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若给出关系Student(S_no, Sname, Sage, S_sex, SD_name, S_add, S_tel),并用SQL语言定义Student关系如下: CREATE Student(S_no CHAR(6), Sname CHAR(30)NOT NULL, Sage CHAR(30), S_sex CHAR(1), SD_name CHAR(20), S_add CHAR(30), S_tel CHAR(20), PRIMARY KEY(S_no)); 采用(45)向Student中插入记录能被正确地执行。

A.INSERT INTO Student(S_no, Sname, Sage, S_sex, SD_name, S_add, S_tel) VALUES('010456', '黎敏', '18', ", ", ", ")

B.INSERT INTO Student(S_no, Sname, Sage, S_sex, SD_name, S_add, S_tel) VALUES('010456', '黎敏', '18', '男', '计算机学院', '北京', '88661200')

C.INSERT INTO Student(S_no, Sname, Sage, S_sex, SD_name, S_add, S_tel) VALUES(, '黎敏', '18', 'F', '计算机学院', '北京', '88661200')

D.INSERT INTO Student(S_no, Sname, Sage, S_sex, SD_name, S_add, S_tel) VALUES('010456', , '18', 'F', '计算机学院', '北京', '88661200')


参考答案

更多 “ 若给出关系Student(S_no, Sname, Sage, S_sex, SD_name, S_add, S_tel),并用SQL语言定义Student关系如下: CREATE Student(S_no CHAR(6), Sname CHAR(30)NOT NULL, Sage CHAR(30), S_sex CHAR(1), SD_name CHAR(20), S_add CHAR(30), S_tel CHAR(20), PRIMARY KEY(S_no)); 采用(45)向Student中插入记录能被正确地执行。A.INSERT INTO Student(S_no, Sname, Sage, S_sex, SD_name, S_add, S_tel) VALUES('010456', '黎敏', '18', ", ", ", ")B.INSERT INTO Student(S_no, Sname, Sage, S_sex, SD_name, S_add, S_tel) VALUES('010456', '黎敏', '18', '男', '计算机学院', '北京', '88661200')C.INSERT INTO Student(S_no, Sname, Sage, S_sex, SD_name, S_add, S_tel) VALUES(, '黎敏', '18', 'F', '计算机学院', '北京', '88661200')D.INSERT INTO Student(S_no, Sname, Sage, S_sex, SD_name, S_add, S_tel) VALUES('010456', , '18', 'F', '计算机学院', '北京', '88661200') ” 相关考题
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