Question

问题是：我使用条件从我的数据库中获取＆＃34; startDate＆＃34;的最大值。和＆＃34; endDate＆＃34;。现在我必须找到数据库时间戳中这两个对象之间的区别。

有谁知道怎么做？

taskTimes.java

public Object taskTimes() {
        Session session = sessionFactory.openSession();
        session.beginTransaction();
        Criteria criteria = session.createCriteria(Task.class).setProjection(Projections.max("endDate"));
        Object avgDate = (Object) criteria.uniqueResult();

        Criteria criteria1 = session.createCriteria(Task.class).setProjection(Projections.max("startDate"));
        Object avgDate1 = (Object) criteria1.uniqueResult();
        System.out.println(avgDate);
        System.out.println(avgDate1);
        session.close();

        return avgDate;
    }

Task.java

@Entity
public class Task {

    @Id
    @GeneratedValue(strategy = GenerationType.AUTO)
    private Long id;

    @Column(nullable = false)
    private int batch;

    @Column
    private java.sql.Timestamp startDate;

    @Column
    private java.sql.Timestamp endDate;

    @ManyToOne
    private Student student;

    @ManyToOne
    private Job job;

    @OneToMany(mappedBy="task")
    private List<Result> results;

    public Long getId() {
        return id;
    }

    public void setId(Long id) {
        this.id = id;
    }

    public Student getStudent() {
        return student;
    }

    public void setStudent(Student student) {
        this.student = student;
    }

    public Job getJob() {
        return job;
    }

    public void setJob(Job job) {
        this.job = job;
    }

    public List<Result> getResults() {
        return results;
    }

    public void setResults(List<Result> results) {
        this.results = results;
    }

    public Task(Long id, int batch, Timestamp startDate, Timestamp endDate, Student student, Job job,
            List<Result> results) {
        super();
        this.id = id;
        this.batch = batch;
        this.startDate = startDate;
        this.endDate = endDate;
        this.student = student;
        this.job = job;
        this.results = results;
    }

    public java.sql.Timestamp getStartDate() {
        return startDate;
    }

    public void setStartDate(java.sql.Timestamp startDate) {
        this.startDate = startDate;
    }

    public java.sql.Timestamp getEndDate() {
        return endDate;
    }

    public void setEndDate(java.sql.Timestamp endDate) {
        this.endDate = endDate;
    }

    public Task() {

    }

    public int getBatch() {
        return batch;
    }

    public void setBatch(int batch) {
        this.batch = batch;
    }

    @Override
    public String toString() {
        return "Task [id=" + id;
    }

Answer 1

根据java.sql.Timestamp的API，它提供了一个方法getTime（）。

返回自此Timestamp对象表示的1970年1月1日00:00:00 GMT以来的毫秒数。

要计算差异，您可以结合使用此方法和#include <stdio.h> int main (void) { static int array[10][10]; int i, j, m, n, sum = 0; printf("Enter the order of the matrix\n"); scanf("%d %d", &m, &n); printf("Enter the co-efficients of the matrix\n"); for (i = 0; i < m; ++i) { for (j = 0; j < n; ++j) { scanf("%d", &array[i][j]); } } for (i = 0; i < m; ++i) { for (j = 0; j < n; ++j) { sum = sum + array[i][j] ; } printf("Sum of the %d row is = %d\n", i, sum); sum = 0; } }。

Answer 2

如果对象的类型为Timestamp，则将它们转换为Timestamp

Timestamp avgDateTs = (Timestamp)avgDate;
Timestamp avgDateTs1 = (Timestamp)avgDate1;

然后以毫秒为单位获取时间。

long timeInMills = avgDateTs.getTime();
long timeInMills1 = avgDateTs1.getTime();

执行毫秒值的差异，这会产生毫秒差异。

然后您可以除以1000以获得以秒为单位的差异。

long differenceInSecs = (timeInMills1-timeInMills)/1000;

计算两个时间戳对象java之间的差异

2 个答案: