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());
}
}
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);