我有两个班
MyItem对象:
@Entity
public class MyItem implements Serializable {
@Id
private Integer id;
@ManyToOne(cascade = {CascadeType.PERSIST, CascadeType.MERGE})
private Component defaultComponent;
@ManyToOne(cascade = {CascadeType.PERSIST, CascadeType.MERGE})
private Component masterComponent;
//default constructor, getter, setter, equals and hashCode
}
组件对象:
@Entity
public class Component implements Serializable {
@Id
private String name;
//again, default constructor, getter, setter, equals and hashCode
}
我想用以下代码坚持下去:
public class Test {
public static void main(String[] args) {
Component c1 = new Component();
c1.setName("comp");
Component c2 = new Component();
c2.setName("comp");
System.out.println(c1.equals(c2)); …在这部分代码中:
System.out.println("Alunos aprovados:");
String[] aprovados = {"d", "a", "c", "b"};
List<String> list = new ArrayList();
for (int i = 0; i < aprovados.length; i++) {
if (aprovados[i] != null) {
list.add(aprovados[i]);
}
}
aprovados = list.toArray(new String[list.size()]);
Arrays.sort(aprovados);
System.out.println(Arrays.asList(aprovados));
System.out.println的示例结果是:
[A B C D]
如果我想要下面的结果,我怎么能修改上面的代码？
一个
b
C
d
或至少:
一个,
b,
C,
d
我正在使用a ThreadPoolExecutor来运行任务.后端是一个SynchronousQueue,所以如果执行程序已经执行了一个任务,它会抛出一个任务RejectedExecutionException.这是一个简单的测试用例:
public class ExecutorTest {
final static Worker worker = new Worker();
public static void main(String[] args) {
ThreadPoolExecutor executor = new ThreadPoolExecutor(1, 1, 0L, TimeUnit.MILLISECONDS, new SynchronousQueue<Runnable>());
while (true) {
try {
executor.execute(worker);
}catch (RejectedExecutionException e) {
}
}
}
static class Worker implements Runnable {
private int i = 0;
private long start = System.currentTimeMillis();
@Override
public void run() {
try {
Thread.sleep(1000);
System.out.println(++i + " " + (System.currentTimeMillis() - start));
} …从这个问题派生出来,是否可以将HQL或Criteria用于以下SQL语句:
SELECT
e.type,
count(e),
count(d),
count (case when gender = 'male' then 1 else NULL end) AS NumberOfMaleEmployees
from Department d
JOIN d.employees e
WHERE e.dead = 'maybe'
GROUP BY e.type
虽然谷歌提出了一些HQL支持CASE语句的命中,但Hibernate 3.6.6失败了
QuerySyntaxException:意外令牌:CASE
当我在EntityManager实例上创建上面的查询时.
为每个e.type创建另一个查询来手动确定男性的数量,例如每个e.type,这有多糟糕.
SELECT
count(e),
from Department d
JOIN d.employees e
WHERE e.dead = 'maybe', e.type = ugly
由于可能有很多类型,这可能很慢.我想数据库为我做的工作.
如果实例化一个新的 HashSet，之后通常会使用 Set 接口来使用它。就像
Set<T> set = new HashSet();
那么显式指定 HashSet 的类型有什么用呢？例如：
Set<T> set = new HashSet<T>();
我在好几本书上见过它，但我根本想不出有什么用处。如果您需要访问该集合，无论如何您都将使用该界面（已参数化）。
public static void main(String args[]) {
System.out.println(reverseString("His"));
}
public static String reverseString(String s) {
if (s.length() <= 1) {
return s;
} else {
char c = s.charAt(0);
return reverseString(s.substring(1)) + c;
}
}
有人可以详细解释我这个方法是如何工作的