SELECT * FROM examiner ORDER BY erdepa ASC，ersalary DESC； 该语句的功能是查询（）。

A、各个学院考官的平均年龄，按平均年龄升序排列

B、各个学院考官的平均年龄，按平均年龄降序排列

C、所有考官的平均年龄，按平均年龄降序排列

D、所有考官的平均年龄，按平均年龄升序排列

A、报考了3门以上试卷的报考号和报考门数，按报考门数降序排列。

B、报考了3门以上试卷的报考号，按报考门数升序排列。

C、报考了3门以下试卷的报考号，按报考门数降序排列。

D、这个语句不合法

Select emp_id，emp_name，sex，title，wage from employee order by emp_name句子得到的结果集是按()排序。

A.emp_id

B.emp_name

C.sex

D.wage

一个完整的SELECT语句包括()子句。

A、FROM

B、WHERE

C、SELECT

D、GROUPBY

E、ORDERBY

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;

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);

●设员工关系Emp(E_no，E_name，E_sex，D_name，E_age，E_Add)，关系Emp中的属性分别表示员工的员工号、姓名、性别、所在部门、年龄和通信地址；其中D_name是部门关系Dept的主键。查询各个部门员工的最大年龄、最小年龄，以及最大年龄与最小年龄之间年龄差的SQL语句如下：

SELECT D_name, MAX(E_age),MIN(E_age), (1)

FROM Emp

(2)，

(1) A． MAX(E_age)-MIN(E_age)IN 年龄差

B．年龄差IN MAX(E_age)-MIN(E_age)

C．MAX(E_age)-MIN(E_age)AS 年龄差

D．年龄差AS MAX(E_age)-MIN(E_age)

(2) A. GROUPBYE_name

B. GROUP BYD_name

C. ORDER BY E_name

D. ORDER BYD_name

设员工关系Emp（E_no，E_name，E_sex，D_name，E_age，E_Add），关系Emp中的属性分别表示员工的员工号、姓名、性别、所在部门、年龄和通信地址；其中D_name是部门关系Dept的主键。查询各个部门员工的最大年龄、最小年龄，以及最大年龄与最小年龄之间年龄差的SQL语句如下：

SELECT D_name, MAX（E_age）,MIN（E_age）, （61） FROM Emp （62），

（61）

A. MAX（E_age）-MIN（E_age）IN 年龄差

B. 年龄差IN MAX（E_age）-MIN（E_age）

C. MAX（E_age）-MIN（E_age）AS 年龄差

D. 年龄差AS MAX（E_age）-MIN（E_age）

（62）

A. GROUPBYE_name

B. GROUP BYD_name

C. ORDER BY E_name

D. ORDER BYD_name

Select emp_id,emp_name,sex,title,wage from Yuangong order by emp_name句子得到的结果集是按( )排序。

A、emp_id

B、emp_name

C、sex

D、wage

Your network contains a DNS server named DNS1 that runs Windows Server 2008 R2.You need to be notified by e-mail if the DNS service logs errors or warnings. The solution must minimize the number of e-mail notifications you receive.What should you do？（）

A. Create an alert in Performance Monitor.

B. Run the Configure a DNS Server Wizard.

C. Select the DNS Server log from Event Viewer and attach a task to the log.

D. Create a custom view from Event Viewer and attach a task to the custom view.