我的理解是静态变量和静态方法属于类,而不是类对象.因此Override,静态方法在Java中不起作用,因为重写我们需要创建一个类的实例.但我今天尝试的东西与我的Java知识相矛盾.
请遵循以下代码:
class Parent{
public static void doIt(){
System.out.println("In static method 'doit' of class Parent ");
}
}
class Child extends Parent{
public static void doIt(){
System.out.println("In static method 'doit' of class Child ");
}
}
public class StaticPractise{
public static void main(String[] args){
Parent.doIt();
Child.doIt();
}
}
上述实现的输出是:
D:\Rahul Shivsharan\MyPractise\JAVA>java StaticPractise
In static method 'doit' of class Parent
In static method 'doit' of class Child
从这个输出中我可以理解,尽管Child类扩展了Parent类,但这些doit方法对于每个类都是个体的static.所以不允许覆盖它们.
现在请按照下面的代码,@Override添加到孩子的 …
我参与了许多J2EE项目,其中视图层是JSP.在大多数项目中,我已经看到我们使用scriptlet中的contextPath引用外部资源,即images,javascript,jsp,css等.
代码如下,
<%@page contentType="text/html" pageEncoding="UTF-8"%>
<!DOCTYPE html>
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8">
<title>GC Demo Using HandlebarsJS</title>
<script type="text/javascript" src="<%=request.getContextPath()%>/js/jqueryUI-AutoComplete/jquery-1.9.1.js"></script>
<script type="text/javascript" src="<%=request.getContextPath()%>/js/jqueryUI-AutoComplete/jquery-ui-1.10.3.custom.js"></script>
<script type="text/javascript" src="<%=request.getContextPath()%>/js/handlebarsJS/handlebars.js"></script>
<link rel="stylesheet" type="text/css" href="${pageContext.servletContext.contextPath}/js/jqueryUI-AutoComplete/jquery-ui-1.10.3.custom.css">
从上面的jsp中,我在这里导入外部资源,这些资源位于我的同一个项目包中,即在我的战争中.
现在上面的JSP可以写成如下代码,
<%@page contentType="text/html" pageEncoding="UTF-8"%>
<!DOCTYPE html>
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8">
<title>GC Demo Using HandlebarsJS</title>
<script type="text/javascript" src="../js/jqueryUI-AutoComplete/jquery-1.9.1.js"></script>
<script type="text/javascript" src="../js/jqueryUI-AutoComplete/jquery-ui-1.10.3.custom.js"></script>
<script type="text/javascript" src="../js/handlebarsJS/handlebars.js"></script>
<link rel="stylesheet" type="text/css" href="../js/jqueryUI-AutoComplete/jquery-ui-1.10.3.custom.css">
在第二个例子中,我也引用了战争中的资源.
现在考虑上述两种情况,第一种情况作为最佳实践被赋予更多意义.
为什么？
使用第二种情况有什么缺点？
使用第二种情况,我们的项目是否与contextpath更紧密地结合在一起？
请向我解释.
我是这个框架的新手,因此练习Angularjs并遵循网站上提供的教程.
有一个例子我们可以搜索表中的数据,例子如下,
<!DOCTYPE html>
<html ng-app="SmartPhoneApp">
<head>
<title>Smart phone Angular</title>
<script type="text/javascript" src="D:/Rahul Shivsharan/Installers/AngularJS/angular.js"></script>
<script type="text/javascript">
var smartPhoneApp = angular.module("SmartPhoneApp",[]);
smartPhoneApp.controller("phoneCtrl",function($scope){
$scope.phones = [
{
"modelName" : "LUMIA 520",
"companyName" : "NOKIA"
},
{
"modelName" : "GALAXY S Series",
"companyName" : "SAMSUNG"
},
{
"modelName" : "CANVAS",
"companyName" : "MICROMAX"
},
{
"modelName" : "OPTIMUS",
"companyName" : "LG"
}
];
});
</script>
</head>
<body>
Search by Model Name : <input ng-model="comp.modelName" />
Search by Company : <input ng-model="comp.companyName" />
<div …以下是"practise01.js"文件中的代码,
function fn(name){
return f;
function f(){
var n = name;
console.log("Next TICK "+n+", ");
}
}
function myTimeout(time,msg){
setTimeout(function(){
console.log("TIMEOUT "+msg);
},time);
}
process.nextTick(fn("ONE"));
myTimeout(500,"AFTER-ONE");
process.nextTick(fn("TWO"));
myTimeout(500,"AFTER-TWO");
process.nextTick(fn("THREE"));
myTimeout(500,"AFTER-THREE");
process.nextTick(fn("FOUR"));
运行上面代码的输出是
rahul@rahul:~/myPractise/PlainNodeJSPractise01/Process$ node practise01.js
Next TICK ONE,
Next TICK TWO,
Next TICK THREE,
Next TICK FOUR,
TIMEOUT AFTER-ONE
TIMEOUT AFTER-TWO
TIMEOUT AFTER-THREE
现在我在"practise02.js"中不使用process.nextTick编写代码,如下所示,
function myTimeout(time,msg){
setTimeout(function(){
console.log("TIMEOUT "+msg);
},time);
}
function fn(name){
return f;
function f(){
var n = name;
console.log("Next TICK "+n+", ");
}
}
fn("ONE")();
myTimeout(500,"AFTER-ONE");
fn("TWO")(); …我在使用嵌入式注释的JPA中尝试关系,但我无法成功运行它,
这里我的数据库sql脚本如下,
create table TBL_COLLEGE(
id integer primary key generated always as identity (start with 1000, increment by 5),
name varchar(50)
)
create table TBL_COURSE(
Id integer primary key generated always as identity (start with 10, increment by 1),
college_Id integer references TBL_COLLEGE,
name varchar(50)
)
以下是JPA的代码,
@Embeddable
public class Course {
...
