prnt sc

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. sc

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

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.

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 #

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 a

50 commonly used test and interview SQL statements Postedon50 commonly used test and interview SQL statements 2009-12-1715: 05 Student (S #, Sname, Sage, Ssex) student table Course (C #, Cname, T #) Course list SC (S #, C #, score) Instructor table question: 1. query all students whose 001 course scores are higher than 002. 50 commonly used test and interview SQL statements Posted on 50 commonly used test, interview SQL statement Student (S #, Sname,

50 Common SQL statements Student (s#,sname,sage,ssex) Student table Course (c#,cname,t#) timetable SC (s#,c#,score) score sheet Teacher (t#,tname) Teacher's Table Problem: 1, Query the "001" Course than "002" course scores of all students of the school number; Select a.s# from (select S#,score from SC where c#= ' 001 ') A, (select S#,score From SC where c#= ' 0

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

SQL database Interview Questions and answers (22:52:54) ReprintedBytes Summary of course number of three c courses by Ye Ping Category: Program World 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 #: Instruc

I. Data Table Student (S #, Sname, Sage, Ssex) Student table S #: Student ID; Sname: Student name; Sage: Student age; Ssex: Student gender Course (C #, cname, T #) curriculum C #, course number; Cname: Course name; T #: Instructor id SC (S #, C #, score) Student table S #: Student ID; C #, course No.; score: score Teacher ( I. Data Table Student (S #, Sname, Sage, Ssex) Student table S #: Student ID; Sname: Student name; Sage: Student age; Ssex: Stude

Http://blog.csdn.net/workingon/archive/2005/08/31/468684.aspx /Proc/bus/USB file system output====================================(Version 200000008.15) /USB devices in the proc file system/Proc/bus/USB/drivers and/proc/bus/USB/devices. /Proc/bus/USB/drivers list registered drivers,One driver per line. If possible, the range of the node number of each driver's USB sub-device is given. ** Note **: If/proc/bus/USB is empty, mount the file system as a root user:Mount-T usbdevfs NONE/proc/bus/USB O

Student (s#,sname,sage,ssex) Student tableCourse (c#,cname,t#) timetableSC (s#,c#,score) score tableTeacher (t#,tname) Teacher tableProblem:1, the inquiry "001" course is higher than "002" of all students of the school number;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, the query average score is more than

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

', n' male ')InsertInto student values ('05 ', n' Zhou mei', '2017-12-01', n' female ')InsertInto student values ('06', n'huland', '2017-03-01 ', n'female ')InsertInto student values ('07 ', n' Zheng Zhu', '2017-07-01 ', n' female ')InsertInto student values ('08', n'wangju', '2017-01-20 ', n'female ')CreateTable course (C # varchar (10), Cname nvarchar (10), T # varchar (10))InsertInto course values ('01', n' ', '02 ')InsertInto course values ('02', n' math ', '01 ')InsertInto course values ('0

Copy codeThe Code is as follows:-- Basic SQL operations -- Create a database Create database Studets -- Create a table Create table student (sno char (5), sname char (20), ssex char (2), sage smallint, sdept char (15 )) Create table course (cno char (3), cname char (30), cpno char (3), ccredit smallint) Create table SC (sno char (5), cno char (3), grade int) -- View table information Select * from student select sno as student ID from student select *

Tags: topic insertion percentage distinct like will design type learning (1) distinguish between having and where: The HAVING clause enables you to specify a filter condition that controls which groups in the query result can appear in the final result. The WHERE clause imposes conditions on the selected column, while the HAVING clause imposes conditions on the groups produced by the GROUP BY clause. The following SQL statements are based on the MySQL5.6.30 version.1. Check the number of all