小编sag*_*mbu的帖子

原始服务器没有找到目标资源的当前表示,或者不愿意透露存在该目标资源

在此输入图像描述

web.xml中

    <?xml version="1.0" encoding="UTF-8"?>
     <web-app xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns="http://xmlns.jcp.org/xml/ns/javaee" xsi:schemaLocation="http://xmlns.jcp.org/xml/ns/javaee http://xmlns.jcp.org/xml/ns/javaee/web-app_3_1.xsd" id="WebApp_ID" version="3.1">
  <display-name>springsecuritydemo</display-name>
<!--   <welcome-file-list>
    <welcome-file>index.html</welcome-file>
    <welcome-file>index.htm</welcome-file>
    <welcome-file>index.jsp</welcome-file>
    <welcome-file>default.html</welcome-file>
    <welcome-file>default.htm</welcome-file>
    <welcome-file>default.jsp</welcome-file>
  </welcome-file-list> -->
  <servlet>
    <description></description>
    <display-name>offers</display-name>
    <servlet-name>offers</servlet-name>
    <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
   <load-on-startup>1</load-on-startup>
  </servlet>
  <servlet-mapping>
    <servlet-name>offers</servlet-name>
    <url-pattern>/DispatcherServlet</url-pattern>
  </servlet-mapping>
</web-app>
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报价-sevlet.xml

    <?xml version="1.0" encoding="UTF-8"?>
<beans xmlns = "http://www.springframework.org/schema/beans"
   xmlns:context = "http://www.springframework.org/schema/context"
   xmlns:xsi = "http://www.w3.org/2001/XMLSchema-instance"
   xmlns:mvc="http://www.springframework.org/schema/mvc"
   xsi:schemaLocation = "http://www.springframework.org/schema/beans     
   http://www.springframework.org/schema/beans/spring-beans-4.3.xsd
   http://www.springframework.org/schema/context 
   http://www.springframework.org/schema/context/spring-context-4.3.xsd
   http://www.springframework.org/schema/mvc
   http://www.springframework.org/schema/mvc/spring-mvc.xsd">

   <context:component-scan base-package="com.spring.security.web"></context:component-scan> 
   <mvc:annotation-driven></mvc:annotation-driven>

   <bean name="jspViewResolver" class="org.springframework.web.servlet.view.InternalResourceViewResolver">
    <property name="prefix" value="/WEB-INF/jsps/"></property>
    <property name="suffix" value=".jsp"></property>
   </bean>

</beans>
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这有什么不对?我无法访问home.jsp.我实际上是在春季3.0观看教程,我已经完成了视频中的显示.谁能在这里指出我的错误?

java spring

10
推荐指数
2
解决办法
15万
查看次数

session/entitymanager已关闭

我有这个hibernate dao,它在我的本地机器上测试时工作正常.但对于某些事务,它会抛出IllegalStateException.我相信这是因为多个用户同时打.(我可能错了).

UpdatePaymentDao

@Repository公共类UpdatePaymentImpl实现UpdatePayment {

@Repository
public class UpdatePaymentImpl implements UpdatePayment {


    @Autowired
    SessionFactory sessionFactory;
    Session session;
    Transaction trans;

    private static final long LIMIT = 100000000000L;
    private static final long LIMIT2 = 10000000000L;
    private static long last = 0;


    public static long getUniqueID() {
        // 10 digits.
        long id = (System.currentTimeMillis() % LIMIT) + LIMIT2;
        System.out.println("id"+id);
        System.out.println("system time"+System.currentTimeMillis());
        System.out.println("milssiiiiii=============="
                + System.currentTimeMillis());
        if (id <= last) {
            id = (last + 1) % LIMIT;
        }
        return last = id;
    }

    public …
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oracle spring hibernate

4
推荐指数
1
解决办法
1万
查看次数

如何仅按日期从 postgres 的时间戳列中选择?

我有一张桌子。

在此处输入图片说明

如您所见, created_date 列是时间戳字段。现在选择我只想考虑日期值。例如,如果我想从今天开始选择行,我想做以下事情:-

select * from audit_logs where created_date = '2018-11-28';
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上面的查询返回 null。可以这样选择吗?

sql postgresql

4
推荐指数
1
解决办法
4405
查看次数

Postgres DDL 错误:''user' 处或附近的语法错误'

我正在尝试使用 postgres 数据库设置 Spring Boot 项目。我的实体是:-

用户

@Entity
public class User implements UserDetails {

    @Id
    @GeneratedValue(strategy=GenerationType.AUTO)
    @Column(name="id", nullable = false, updatable = false)
    private Long id;
    private String username;
    private String password;
    private String firstName;
    private String lastName;

    @Column(name="email", nullable = false, updatable = false)
    private String email;
    private String phone;
    private boolean enabled=true;


    @OneToMany(mappedBy = "user", cascade = CascadeType.ALL, fetch = FetchType.EAGER)
    @JsonIgnore
    private Set<UserRole> userRoles = new HashSet<>();

    public Long getId() {
        return id;
    }
    public void setId(Long …
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postgresql spring-data-jpa spring-boot

3
推荐指数
1
解决办法
3077
查看次数

如何处理java流api中的可选对象?