...
..
@Id
@GeneratedValue(strategy= GenerationType.IDENTITY)
@Column(name="ID")
private Integer courseId;
@Column(name="NAME")
private String courseName;
@Column(name="COLLEGE_ID")
private Integer collegeId;
....
// getter and setter
}
这是学院的地图,
@Entity
@Table(name="TBL_COLLEGE")
public …我的id类如下,
public class EmployeeId implements Serializable{
public EmployeeId(){}
public EmployeeId(Integer id, String country){
this.id = id;
this.country = country;
}
private Integer id;
private String country;
@Override
public int hashCode(){
return this.getCountry().hashCode() + getId();
}
@Override
public boolean equals(Object o){
boolean flag = false;
EmployeeId myId = (EmployeeId) o;
if((o instanceof EmployeeId)
&& (this.getCountry().equals(myId.getCountry()))
&& (this.id == myId.getId())){
flag = true;
}
return flag;
}
// rest of the code with getters only
}
以下是我的实体使用
@Entity
@IdClass(EmployeeId.class)
@Table(name="TBL_EMPLOYEE_FOUR")
public class …下面的代码，
console.log([].concat.apply([2],[[99],5,6,[2,3]]));
输出
[ 2, 99, 5, 6, 2, 3 ]
还有下面的代码
console.log([2].concat([99]).concat([5,6,[2,3]]));
输出
[ 2, 99, 5, 6, [ 2, 3 ] ]
我的假设是
console.log([].concat.apply([2],[[99],5,6,[2,3]]));
应该
[2,[99],5,6,[2,3]]
但是，为什么呢？
在这里，我正在尝试在JPA中建立多对多关系，我创建了“ tblcourse”和“ tblStudent”表格，一个学生可以注册许多课程，
create table tblcourse(
id integer primary key,
name varchar(100),
duration integer
);
create table tblcourseStudent(
studentid integer references tblstudent(studentId),
courseId integer references tblcourse(id),
constraint pk_composit_cs primary key(studentid,courseId)
)
Create table tblStudent(
studentId integer primary key,
……..
….
);
上述关系的JPA表示法如下，这是StudentEntity.java的代码，
@Entity
@Table(name="TBLSTUDENT")
public class StudentEntity implements Serializable{
private static final long serialVersionUID = 100034222342L;
@Id
@Column(name="STUDENTID")
private Integer studentId;
@Column(name="STUDENTNAME")
private String studentName;
@Column(name="CONTACTNO")
private String contactNumber;
@Embedded
private StudentAddress address;
@ManyToOne(fetch=FetchType.LAZY)
@JoinColumn(name="DEPTID")
private DeptEntity deptEntity;
@ManyToMany(fetch=FetchType.LAZY)
@JoinTable(name="tblcourseStudent", …今天在一次采访中，我被告知要编写一个程序，该程序将输出未排序数组中的第 n 个最高数字，
我用javascript解决了这个问题，程序如下，
var fn50 = function(){
var reverseSort = function(myArray,highest){
var x = 0,
y = 0,
z = 0,
temp = 0,
totalNum = myArray.length, // total numbers in array
flag = false, // is the numbers sorted in reverse while iteration
isAchieved = false; // whether we achieved the nth highest
while(x < totalNum){
y = x + 1; // start comparing 'yth' number which is next to 'xth' number.
if(y < totalNum){
// start comparing …嗨，有人问我以下问题。
给定两个数组，即array1和array2。他们两个都包含数字的排序顺序。
Array1还包含-1，例如；array2中的数字与array1中的-1一样多。
示例如下，
array1 = [-1,-1,-1,-1,56,78,90,1200];
array2 = [1,4,5,1000]
我需要编写一个程序，将上述数组合并为一个，该程序将按排序顺序包含两个数组中的数字，但-1除外。
这是我的代码如下，
puzzle04([3,6,-1,11,15,-1,23,34,-1,42],[1,12,28]);
puzzle04([3,6,-1,11,15,-1,23,34,-1,42],[7,19,38]);
puzzle04([3,6,11,15,32,34,42,-1,-1,-1,-1],[1,10,17,56]);
puzzle04([-1,-1,-1,-1,3,6,11,15,32,34,42],[1,10,17,56]);
puzzle04([-1,-1,-1,-1,3,6,11,15,32,34,42],[56,78,90,100]);
puzzle04([12,34,65,-1,71,85,90,-1,101,120,-1,200],[24,37,94]);
puzzle04([3,6,-1,11,15,-1,32,34,-1,42,-1],[1,10,17,56]);
puzzle04([-1,-1,-1,56,78,90,112],[1,4,5]);
puzzle04([-1,-1,-1,-1,56,78,90,112],[1,4,5,1000]);
puzzle04([-1,-1,-1,-1,56,78,90,1200],[1,4,5,1000]);
function puzzle04(array1,array2){
var outputArray = [],
array1Counter = 0, // counter for array1
array2Counter = 0, // counter for array2
isArray2NumPlaced = false, // has number from array2 found its position in output array ?
areAllArray2NumsFilled = false; // is number pushed in output array
// iterating through array2 loop
for(array2Counter = 0; array2Counter < array2.length; array2Counter++){
// …javascript ×5
java ×4
jpa ×3
jpa-2.0 ×3
arrays ×2
sorting ×2
algorithm ×1
angularjs ×1
contextpath ×1
ejb-3.0 ×1
hibernate ×1
java-ee ×1
jsp ×1
node.js ×1
overriding ×1