T1.CNAME 课程名称, MAX(T.SCORE) 最高分 ,MIN(T.SCORE) 最低分 FROM SC T
LEFT JOIN COURSE T1 ON T1.CID=T.CID
GROUP BY T.CID,T1.CNAME  ORDER BY 1;

--19、按各科平均成绩从低到高和及格率(60分及以上)的百分数从高到低顺序-★

SELECT T.CID ,ROUND(AVG(T.SCORE),2) AVG_SCORE,
TO_CHAR(ROUND((SUM(CASE WHEN T.SCORE>=60 THEN 1 ELSE 0 END )/COUNT(T.SID))*100,4),'990.99')||'%'  PASS_PERCENTFROM SC T GROUP BY T.CID ORDER BY 2,3;

--20、查询不同老师所教不同课程平均分从高到低显示-★

SELECT  T3.TID,T3.TNAME,T1.CID,ROUND(AVG(T1.SCORE),2)AVG_SCORE FROM SC T1
LEFT JOIN COURSE T2 ON T1.CID=T2.CID
LEFT JOIN TEACHER T3 ON T2.TID=T3.TID
GROUP BY T3.TID,T3.TNAME,T1.CID  ORDER BY 4 DESC ,1,3 ;

--21、查询各科成绩前三名的记录:（不考虑并列情况）-★★

重点:DENSE_RANK() OVER(PARTITION BY T.CID ORDER BY T.SCORE DESC NULLS LAST)

SELECT A.CID, A.SCORE, A.RANKFROM (SELECT T.SID,T.CID,T.SCORE,DENSE_RANK() OVER(PARTITION BY T.CID ORDER BY T.SCORE DESC NULLS LAST) AS RANKFROM SC T --不考虑并列情况) AWHERE A.RANK <= 3;

--22、查询如下课程成绩第3名到第6名的学生成绩单：企业管理（001),马克思（002）,UML （003）,数据库（004),

按如下形式显示：[学生ID],[学生姓名],企业管理,马克思,UML,数据库,平均成绩-★★★★

SELECT  C.SID 学生ID,C.SNAME 学生姓名,B.A001 企业管理,B.A002 马克思,B.A003 UML,B.A004 数据库,(NVL(B.A001,0)+NVL(B.A002,0)+NVL(B.A003,0)+NVL(B.A004,0))/4 平均成绩
FROM (--把以下查询结果看成 临时表B
WITH TEMP_RANK AS
(SELECT * FROM (
SELECT T.SID,T.CID,T.SCORE, RANK() OVER(PARTITION BY T.CID ORDER BY T.SCORE DESC NULLS LAST ) RANK FROM SC T WHERE T.CID IN('001','002','003','004') ) A WHERE A.RANK BETWEEN 3 AND 6 )SELECT * FROM TEMP_RANK B PIVOT ( MAX(B.SCORE) FOR  CID IN ( --注意CID 只能使用CID 不能使用表别名.CID，否则会报[ORA-01748：此处只允许简单的列名]错误'001' AS A001,'002' AS A002,'003' AS A003,'004' AS A004))) B LEFT JOIN STUDENT C ON B.SID=C.SID ;

--23、统计列印各科成绩,各分数段人数:课程ID,课程名称,A [100-85],B[85-70] ,C[70-60] ,D[<60]-★★

SELECT T1.CID,T2.CNAME ,SUM (CASE WHEN  T1.SCORE  BETWEEN 85 AND 100 THEN 1 ELSE 0 END) A ,SUM (CASE WHEN  T1.SCORE  BETWEEN 70 AND 84 THEN 1 ELSE 0 END) B ,SUM (CASE WHEN  T1.SCORE  BETWEEN 60 AND 69 THEN 1 ELSE 0 END) C ,SUM (CASE WHEN  T1.SCORE <60  THEN 1 ELSE 0 END) D
FROM SC T1
LEFT JOIN COURSE T2 ON T1.CID=T2.CID
GROUP BY T1.CID,T2.CNAME
ORDER BY 1;