我在做简单的测试。

@Test
    public void whenFilterEmployees_thenGetFilteredEmployees(){
        Integer[] empIds = {1,2,3};
        List<Optional<Employee>> employees = Stream.of(empIds)
                .map(employeeRepository::findById)
                .collect(Collectors.toList());

        List<Employee> employeeList = employees
                .stream().filter(Optional::isPresent)
                .map(Optional::get)
                .filter(e->e !=null)
                .filter(e->e.getSalary()>200000)
                .collect(Collectors.toList());

        assertEquals(Arrays.asList(arrayOfEmps[2]), employeeList);


    }
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我的员工表包含数据:

1   Jeff Bezos  100000
2   Bill Gates  200000
3   Mark Zuckerberg 300000
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当前测试成功运行。

如您所见,我准备了两个员工列表,即员工和员工列表

我这样做是因为findById方法返回 Optional。我如何使用流 api 以便我可以简单地获取员工列表

List<Employee> employeeList= ....
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java java-stream

2
推荐指数
1
解决办法
2659
查看次数

Thymeleaf:org.thymeleaf.exceptions.TemplateInputException

我正在尝试使用3.0.7.RELEASE基于spring mvc java的配置的thymeleaf .

@Configuration
@EnableWebMvc
@ComponentScan(basePackages = "com.sagar")
public class MvcConfig extends WebMvcConfigurerAdapter implements ApplicationContextAware{

    @Autowired
    RoleToUserProfileConverter roleToUserProfileConverter;

    private ApplicationContext applicationContext;

    public void setApplicationContext(ApplicationContext applicationContext) {
        this.applicationContext = applicationContext;
    }

    @Bean
    public ViewResolver viewResolver() {
        ThymeleafViewResolver resolver = new ThymeleafViewResolver();
        resolver.setTemplateEngine(templateEngine());
        resolver.setCharacterEncoding("UTF-8");
        return resolver;
    }

    @Bean
    public TemplateEngine templateEngine() {
        SpringTemplateEngine engine = new SpringTemplateEngine();
        engine.setEnableSpringELCompiler(true);
        engine.setTemplateResolver(templateResolver());
        return engine;
    }

    private ITemplateResolver templateResolver() {
        SpringResourceTemplateResolver resolver = new SpringResourceTemplateResolver();
        resolver.setApplicationContext(applicationContext);
        resolver.setPrefix("/WEB-INF/templates/");
        resolver.setTemplateMode(TemplateMode.HTML);
        return resolver;
    }

    public void …
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java spring thymeleaf

1
推荐指数
1
解决办法
8336
查看次数

使用oracle作为数据库打包spring boot应用程序

我正在尝试打包我的spring boot应用程序,以便可以将其部署在tomcat服务器上。观看完youtube视频后,我扩展了SpringBootServletInitializer类并进行了一些更改。

 @ComponentScan("com.infodev.loksewa")
    @SpringBootApplication
    public class LoksewaApplication extends SpringBootServletInitializer {

    public static void main(String[] args) {
        SpringApplication.run(LoksewaApplication.class, args);
    }

    @Override
    protected SpringApplicationBuilder configure(SpringApplicationBuilder builder) {
        return builder.sources(LoksewaApplication.class);
    }
}
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pox.xml

<?xml version="1.0" encoding="UTF-8"?>
<project xmlns="http://maven.apache.org/POM/4.0.0"
         xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
         xsi:schemaLocation="http://maven.apache.org/POM/4.0.0 http://maven.apache.org/xsd/maven-4.0.0.xsd">
    <modelVersion>4.0.0</modelVersion>

    <groupId>com.infodev</groupId>
    <artifactId>loksewa</artifactId>
    <version>0.0.1-SNAPSHOT</version>
    <packaging>war</packaging>

    <name>loksewa</name>
    <description>Demo project for Spring Boot</description>

    <parent>
        <groupId>org.springframework.boot</groupId>
        <artifactId>spring-boot-starter-parent</artifactId>
        <version>2.0.2.RELEASE</version>
        <relativePath/> <!-- lookup parent from repository -->
    </parent>

    <properties>
        <project.build.sourceEncoding>UTF-8</project.build.sourceEncoding>
        <project.reporting.outputEncoding>UTF-8</project.reporting.outputEncoding>
        <java.version>1.8</java.version>
    </properties>

    <dependencies>
        <dependency>
            <groupId>org.springframework.boot</groupId>
            <artifactId>spring-boot-starter-data-jpa</artifactId>
        </dependency>
        <dependency>
            <groupId>org.springframework.boot</groupId>
            <artifactId>spring-boot-starter-thymeleaf</artifactId>
        </dependency>
        <dependency>
            <groupId>org.springframework.boot</groupId>
            <artifactId>spring-boot-starter-web</artifactId>
        </dependency>
        <dependency> …
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java intellij-idea spring-boot

1
推荐指数
1
解决办法
1789
查看次数

更新资源时Intellij不更新html文件

每当我在html文件上更改某些内容时,我必须重新启动整个项目.我试过了:-

- >编辑配置 - >服务器并更改帧检测以更新资源.

在此输入图像描述

在我的服务器选项卡上,还有关于帧检测的更新资源.但仍然无法更新我的html文件,我必须重新启动服务器以反映我的更改.

在此输入图像描述

Intellij Idea版本是2017年.

java thymeleaf

0
推荐指数
1
解决办法
797
查看次数

发送不同对象的列表并返回其属性的总和

我必须上课

public class Consumer{


private String name;
private int salary;


public String getName() {
    return name;
}

public void setName(String name) {
    this.name = name;
}

public int getSalary() {
    return salary;
}

public void setSalary(int salary) {
    this.salary = salary;
}
}
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接下来

public class Donor {

private String name;
private int amount;
private String location;



public String getName() {
    return name;
}

public void setName(String name) {
    this.name = name;
}

public int getAmount() {
    return amount;
} …
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java

0
推荐指数
1
解决办法
47
查看次数