generic name
第1题:
A.Object
B.Generic KeyValuePairt object
C.Key
D.Value
第2题:
有如下程序:
#include<iostream>
using flamespace std;
class Name{
char name[20];
public:
Name(){
strcpy(name,“”); cout<<‘?’;
}
Name(char*fname){
strcpy(name,fname); cout<<‘?’;
}
};
int main(){
Name names[3]={Name(”张三”),Name(”李四”)};
return 0;
}
运行此程序输出符号?的个数是
A.0
B.1
C.2
D.3
第3题:
有如下程序: #include<iostream> using namespace std; class Name{ char name[20]; public: Name( ){strepy(name," ");tout<<'?';} Name(char*fname){strcpy(name,fname);cout<<'?';} }; int main( ){ Name name[3]={Name("张三"),Name("李四")}; return 0; } 运行此程序输出符号“?”的个数是
A.0
B.1
C.2
D.3
第4题:
具有属类索引(generic index)的是( )。
第5题:
BWP-Generic-parameter包括()
第6题:
证件上同时出现“Name”、“Full name”和“First name”,且无护照资料页下方的机读码可供参照的,则()。
第7题:
html语言中,创建一个位于文档内部的锚接的标记是?()
第8题:
已知httpClient连接网络的url,通过Post方式访问时要传递name参数,下列方式正确的是()。
第9题:
Dictionary枚举数返回哪种对象?()
第10题:
ALTER TABLE table_name ENABLE constraint_name;
ALTER TABLE table_name STATUS = ENABLE CONSTRAINT constraint _ name;
ALTER TABLE table_name ENABLE CONSTRAINT constraint _ name;
ALTER TABLE table_name TURN ON CONSTRAINT constraint _ name;
第11题:
ALTER TABLE customers ADD CONSTRAINT cust_name_nn CHECK customer_name IS NOT NULL;
ALTER TABLE customers MODIFY CONSTRAINT cust_name_nn CHECK customer_name IS NOT NULL;
ALTER TABLE customers MODIFY customer_name CONSTRAINT cust_name_nn NOT NULL;
ALTER TABLE customers MODIFY customer_name CONSTRAINT cust_name_nn IS NOT NULL;
ALTER TABLE customers MODIFY name CONSTRAINT cust_name_nn NOT NULL;
ALTER TABLE customers ADD CONSTRAINT cust_name_nn CHECK customer_name NOT NULL;
第12题:
CREATE INDEX NAME _IDX (first_name, last_name);
CREATE INDEX NAME _IDX (first_name, AND last_name)
CREATE INDEX NAME_IDX ON (First_name, last_name);
CREATE INDEX NAME_IDX ON employees (First_name, AND last_name);
CREATE INDEX NAME_IDX ON employees (First_name, last_name);
CREATE INDEX NAME_IDX FOR employees (First_name, last_name);
第13题:
Examine the structure of the EMPLOYEES table:Column name Data type RemarksEMPLOYEE_ID NUMBER NOT NULL, Primary KeyLAST_NAME VARCNAR2(30)FIRST_NAME VARCNAR2(30)JOB_ID NUMBERSAL NUMBERMGR_ID NUMBER References EMPLOYEE_ID column DEPARTMENT_ID NUMBERYou need to create an index called NAME_IDX on the first name and last name fields of the EMPLOYEES table. Which SQL statement would you use to perform this task? ()
A. CREATE INDEX NAME _IDX (first_name, last_name);
B. CREATE INDEX NAME _IDX (first_name, AND last_name)
C. CREATE INDEX NAME_IDX ON (First_name, last_name);
D. CREATE INDEX NAME_IDX ON employees (First_name, AND last_name);
E. CREATE INDEX NAME_IDX ON employees (First_name, last_name);
F. CREATE INDEX NAME_IDX FOR employees (First_name, last_name);
第14题:
html语言中,创建一个位于文档内部的锚点的语句是()。
A.<name=“NAME”>
B.<name=“NAME”></name>
C.<aname=“NAME”></a>
D.<aname=“NAME”></a>
第15题:
康唑类药品的药名词干是
A.#NAME?
B.#NAME?
C.#NAME?
D.#NAME?
E.#NAME?
第16题:
A. Generic Router Encapulation
B. Generic Routing Encapulation
C. General Routing Encapulation
D. General Router Encapulation
第17题:
html语言中,创建一个位于文档内部的靶位的标记是?()
第18题:
Which syntax turns an existing constraint on?()
第19题:
Generic drug application
第20题:
What describes the difference between NPIV and vSCSI in the way fibre-channel attached disk and tape are presented when VIO Server is used?()
第21题:
Object
Generic
Key
Value
第22题:
generic drugs cost half as much as buying name-brand drugs
buying generic drugs costs half as much as name-brand drugs
generic drugs cost half as much as name-brand drugs
buying generic drugs cost half as much as buying name-brand drugs
to buy generic drugs costs half as much as buying name-brand drugs
第23题:
ALTER TABLE table_name ENABLE constraint_name;
ALTER TABLE table_name STATUS = ENABLE CONSTRAINT constraint_name;
ALTER TABLE table_name ENABLE CONSTRAINT constraint_name;
ALTER TABLE table_name STATUS ENABLE CONSTRAINT constraint_name;
ALTER TABLE table_name TURN ON CONSTRAINT constraint_name;