题目内容
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【】是指一个SELECT-FROM-WHERE查询块可以嵌入在另一个查询块之中。
【 】是指一个SELECT-FROM-WHERE查询块可以嵌入在另一个查询块之中。
提问人：网友idler_cat
发布时间：2022-01-07
题目内容
（请给出正确答案）
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更多“【】是指一个SELECT-FROM-WHERE查询块可以嵌入…”相关的问题
第2题
用UNION合并两个SELECT查询的结果时，下列叙述中错误的是（ ）。
A、被组合的每个查询都可以使用ORDER BY子句
B、两个SELECT语句必须输出同样的列数
C、两个表各相应列的数据类型必须相同
D、将来自不同查询的数据组合起来
第5题
执行以下SQL语句：
SELECT 产地，COUNT(*) 提供的商品种类数；
FROM 商品表；
WHERE 单价＞200;
GROUP BY 产地 HAVING COUNT(*)＞=2;
ORDER BY 2 DESC
查询结果的第一条记录的产地和提供的商品种类数是( )。
A．北京，1 B．上海，2 C．广东，5 D．广东，7
第7题
检索全部考生都报考的试卷号和试卷名，查询语句就可以写成（ ）。
A、SELECT eid,ename FROM exampaper WHERE eid in (SELECT eid FROM eeexam GROUP BY eid HAVING COUNT(eeid)=(SELECT COUNT(eeid) FROM examinee) );
B、SELECT eid,ename FROM exampaper WHERE NOT EXIST (SELECT * FROM examinee WHERE NOT EXIST (SELECT * FROM eeexam WHERE examinee.eeid=eeid and exampaper.eid=eid) );
C、SELECT eid，ename FROM exampaper WHERE NOT EXISTS （（SELECT eeid FROM examinee） EXCEPT （SELECT eeid FROM eeexam WHERE exampaper.eid=eid））;
D、SELECT * FROM examinee ORDER BY eeid limit 2 offset ((SELECT count(*) FROM examinee)-2);
E、SELECT * FROM examinee ORDER BY eeid LIMIT (SELECT COUNT(*) FROM eeexam);
F、SELECT * FROM department WHERE EXISTS (SELECT * FROM examiner WHERE erdepa=department. dname AND eid= ername= '范冰冰');
G、SELECT eedepa, AVG(eeage) FROM examinee GROUP BY eedepa HAVING eedepa IN (SELECT eedepa FROM examinee WHERE eename='刘诗诗');
H、SELECT dname, (SELECT COUNT(*) FROM examiner WHERE department.dname = examiner.erdepa) FROM department;
I、SELECT (SELECT eedepa FROM examinee WHERE eeexam.eeid=examinee.eeid), AVG(achieve) FROM eeexam GROUP BY (SELECT eedepa FROM examinee WHERE eeexam.eeid=examinee.eeid);
第9题
查询平均成绩良好，也就是≥80分的考生人数，查询语句就可以写成（ ）。
A、WITH avgach (eeid,avgachieve) AS ( SELECT eeid,AVG(achieve) FROM eeexam GROUP BY eeid ) SELECT COUNT(*) FROM avgach WHERE avgachieve>=80;
B、WITH avgach (eeid,avgachieve) ( SELECT eeid,AVG(achieve) FROM eeexam GROUP BY eeid ) SELECT COUNT(*) FROM avgach WHERE avgachieve>=80;
C、SELECT COUNT(*) FROM (SELECT eeid,avg(achieve) FROM eeexam GROUP BY eeid )AS avgach (eeid,avgachieve) WHERE avgachieve>=80;
D、SELECT COUNT(*) FROM (SELECT eeid,avg(achieve) FROM eeexam GROUP BY eeid ) avgach (eeid,avgachieve) WHERE avgachieve>=80;
E、SELECT dname, (SELECT COUNT(*) FROM examiner WHERE department.dname = examiner.erdepa) FROM department WHERE avgachieve>=80;
F、SELECT * FROM erexam NATURAL LEFT OUTER JOIN exampaper;
