小编Evg*_*rov的帖子

如果java.net.URL在Spring Boot应用程序中使用,使用classpath协议,它会按预期工作,因为Spring Boot注册URLStreamHandlerFactory.例如new URL("classpath: someFile.whatever").

但是当执行此代码时,java.net.MalformedURLException: unknown protocol: classpath将抛出JUnit测试异常.

URLStreamHandlerFactory在为JUnit测试初始化​​Spring上下文时,似乎没有注册适当的.

重现步骤:

1)创建Spring Boot Starter项目(例如,仅使用入门Web).

2)添加test.json文件src/main/resources

3)添加以下bean:

@Component
public class MyComponent {
    public MyComponent() throws MalformedURLException {
        URL testJson = new URL("classpath: test.json");
        System.out.println(testJson);
    }
}

4)启动应用程序作为Java应用程序工作正常

5)运行默认的"contextLoads"测试:

@RunWith(SpringRunner.class)
@SpringBootTest
public class SpringUrlTestApplicationTests {

    @Test
    public void contextLoads() {
    }

}

java.net.MalformedURLException: unknown protocol: classpath抛出异常.

在JUnit测试中使用URL与classpath资源的合适方法是什么？

在真实的用例中,我无法改变new URL("classpath: test.json")它,因为它来自第三方库.

试图复制test.json中src/test/resources,以测试是否错误可能由缺少的资源造成的-没有成功.

我正在尝试编写一个简单的Spring AOP应用程序,但我遇到了xml配置问题.

我的xml:

<?xml version="1.0" encoding="UTF-8"?>


<beans xmlns="http://www.springframework.org/schema/beans"
    xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" 
    xmlns:aop="http://www.springframework.org/schema/aop"
    xsi:schemaLocation="http://www.springframework.org/schema/beans 
    http://www.springframework.org/schema/beans/spring-beans-3.0.xsd 
    http://www.springframework.org/schema/aop">

<bean id="audience" class="springaop.Audience">
</bean>

<bean id="sam" class="springaop.Singer">
    <property name="id" value="1"></property>
</bean>

<aop:config>
    <aop:aspect ref="audience">

        <aop:before pointcut="* springaop.Singer.perform(..)" 
        method="takeSeats"></aop:before>

    </aop:aspect>
</aop:config>

</beans>

我得到这个警告和例外:

WARNING: Ignored XML validation warning
    org.xml.sax.SAXParseException: SchemaLocation: schemaLocation value = 'http://www.springframework.org/schema/beans  http://www.springframework.org/schema/beans/spring-beans-3.0.xsd   http://www.springframework.org/schema/aop' must have even number of URI's.

Exception: Line 18 in XML document from class path resource [aop-conf.xml] is invalid; nested exception is org.xml.sax.SAXParseException: cvc-complex-type.2.4.c: The matching wildcard is strict, but no declaration …

我试图从spring网站运行这个例子: 教程 除了Spring Boot部分.

在web.xml

<web-app>
    <display-name>Archetype Created Web Application</display-name>

    <servlet>
        <servlet-name>sample</servlet-name>
        <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
        <init-param>
            <param-name>contextClass</param-name>
            <param-value>
                org.springframework.web.context.support.AnnotationConfigWebApplicationContext
            </param-value>
        </init-param>
        <init-param>
            <param-name>contextConfigLocation</param-name>
            <param-value>
                com.evgeni.websock.WebSocketConfig
            </param-value>
        </init-param>
    </servlet>

    <servlet-mapping>
        <servlet-name>sample</servlet-name>
        <url-pattern>/</url-pattern>
    </servlet-mapping>
</web-app>

Java配置:

@Configuration
@ComponentScan(basePackages = {"com.evgeni.controller"})
@EnableWebSocketMessageBroker
@EnableWebMvc
public class WebSocketConfig extends WebMvcConfigurerAdapter implements WebSocketMessageBrokerConfigurer  {

    public void registerStompEndpoints(StompEndpointRegistry registry) {
        registry.addEndpoint("/hello").withSockJS();
    }

    public void configureClientInboundChannel(ChannelRegistration registration) {
        // TODO Auto-generated method stub

    }

    public void configureClientOutboundChannel(ChannelRegistration registration) {
        // TODO Auto-generated method stub

    }

    public void configureMessageBroker(MessageBrokerRegistry registry) { …

我设法用Spring 4和Stomp创建简单的Websocket应用程序.在这里查看我的上一个问题 然后我尝试使用远程消息代理(ActiveMQ).我刚开始经纪人并改变了

registry.enableSimpleBroker("/topic");

至

registry.enableStompBrokerRelay("/topic");

它起作用了.

问题是如何配置代理？据我所知,在这种情况下,应用程序自动发现localhost上的代理:defaultport,如果我需要将应用程序指向其他机器上的其他代理,该怎么办？

如果您知道父 ID，如何使用 Spring data JPA 存储库保存子实体？

例如，如果我们有一对多关系：

@Entity
public class Customer {
    @Id @GeneratedValue(strategy=GenerationType.AUTO)
    private Long id;
    private String firstName;
    private String lastName;
    @ManyToOne
    private CustomerCategory category;
}

@Entity
public class CustomerCategory {
    @Id @GeneratedValue(strategy=GenerationType.AUTO)
    private Long id;
    private String name;
}

人们可以做这样的事情：

CustomerCategory someCategory = customerRepository.findOne(1L);// how to skip that line. The id should be enough. 
Customer cust = new Customer("Evgeni", "Dimitrov", someCategory);           
customerRepository.save(cust);

JPA 可以做到load（只需创建代理并设置 Id）而不是“获取”（从数据库中选择）。Spring Data JPA 可以实现这一点吗？

我将创建将在数据库上运行的类.该类将具有addRecord(),getAllRecords()等函数.我正在寻找一种设计课程的好方法.我应该:1)为每个功能创建新的连接.像这样:

void readRecords(){
    try {
        Connection con = DriverManager.getConnection (connectionURL);

        Statement stmt = con.createStatement();

        ResultSet rs = stmd.executeQuery("select moviename, releasedate from movies");

        while (rs.next())
            System.out.println("Name= " + rs.getString("moviename") + " Date= " + rs.getString("releasedate");
    }
    catch (SQLException e) {
        e.printStackTrace();
    }
    catch (Exception e) {
        e.printStackTrace();
    }
    finally {
        con.close();
    }
}

要么

2)最好将一个连接作为一个memeber变量

class MyClass{
     private  Connection con;

     public MyClass(){
          con = DriverManager.getConnection (connectionURL);
     }
}

并为每个函数创建语句.

