使用Spring注入的InstantiationException

时间:2013-01-11 17:43:43

标签: java spring tomcat code-injection

我第一次尝试使用Spring进样。我肯定会忘记一些明显的东西,但我不知道它是什么。

在src / main / java下,我有一个包含Hello,Animal,Cat。

的包'example'

在src / main / webapp / WEB-INF下,我有web.xml和springapp-servlet.xml。

当我使用Tomcat部署我的应用程序时,我得到了一个:

javax.servlet.ServletException: Error instantiating servlet class example.Hello
    org.apache.catalina.authenticator.AuthenticatorBase.invoke(AuthenticatorBase.java:472)

注射工作我错过了什么?

来源:

Hello.java

package example;

import java.io.IOException;

import javax.servlet.ServletException;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;

import org.springframework.beans.factory.annotation.Autowired;
import org.springframework.stereotype.Service;

@Service
public class Hello extends HttpServlet {
  private final Animal animal;

  @Autowired
  public Hello(final  Animal animal) {
    this.animal = animal;
  }

  @Override
  protected void doGet(final HttpServletRequest req,
          final HttpServletResponse resp) throws ServletException, IOException {
    resp.getWriter().write(animal.sound());
  }
}

Cat.java

package example;

import org.springframework.stereotype.Service;

@Service
public class Cat implements Animal {
  public String sound() {
    return "Miaou";
  }
}

Animal.java

package example;

public interface Animal {
  public String sound() ;
}

web.xml

<?xml version="1.0" encoding="ISO-8859-1"?>
<web-app xmlns="http://java.sun.com/xml/ns/javaee" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
    xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd"
    version="2.5">

    <context-param>
        <param-name>contextConfigLocation</param-name>
        <param-value>/WEB-INF/springapp-servlet.xml</param-value>
    </context-param>

    <listener>
        <listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
    </listener>

    <servlet>
        <servlet-name>Hello</servlet-name>
        <servlet-class>example.Hello</servlet-class>
    </servlet>
    <servlet-mapping>
        <servlet-name>Hello</servlet-name>
        <url-pattern>/hello</url-pattern>
    </servlet-mapping>

</web-app>

springapp-servlet.xml中

<?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:mvc="http://www.springframework.org/schema/mvc"
  xmlns:context="http://www.springframework.org/schema/context"
  xsi:schemaLocation="
        http://www.springframework.org/schema/mvc http://www.springframework.org/schema/mvc/spring-mvc-3.0.xsd
        http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans-3.0.xsd
        http://www.springframework.org/schema/context http://www.springframework.org/schema/context/spring-context-3.0.xsd">

  <context:component-scan base-package="example" />
    <mvc:annotation-driven />

</beans>

我最初认为也许我的springapp-servlet.xml甚至没有读过,但是如果我在web.xml中的名字springapp-servlet.xml上输入错误,我在部署时遇到错误,所以我显然有springapp-servlet.xml的正确路径。它正在但是注射不起作用。

更新:

由于下面的答案,我在下面显示了对我有用的解决方案。除了Hello:

之外,所有代码都保持不变

Hello.java

public class Hello extends HttpServlet {

  @Inject
  private Animal animal;

  @Override
  public void init(final ServletConfig config) throws ServletException {
    super.init(config);
    SpringBeanAutowiringSupport.processInjectionBasedOnServletContext(this,
            config.getServletContext());
  }

  @Override
  protected void doGet(final HttpServletRequest req,
          final HttpServletResponse resp) throws ServletException, IOException {
    resp.getWriter().write(animal.sound());
  }    
}

1 个答案:

答案 0 :(得分:2)

这是错误的:

@Service
public class Hello extends HttpServlet {

Servlet的生命周期由servlet容器控制,而不是由Spring控制。因此,您无法将Spring bean直接自动装配到servlet。 Spring根本不应该创建servlet。基本上Spring不知道你的servlet,它试图实例化它,但它不是由servlet容器创建的用于处理请求的实例。

最后,您的servlet没有no-arg构造函数。这样的构造函数是必需的,但它不会使你的例子通过。

解决方案是直接从已注册的Web应用程序上下文中获取所需的Spring bean:

WebApplicationContext context = WebApplicationContextUtils.getRequiredWebApplicationContext(getServletContext());
Animal animal = context.getBean(Animal.class);

另见(其他解决方案)