小编geo*_*van的帖子

我正在尝试按照我老师的文档提供的信息建立一个简单的jpa 2.0项目.我已经在这几个小时了,但无论我做什么,当我尝试创建一个EntityManagerFactory时,我总是得到这个例外:我发现了很多关于这个例外的类似问题,但没有解决方案,我能够开始工作.我在这做错了什么？

我从Eclipse创建了这个项目(没有命令提示符)

Exception in thread "main" javax.persistence.PersistenceException: No Persistence provider for EntityManager named course
    at javax.persistence.Persistence.createEntityManagerFactory(Persistence.java:56)
    at javax.persistence.Persistence.createEntityManagerFactory(Persistence.java:34)
    at message.SaveMessage.main(SaveMessage.java:8)

目录结构

我的persistence.xml

<persistence xmlns="http://java.sun.com/xml/ns/persistence"
    xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
    xsi:schemaLocation="http://java.sun.com/xml/ns/persistence http://java.sun.com/xml/ns/persistence/persistence_2_0.xsd"
    version="2.0">
    <persistence-unit name="course" transaction-type="RESOURCE_LOCAL">

        <provider>org.hibernate.ejb.HibernatePersistence</provider>


        <properties>
            <property name="hibernate.dialect" value="org.hibernate.dialect.MySQL5InnoDBDialect" />
            <property name="hibernate.hbm2ddl.auto" value="update" />

            <property name="javax.persistence.jdbc.driver" value="com.mysql.jdbc.Driver" />
            <property name="javax.persistence.jdbc.url" value="jdbc:mysql://localhost:3306/StudentDB" />
            <property name="javax.persistence.jdbc.user" value="root" />
            <property name="javax.persistence.jdbc.password" value="pasapas2005" />
        </properties>
    </persistence-unit>
</persistence>

我的课

package message;

import java.io.Serializable;

import javax.persistence.*;

@Entity
public class Message implements Serializable {

    private long id;
    private …

我正在学习使用 Spring Boot 2.0 和 React 进行全栈开发。身份验证和授权由 JWT 管理，应用程序按预期工作，只是刷新浏览器后必须重新登录。即使浏览器刷新后如何维护 JWT 令牌？

import React, { Component } from 'react';
    import TextField from '@material-ui/core/TextField';
    import Button from '@material-ui/core/Button';
    import Snackbar from '@material-ui/core/Snackbar';
    import Carlist from './Carlist';
    import {SERVER_URL} from '../constants.js';

    class Login extends Component {
      constructor(props) {
        super(props);
        this.state = {username: '', password: '', isAuthenticated: false, open: false};
      }

      logout = () => {
        sessionStorage.removeItem("jwt");
        this.setState({isAuthenticated: false});
    }

      login = () => {
        const user = {username: this.state.username, password: this.state.password}; …

从char转换为String应导致以下错误:此代码:

char [] arr = {'H', 'e', 'l', 'l', 'o'};
        String c = arr[1]; 

错误:类型不匹配:无法从char转换为String

这段代码:

char [] arr = {'H', 'e', 'l', 'l', 'o'};
String c = "";
for(char i : arr) {
    c += i;
}

作品.

此代码在文件夹目录中创建一个.txt(它可以工作),但是当有时间删除整个目录或使用delete()方法的.txt文件时,没有任何反应.delete()仅当我将.txt文件替换为普通文件夹时,该方法才有效

import java.io.*;

public class Filemkdir {
        public static void main(String[] args) throws Exception {
                File f = new File("C:/Temp/Java/secret.txt");


                FileWriter fSecret = new FileWriter(f);
                f.mkdir();

                f.delete();
        }
}

可能重复:
如何比较Java中的字符串？

我可能在某个地方犯了一个逻辑错误,但我不知道在哪里.即使条件似乎为TRUE,输出也始终为FALSE

public class Test {
  public static void main(String[] args) {

    String str1 ="Hello world";
    String str2 ="Hello world";

    if (checkSubstring(str1,str2)){
         System.out.println("Cool");
    }
    else 
         System.out.println("Not cool");
 }

 static boolean checkSubstring(String str1, String str2) {

    String s1 = str1;
    String s2 = str2;
    if (s1.substring(4)== s2.substring(4)){
      return true;  
    }
    else
    return false;
 }
}

我正在尝试使用JSP/JSTL迭代随机生成的整数的myList(数组).生成和存储整数的代码snipet位于我的servlet中.

另一方面,遍历字符串的数组列表(参见下面的代码)可以很好地工作,但是当我使用基于相同逻辑的数组进行迭代时,我的网页只显示任何无序的随机整数列表.

谢谢你帮助我

我的Servlet

package be.intec.servlets;

    import java.io.IOException;
    import java.math.BigDecimal;
    import java.util.ArrayList;
    import java.util.Arrays;
    import java.util.LinkedHashMap;
    import java.util.List;
    import java.util.Map;

    import javax.servlet.RequestDispatcher;
    import javax.servlet.ServletException;
    import javax.servlet.annotation.WebServlet;
    import javax.servlet.http.HttpServlet;
    import javax.servlet.http.HttpServletRequest;
    import javax.servlet.http.HttpServletResponse;

    import be.intecbrussel.entities.Auto;

    @WebServlet("/JSTLServlet")
    public class JSTLServlet extends HttpServlet {

        private static final long serialVersionUID = 1L;

        private static final String VIEW = "WEB-INF/JSP/JSTL.jsp";

        protected void doGet(HttpServletRequest request,
                HttpServletResponse response) throws ServletException, IOException {

            RequestDispatcher dispatcher = request.getRequestDispatcher(VIEW);

            //=======below is the code using Array=====================================
            int[] myList = new int[42]; …

我正在尝试构建一个简单的程序,演示如何在switch语句中使用String类,但我收到编译错误,因为此功能需要Java SE 7(我的计算机中已经安装了).

第5行的错误消息:无法打开String类型的值.只允许使用可转换的int值或enum consants

Eclipse是否指向JRE7？=是.Environnement变量指向JDK7 = YES.

java -version = OK(见下图).

我甚至试图用控制台提示符编译代码.看错误:

import javax.swing.JOptionPane;

public class SwitchDemo {
    public static void main(String[] args) {

        String name = "georges";

        switch (name.toLowerCase()) {


        case "Jhon":
            JOptionPane.showMessageDialog(null, "Good morning, Jhon!");
         break;
      case "georges":
             JOptionPane.showMessageDialog(null, "How's it going, georges?");
         break;
      case "sergei":
             JOptionPane.showMessageDialog(null, "sergei, my old sergei!");
          break;
      case "Steph":
             JOptionPane.showMessageDialog(null, "Afternoon lennert, how's the Steph?");
          break;
      default:
             JOptionPane.showMessageDialog(null, "Pleased to meet you, xxxxxx.");
           break;
   }

    }
}