下列不合法的标识符是___________。 A: _name B: student_name C: 18name D: NAME3

若NAME是一个三行四列的矩阵，则 A: NAME(3, B: =[]表示把NAME中第三行元素删除 C: NAME(3, D: =[]表示把NAME中第三行元素变为0 E: NAME(3, F: =[]表示把NAME中第三列元素删除

对象设置属性和属性值的方法正确的是 A: obj.name=3 B: obj[“name”]=3 C: obj{name} = 3 D: Obj->name=3;

$str = "My name is Jack， and what is your name？"；echo substr_count （$str，“name”）；程序的输出结果为（）。 A: 1 B: 2 C: 3 D: 4

In business situation, we can think about 3 most important things, namely, your________. A: name, age, company B: name, title, company C: name, company, marital status D: name, title, marital status

有如下定义‏struct person{char name［9］;int age;};‏struct person class［10］={"John",17,"paul",19,"Mary",18,"Adam",16,};‏则能输出字母M的语句是（ ）。‏​‏ A: cout<<class［3］.name; B: cout<<class［3］.name［1］; C: cout<<class［2］.name［1］; D: cout<<class［2］.name［0];

This is ______ (I ,my) brother. 2、How ______ ( is ,are) you? 3、My ______ (name ,name’s) Zhu Xiaoqing. 4、______ __( What, What‘s) your name,please? 5、Nice to meet ______ ( you, your) .

‍下面SQL语句中不正确的一项是‌ A: select name from reviewer,rating where reviewer.rid=rating.rid group by reviewer.rid B: select name from reviewer,rating where reviewer.rid=rating.rid having count(stars)>=3 C: select name from reviewer,rating group by reviewer.rid having count(stars)>=3 D: select name from reviewer,rating group by reviewer.rid where count(stars)>=3

根据下述定义，能输出字母M的语句是：strcut person{char name&#91;9&#93;;int age;};struct person class&#91;10&#93;={"John",17,"Paul",19,"Mary",18,"Adam",16}; A: class[3].name B: class[2].name[0] C: class[3].name[1] D: class[2].name[1]

Elevenses, just as its name implies, is a cup of tea and biscuits at around 3 p.m.( )