--24、查询学生平均成绩及其名次-★★

SELECT A.SID,ROUND(A.AVG_SCORE, 2) AVG_SCORE,RANK() OVER(ORDER BY A.AVG_SCORE DESC) AVG_RANKFROM (SELECT T.SID, AVG(T.SCORE) AVG_SCORE FROM SC T GROUP BY T.SID) A
ORDER BY A.SID;

--25、查询各科成绩前三名的记录:(考虑成绩并列情况)-★★
--和21题类似。 重点:RANK() OVER(PARTITION BY T.CID ORDER BY T.SCORE DESC NULLS LAST)

SELECT A.CID, A.SCORE, A.RANKFROM (SELECT T.SID,T.CID,T.SCORE,RANK() OVER(PARTITION BY T.CID ORDER BY T.SCORE DESC NULLS LAST) AS RANKFROM SC T --考虑并列情况) AWHERE A.RANK <= 3;

--26、查询每门课程被选修的学生数-★

SELECT T1.CID, T2.CNAME, COUNT(1) AS 选修数量FROM SC T1LEFT JOIN COURSE T2 ON T1.CID = T2.CIDGROUP BY T1.CID, T2.CNAMEORDER BY 1;

--27、查询出只选修了一门课程的全部学生的学号和姓名-★

SELECT T1.SID, T2.SNAME, COUNT(DISTINCT T1.CID) AS 选课数量FROM SC T1LEFT JOIN STUDENT T2 ON T1.SID = T2.SIDGROUP BY T1.SID, T2.SNAME
HAVING COUNT(DISTINCT T1.CID) = 1;

--28、查询男生、女生人数-★

SELECT T.SSEX, COUNT(T.SID) FROM STUDENT T GROUP BY T.SSEX;

--29、查询姓"张"的学生名单-★

SELECT * FROM STUDENT T WHERE T.SNAME LIKE '张%';

--30、查询同名同姓学生名单,并统计同名人数-★

SELECT T.SNAME, COUNT(DISTINCT T.SID)FROM STUDENT T
GROUP BY T.SNAME HAVING COUNT(DISTINCT T.SID) > 1
ORDER BY 1;

--31、查询当前系统时间的年月日（格式如：2021-05-16)

SELECT TO_CHAR(SYSDATE,'yyyy-MM-DD') FORMAT_DATE FROM DUAL;

--32、查询每门课程的平均成绩，结果按平均成绩升序排列，平均成绩相同时，按课程号降序排列-★

SELECT T1.CID, T2.CNAME, ROUND(AVG(T1.SCORE), 2)AVG_SCOREFROM SC T1LEFT JOIN COURSE T2 ON T1.CID = T2.CIDGROUP BY T1.CID, T2.CNAMEORDER BY 3 ASC, 1;

--33、查询平均成绩大于85的所有学生的学号、姓名和平均成绩-★

SELECT T1.SID, T2.SNAME, AVG(T1.SCORE)AVG_SCOREFROM SC T1LEFT JOIN STUDENT T2 ON T1.SID = T2.SIDGROUP BY T1.SID, T2.SNAME HAVING AVG(T1.SCORE) > 85ORDER BY 1;

--34、查询课程名称为"数据库"，且分数低于60的学生姓名和分数-★

SELECT T.SID, T2.SNAME, T.SCOREFROM SC TLEFT JOIN STUDENT T2ON T.SID = T2.SIDWHERE T.CID = (SELECT T2.CID FROM COURSE T2 WHERE T2.CNAME = '数据库')AND T.SCORE < 60ORDER BY 1;

--35、查询所有学生的选课情况-★

SELECT T3.SID,T3.SNAME,T2.CID, T2.CNAME ,COUNT(DISTINCT T1.CID) 选课情况  FROM SC T1LEFT JOIN COURSE T2 ON T1.CID=T2.CID LEFT JOIN STUDENT T3 ON T1.SID=T3.SID
GROUP BY T3.SID,T3.SNAME,T2.CID, T2.CNAME   ORDER BY 1,3;

