avg or norton

Tags: ASE set job Finance Department GROUP by nbsp own case Count--1 Group Statistics The average wage of employees under each department's salary >2000Select AVG (s.sal) from(SELECT * from emp e where e.sal>2000) sGroup BY Deptno; --2 statistics departments with average wage greater than 1600Select E.deptno,avg (e.sal) from EMP e GROUP by E.DEPTNO have avg (e.sa

generated by a process with those generated by the system to determine the impact of the process on system Page faults. Pages per second: the number of Pages retrieved per second. This number should be less than one Working set per second: The Memory Page recently used by the thread, reflecting the number of memory pages used by each process. If the server has enough idle memory, the page will be left in the work set. When the free memory is less than a specific threshold, the page will be clea

event Execution:14.5616s Response Time: min: 1.40ms avg: 1.46ms Max: 10.59ms Approx. Percentile: 1.52msThreads Fairness: Events (Avg/stddev): 10000.0000/0.00 Execution Time (avg /stddev): 14.5616

implementation, but it is much easier to implement it with ref and out. Ref: 1 static int GetIntResult(int[] arry, ref float avg, ref int max, ref int min) 2 { 3 int sum = 0; 4 max = arry[0]; 5 min = arry[0]; 6 for (int i = 0; i Then we tried to call this method in the console: 1 static void Main(string[] args)2 {3 int[] arr = { 1,2,3,4,5,6,7,8,9};4 flo

50 common SQL statementsStudent (S #, Sname, Sage, Ssex) Student tableCourse (C #, Cname, T #) CurriculumSC (S #, C #, score) Orders tableTeacher (T #, Tname) Instructor tableProblem:1. query the student ID of all students whose score is higher than that of the "002" course;Select a. S # from (select s #, score from SC where C # = '001') a, (select s #, scoreFrom SC where C # = '002') BWhere a. score> B. score and a. s # = B. s #;2. query the student ID and average score of students whose averag

use Iostat to view disk usage in more detail.IostatIostat is the ideal tool for determining if the system has I/O bottlenecks. It shows the read and write speed for all disks. This makes it an ideal tool to determine whether you need to "scatter" disk workloads across multiple disks. The tool also reports the same CPU activity as Vmstat.When your application is running, start from the simple iostat -s to determine what the system is doing on the whole. This command prints the following: TTY:

chinese_prc_ci_as null ) On [primary] Go Shows the input data: Enter the following SQL statement: Select * From tb_stuachievement07 Select class, gender, AVG (Chinese) as average Chinese score, AVG (algebra) as algebra average score, AVG (ry) as geometric average score, AVG (English) as average English score From

Student (s #, sname, sage, ssex) Student table Course (C #, cname, T #) Curriculum SC (s #, C #, score) Orders table Teacher (T #, tname) Instructor table Problem: 1. query the student IDs of all students whose "course scores are higher than" course scores; Select a. s # From (select s #, score from SC where C # = '001'), (Select s #, score from SC where C # = '002 ') B Where a. score> B. score and a. s # = B. S #; 2. query the student ID and average score of students whose average scor

50 SQL Query Techniques, query statement examples, 50 SQL statements Student (S #, Sname, Sage, Ssex) Student tableCourse (C #, Cname, T #) CurriculumSC (S #, C #, score) Orders tableTeacher (T #, Tname) Instructor tableProblem:1. query the student ID of all students whose score is higher than that of the "002" course;Copy codeThe Code is as follows:Select a. S # from (select s #, score from SC where C # = '001') a, (select s #, scoreFrom SC where C # = '002') BWhere a. score> B. score and a. s

Basic Database Operations, databasesI. Data Table Student (S #, Sname, Sage, Ssex) Student table S #: Student ID; Sname: Student name; Sage: Student age; Ssex: Student genderCourse (C #, Cname, T #) curriculum C #, Course No.; Cname: Course name; T #: Instructor No.SC (S #, C #, score) Student table S #: Student ID; C #, course number; score: scoreTeacher (T #, Tname) Instructor table T #: Instructor ID; Tname: Instructor name Ii. Basic operations:1. query the student ID of all students whose sc

50 common SQL statements Student (S #, Sname, Sage, Ssex) Student tableCourse (C #, Cname, T #) CurriculumSC (S #, C #, score) Orders tableTeacher (T #, Tname) Instructor table Problem:1. query the student ID of all students whose score is higher than that of the "002" course;Select a. S # from (select s #, score from SC where C # = '001') a, (select s #, scoreFrom SC where C # = '002') BWhere a. score> B. score and a. s # = B. s #;2. query the student ID and average score of students whose ave

