我无法指出getFreeSpace()和阶级getUsableSpace()方法之间的确切区别File.当我运行以下代码时,得到相同的o/p.
Class Start {
public static void main(String [] args) {
File myfile = new File("C:\\html\abc.txt");
myfile.createNewFile();
Systyem.out.println("free space"+myfile.getFreeSpace()+"usable space"+myfile.getUsableSpace());")
}
}
O/P是
免费space445074731008可用空间445074731008
谁能告诉我究竟有什么区别？
get()和load()方法有什么区别？关于数据获取方法
public static void main(String[] args) {
SessionFactory factory= new Configuration().configure().buildSessionFactory();
Session session = factory.openSession();
Transaction tx = null;
tx = session.beginTransaction();
System.out.println("1 st time calling load method");
Account acc =
(Account)session.load(Account.class, 180);
System.out.println("bal"+acc.getBalance());
System.out.println("2nd time calling load method");
Account acc1=(Account)session.load(Account.class, 180);
System.out.println("bal"+acc1.getBalance());
System.out.println("1 st time calling get method");
Account acc2= (Account) session.get(Account.class, accId);
System.out.println("bal"+acc2.getBalance());
System.out.println("2 st time calling get method");
Account acc2= (Account) session.get(Account.class, accId);
System.out.println("bal"+acc2.getBalance());
tx.commit();
session.close();
}
我得到了以下输出
1 st time calling load method
Hibernate:
/* load …我试图编译以下代码但是出错了
static void test(long... x)
{
System.out.println("long...");
}
static void test(Integer... x)
{
System.out.println("Integer...");
}
public static void main(String [] args) {
int no=5;
test(no,no);//getting error at this point in eclipse 'The method test(long[]) is ambiguous '
}
我不知道为什么它含糊不清.意味着如果我传递一个int值它应该自动框并test(Integer..x)应该被调用..类似的行test(long..x )应该被调用..这是我的理解..有人可以解释为什么它是模棱两可的？
我想创建自定义文件上传组件.我在html中执行了以下代码
HTML代码
<input id="upload" type="file" style="display: none;">// don`t want to render default
<button class="parimarybtnVD" type="button" ng-click="clickUpload()">Browse</button>
JS代码
$scope.clickUpload = function() {
angular.element('#upload').trigger('click');
};
但是当我点击"按钮"时出现以下错误.
Error: [$rootScope:inprog] http://errors.angularjs.org/1.2.16/$rootScope/inprog?p0=%24apply
at Error (<anonymous>)
at http://localhost:7001/RightsWeb/scripts/resource/angular.min.js:6:450
at l (http://localhost:7001/RightsWeb/scripts/resource/angular.min.js:102:171)
at h.$digest (http://localhost:7001/RightsWeb/scripts/resource/angular.min.js:105:497)
at HTMLDocument.D (http://localhost:7001/RightsWeb/scripts/utill/ui-bootstrap-tpls-0.11.0.min.js:9:14775)
at HTMLDocument.f.event.dispatch (http://localhost:7001/RightsWeb/extResources/jquery/jquery-1.7.1.min.js:3:4351)
at HTMLDocument.h.handle.i (http://localhost:7001/RightsWeb/extResources/jquery/jquery-1.7.1.min.js:3:328)
at Object.f.event.trigger (http://localhost:7001/RightsWeb/extResources/jquery/jquery-1.7.1.min.js:3:3038)
at <error: TypeError: Accessing selectionDirection on an input element that cannot have a selection.>
at Function.e.extend.each (http://localhost:7001/RightsWeb/extResources/jquery/jquery-1.7.1.min.js:2:11937)
谁能告诉我为什么我会收到这个错误？如果有更好的方法在angularjs上进行自定义文件上传请告诉.谢谢你提前.
我是angularjs的新手.我正面临一个问题.我有一个带一个单选按钮的桌子.当用户选择任何单选按钮并单击提交按钮时,我需要显示所选行的详细信息.
Html代码:
<table>
<thead>
<tr>
<th>Select</th>
<th>First Name</th>
<th>Last Name</th>
<th>Unique Reference ID</th>
<th>Country</th>
<th>Branch</th>
<th>Card No</th>
</tr>
</thead>
<tr ng-repeat="d in Employees">
<td><input type="radio"></td>
<td>{{d.fname}}</td>
<td>{{d.lname}}</td>
<td>{{d.uniqueid}}</td>
<td>{{d.country.name}}</td>
<td> {{d.branch.name}}</td>
<td>{{d.cardno}}</td>
</tr>
</table>
<table>
<tr>
<td><input type="button" value="Edit" ng-click="Edit()"></td>
</tr>
</table>
控制器代码:
$scope.Edit=function()
{
//need id of selected row ?
};
我对String的实习方法没有很好的理解.
String s1="java"; // should create one object in String Constant pool
String ss="java"; // no object is created (java is already in String pool)..it refers to object in String constant pool
String s2= new String("Android").intern(); // should create 2 objects one in heap and second in String constant pool
String s3= new String("java").intern()// i guess only one object is created on heap and s3 will point to object in String constant pool (as 'java' already exist).so the object in …我想知道从链接哈希集中删除元素的不同方法.我试过以下代码
LinkedHashSet<String> lhs = new LinkedHashSet<String>();
for(int i=0;i<10;i++)
lhs.add(String.valueOf(i));
Iterator<String> it=lhs.iterator();
System.out.println("removed?=="+lhs.remove("1"));
while(it.hasNext())
{
System.out.println("lhs"+it.next());
}
我得到了以下输出
removed?==true
Exception in thread "main" java.util.ConcurrentModificationException
at java.util.LinkedHashMap$LinkedHashIterator.nextEntry(Unknown Source)
at java.util.LinkedHashMap$KeyIterator.next(Unknown Source)
at preac.chapter1.Start.main(Start.java:321)
我想念的是什么？提前致谢.
PS我也试过iterator.remove()方法,但得到了非法状态异常
编辑
我刚才知道我必须使用iterator remove方法.然后使用Link Hash Set删除方法是什么？在哪些情况下我们应该使用这种方法？
我知道'between'运算符包含指定的范围.但在下面的情况下,它的工作方式不同
我有表客户具有以下属性.
customer
{
customername varchar2(30),
custid integer(10,0)
}
询问
select * from customer c where c.customername between 'a' and 'b';
以上查询仅提取数据,以"a"开头的客户名称.但是当我们使用'between'运算符和数字时,两者都是包容性的.任何人都可以向我解释这个行为.