--36、查询任何一门课程成绩在70分以上的姓名、课程名称和分数-★

SELECT T1.SID,T3.SNAME,T2.CNAME,T1.SCORE FROM SC T1 LEFT JOIN COURSE T2 ON T1.CID=T2.CIDLEFT JOIN STUDENT T3 ON T1.SID=T3.SID
WHERE T1.SCORE>70 ORDER BY 4 DESC;

--37、查询不及格的课程，并按课程号从大到小排列-★

SELECT * FROM SC T WHERE T.SCORE<60 ORDER BY T.CID DESC ;

--38、查询课程编号为003且课程成绩在80分以上的学生的学号和姓名-★

SELECT T1.SID, T2.SNAME, T1.SCOREFROM SC T1LEFT JOIN STUDENT T2 ON T1.SID = T2.SIDWHERE T1.CID = '001'AND T1.SCORE > 80ORDER BY 1;

--39、求选了课程的学生人数-★

SELECT  COUNT(DISTINCT T.SID) 学生人数   FROM SC T ;

--40、查询选修"叶平"老师所授课程的学生中，成绩最高的学生姓名及其成绩-★★

WITH TEMP_SCORE AS(
SELECT *   FROM SC T1
WHERE T1.CID  IN(
SELECT T3.CID FROM TEACHER T2 ,COURSE T3 WHERE T2.TNAME='叶平' AND T2.TID=T3.TID )
)
SELECT A.SID,B.SNAME,A.SCORE FROM SC A
LEFT JOIN STUDENT B ON A.SID=B.SID
LEFT JOIN TEMP_SCORE C ON C.SID=A.SID AND C.CID=A.CID
WHERE A.SCORE IN (SELECT MAX(D.SCORE) FROM TEMP_SCORE D);

--41、查询各个课程及相应的选修人数-★

SELECT T.CID,COUNT(DISTINCT T.SID)选修人数 FROM SC T GROUP BY T.CID ORDER BY 1 ;

--42、查询不同课程成绩相同的学生的学号、课程号、学生成绩-★★

SELECT * FROM SC T3WHERE T3.SCORE IN (SELECT T1.SCORE  FROM SC T1 GROUP BY T1.SCORE HAVING COUNT(DISTINCT T1.CID)>1) ORDER BY 3 ;

--43、查询每门功成绩最好的前两名-★★★

SELECT *FROM (SELECT T1.SID,T1.CID,T1.SCORE,RANK() OVER(PARTITION BY T1.CID ORDER BY T1.SCORE DESC NULLS LAST) RANKFROM SC T1) AWHERE A.RANK <= 2;

--44、统计每门课程的学生选修人数（超过3人的课程才统计）要求输出课程号和选修人数，查询结果按人数降序排列，查询结果按人数降序排列，若人数相同，按课程号升序排列-★★★

SELECT T1.CID, COUNT(T1.SID) CID_COUNTFROM SC T1GROUP BY T1.CID HAVING COUNT(T1.SID) > 3ORDER BY 2 DESC, 1 ASC;

--45、检索至少选修两门课程的学生学号-★

SELECT T1.SID, T2.SNAME--, COUNT(DISTINCT T1.CID)FROM SC T1LEFT JOIN STUDENT T2ON T1.SID = T2.SIDGROUP BY T1.SID, SNAME
HAVING COUNT(DISTINCT T1.CID) >= 2ORDER BY 1;

--46、查询全部学生都选修的课程的课程号和课程名-★

SELECT T1.CID, COUNT(DISTINCT T1.SID)FROM SC T1GROUP BY T1.CID
HAVING COUNT(DISTINCT T1.SID) = (SELECT COUNT(DISTINCT T2.SID)  FROM STUDENT T2);

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