Student (s #, sname, sage, ssex) Student tableCourse (C #, cname, T #) CurriculumSC (s #, C #, score) Orders tableTeacher (T #, tname) Instructor table Problem:1. query the student ID of all students whose score is higher than that of the "002" course;Select a. s # From (select s #, score from SC where C # = '001') A, (select s #, scoreFrom SC where C # = '002') BWhere a. score> B. score and a. s # = B. S #;2. query the student ID and average score of students whose average score is greater than

Student (s#,sname,sage,ssex) Student tableCourse (c#,cname,t#) timetableSC (s#,c#,score) score sheetTeacher (t#,tname) Teacher's TableProblem:1, Query the "001" Course than "002" course scores of all students of the school number; Copy Code code as follows: Select a.s# from (select S#,score from SC where c#= ' 001 ') A, (select S#,score From SC where c#= ' 002 ') b where A.score>b.score and a.s#=b.s#; 2, the average score is greater than 60 students of the school numb

Parameters (rates in bytes/sec) In PMP SVC 0/193 47000580ffe1000000f21a2e180020481a2e180b Tx:ubr,peak 1516,avg 0,MAXSDU 0 Rx:ubr,peak 16953936,avg 16953936,MAXSDU 1516 Out P-p SVC 0/192 47000580ffe1000000f21a2e180020481a2e180b Tx:ubr,peak 1516,avg 16953936,MAXSDU 16953936 Rx:ubr,peak 16953936,avg 16953936,MAXSDU 1516

Article Source: http://www.ynpxrz.com/n822738c2024.aspxT_tudent (sid,sname,sage,ssex,sdept) Student tableT_course (Cid,cname,tid) timetableT_score (scid,sid,cid,grade) score tableT_teacher (tid,tname) Teacher tableProblem:1, the inquiry "001" course is higher than "002" of all students of the school number;SelectT1.sid from(SelectSid,grade fromT_scorewhereCid= '001') T1, (SelectSid,grade fromT_scorewhereCid= '002') T2whereT1.grade>T2.grade andT1.sid=T2.sid;SelectT2.SID,T2.CA,T2.CB (SelectT1.sid,

Tags: Tween group by order Search show amp ASP other totalHttp://www.ynpxrz.com/n822738c2024.aspx T_tudent (sid,sname,sage,ssex,sdept) Student table T_course (Cid,cname,tid) timetable T_score (scid,sid,cid,grade) score table T_teacher (tid,tname) Teacher table Problem: 1, the inquiry "001" course is higher than "002" of all students of the school number; Select T1.sid from (select Sid,grade from t_score WHERE cid = ' 001 ') T1, (select Sid,grade from t_score WHERE cid = ' 002 ') T2 where T1.grad

all students whose score is higher than that of the "002" course; select a.S# from (select s#,score from SC where C#='001') a,(select s#,score from SC where C#='002') b where a.score>b.score and a.s#=b.s#;2. query the student ID and average score of students whose average score is greater than 60; select S#,avg(score) from sc group by S# having avg(score) >60;3. query the st

This text has been reprinted online, and some SQL statements are suitable for use in application systems. Some "reports" have a heavier feeling, mainly to review the former. This text has been reprinted online, and some SQL statements are suitable for use in application systems. Some "reports" are more likely to be used, mainly to review the former. 50 common SQL statements Student (S #, Sname, Sage, Ssex) Student table Course (C #, Cname, T #) Curriculum SC (S #, C #, score) Orders table Teac

where C # = '002') BWhere a. score> B. score and a. s # = B. s #;2. query the student ID and average score of students whose average score is greater than 60;Select S #, avg (score)From SCGroup by S # having avg (score)> 60;3. query the student ID, name, number of course selections, and total score of all students;Select Student. S #, Student. Sname, count (SC. C #), sum (score)From Student left Outer join

50 SQL statements involved in one question Student (s #, sname, sage, ssex) Student tableCourse (C #, cname, T #) CurriculumSC (s #, C #, score) Orders tableTeacher (T #, tname) Instructor table Problem:1. query the student ID of all students whose score is higher than that of the "002" course;Select a. s # From (select s #, score from SC where C # = '001') A, (select s #, scoreFrom SC where C # = '002') BWhere a. score> B. score and a. s # = B. S #;2. query the student ID and average score