3)或别的......

我尝试使用2个视图解析器:

<?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.xsd
        http://www.springframework.org/schema/context 
        http://www.springframework.org/schema/context/spring-context.xsd
        http://www.springframework.org/schema/mvc
        http://www.springframework.org/schema/mvc/spring-mvc.xsd">

    <context:component-scan base-package="com.evgeni.dfr.controller" />

    <context:annotation-config />
    <mvc:annotation-driven />

    <bean id="viewResolver"
        class="org.springframework.web.servlet.view.UrlBasedViewResolver">
        <property name="cache" value="false" />
        <property name="viewClass" value="com.evgeni.drf.faces.FacesView" />
        <property name="prefix" value="/WEB-INF/pages/" />
        <property name="suffix" value=".xhtml" />

        <property name="order" value="1" />
    </bean>

    <bean
        class="org.springframework.web.servlet.view.InternalResourceViewResolver">
        <property name="prefix">
            <value>/WEB-INF/views/</value>
        </property>
        <property name="suffix">
            <value>.jsp</value>
        </property>
        <property name="order" value="0" />
    </bean>
</beans>

应用程序始终只使用最低顺序而不是另一个.在目前的情况下,如果我的控制器返回"someView",The requested resource (/MyProject/WEB-INF/views/someView.jsp) is not available.即使有"pages/someView.xhtml" ,应用程序也会响应.

春季版 - 3.2.3

编辑:如果我在控制器中有2个方法,方法A返回"viewA",方法B返回"viewB".我们在'views'文件夹中有viewA.jsp,在'pages'中有viewB.xhtml.

Case1:UrlBasedViewResolver - > …

我试图制作一个简单的Spring Boot Web应用程序:

POM:

    <parent>
        <groupId>org.springframework.boot</groupId>
        <artifactId>spring-boot-starter-parent</artifactId>
        <version>1.0.0.BUILD-SNAPSHOT</version>
    </parent>

    <dependencies>
        <dependency>
            <groupId>org.springframework.boot</groupId>
            <artifactId>spring-boot-starter-actuator</artifactId>
        </dependency>
        <dependency>
            <groupId>org.springframework.boot</groupId>
            <artifactId>spring-boot-starter-web</artifactId>
        </dependency>
        <dependency>
            <groupId>org.springframework.boot</groupId>
            <artifactId>spring-boot-starter-test</artifactId>
            <scope>test</scope>
        </dependency>
    </dependencies>

主要课程:

@ComponentScan(basePackages={"controller"})
@Import(MvcConfig.class)
@EnableAutoConfiguration
public class Application {

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

控制器:

@Controller
@RequestMapping("/home")
public class HomeController {
    @RequestMapping("/hello")
    @ResponseBody
    public String hello(){
        return "hello";
    }
    @RequestMapping("/helloView")
    public String helloView(){
        return "homeView";
    }
}

此外,根据src/main我得到

src
   -main
       -resources
            applicaion.properties
       -webapp
            -WEB-INF
                 -jsp
                     -homeView.jsp

在application.properties我得到:

  spring.view.prefix: /WEB-INF/jsp/
  spring.view.suffix: …

我有一个Java类(MyResponse)由多个RestController方法返回,并有很多字段.

@RequestMapping(value = "offering", method=RequestMethod.POST)
public ResponseEntity<MyResponse> postOffering(...) {}

@RequestMapping(value = "someOtherMethod", method=RequestMethod.POST)
public ResponseEntity<MyResponse> someOtherMethod(...) {}

我想忽略(例如,不序列化)一个方法的属性之一.

我不想忽略该类的空字段,因为它可能对其他字段有副作用.

@JsonInclude(Include.NON_NULL)
public class MyResponse { ... }

该JsonView看起来不错,但据我了解我必须标注在类的所有其他领域有@JsonView不同之处,我想忽略这听起来笨拙的人.如果有办法做"反向JsonView"这样的话会很棒.

关于如何忽略控制器方法的属性的任何想法？

我有Spring-boot-starter-remote-shell的Spring Boot应用程序.当我把这个hello.groovy脚本打印出'hello'时就可以了.

package commands

import org.crsh.cli.Usage
import org.crsh.cli.Command

class hello {

    @Usage("Say Hello")
    @Command
    def main(InvocationContext context) {
        return "hello";
    }

}

但是当我尝试注入一些Spring bean时,它总是为null.

package commands

import org.crsh.cli.Usage
import org.crsh.cli.Command
import org.springframework.beans.factory.annotation.Autowired
import org.springframework.stereotype.Component
import org.springframework.batch.core.launch.JobLauncher

@Component
class hello {
    @Autowired
    JobLauncher jobLauncher;

    @Usage("Say Hello")
    @Command
    def main(InvocationContext context) {
        if(jobLauncher != null){
            return "OK";
        }else{
            return "NULL";
        }
        return "hello j";
    }

}

我有 @ComponentScan(basePackages={"com....", "commands"})