A、$student->$getName（）；
B、$student->name；
C、$student->$name；
D、$student.getName（）；
更多“ 下列对$student使用正确的是（）。<br /> <img src="https://nimg.ppkao.com/2018-08/tianmin/2018081511374434877.jpg?sign=6f1ecaf9e6fd30ab95dd0d8b1499dd34&t=62d310e1" />”相关的问题
A、A.Student student 声明了一个类 B、B.new Student（）创建了Student 对象的一个实例 C、C.Student student 声明了对象Student 的一个引用 D、D.class Student 声明了一个类
A、new Student（）创建了Student对象的一个实例 B、Student student声明了对象Student的一个引用 C、class Student声明了一个类 D、new Student（）创建了一个类 E、Student student 声明了一个类
A、SELECT TOP 4 WITH TIES * FROM Student ORDER BY Sage ASC B、SELECT TOP 4 WITH TIES * FROM Student GROUP BY Sage ASC C、SELECT TOP 4 WITH TIES * FROM Student ORDER BY Sage DESC D、SELECT TOP 4 WITH TIES * FROM Student GROUP BY Sage DESC
A、Select Count（1）From Student B、Select Sum（*）From Student C、Selec tAvg（*）From Student D、Select Count（*）From Student
A、Select email from student where email!=null B、Select email from student where emailnotisnull C、Select email from student where email<>null D、Select email from student where emailisnotnull
A、SELECT*FROM students ORDERBYavg_grade B、SELECT*FROM students GROUPBYavg_grade ASC C、SELECT*FROM students ORDERBYavg_grade DESC D、SELECT*FROM students ORDERbyavg_gradeasc
A、Select* From Student B、Select Name From Student C、Select Name，Age，IDCard From Student D、Select Name，Age，IDCard，Sex From Student
A、SELECT Sname，MIN（Sage） FROM Student B、SELECT Sname，Sage FROM Student WHERE Sage = MIN（Sage） C、SELECT TOP 1 Sname，Sage FROM Student D、SELECT TOP 1 Sname，Sage FROM Student ORDER BY Sage
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