运行一个简单的Javaservlet并不是一个Servlet错误
我有一个简单的Javaservlet项目,我正在通过maven tomcat插件运行它运行一个简单的Javaservlet并不是一个Servlet错误,java,tomcat,servlets,Java,Tomcat,Servlets,我有一个简单的Javaservlet项目,我正在通过maven tomcat插件运行它 javax.servlet.ServletException: Class in.vshukla.MyServlet is not a Servlet java.lang.ClassCastException: in.vshukla.MyServlet cannot be cast to javax.servlet.Servlet servlet类是 import javax.servlet.Servle
javax.servlet.ServletException: Class in.vshukla.MyServlet is not a Servlet
java.lang.ClassCastException: in.vshukla.MyServlet cannot be cast to javax.servlet.Servlet
servlet类是
import javax.servlet.Servlet;
import javax.servlet.ServletConfig;
import javax.servlet.ServletException;
import javax.servlet.ServletRequest;
import javax.servlet.ServletResponse;
import java.io.IOException;
import java.io.PrintWriter;
public class MyServlet implements Servlet {
.
.
.
public void service(ServletRequest servletRequest, ServletResponse servletResponse) throws ServletException, IOException {
PrintWriter out = servletResponse.getWriter();
out.println("<html>");
out.println("<body>");
out.println("<h1>Hello, world!</h1>");
out.println("</body>");
out.println("</html>");
out.close();
}
.
.
.
}
import javax.servlet.servlet;
导入javax.servlet.ServletConfig;
导入javax.servlet.ServletException;
导入javax.servlet.ServletRequest;
导入javax.servlet.ServletResponse;
导入java.io.IOException;
导入java.io.PrintWriter;
公共类MyServlet实现Servlet{
.
.
.
公共void服务(ServletRequest ServletRequest,ServletResponse ServletResponse)抛出ServletException,IOException{
PrintWriter out=servletResponse.getWriter();
out.println(“”);
out.println(“”);
out.println(“你好,世界!”);
out.println(“”);
out.println(“”);
out.close();
}
.
.
.
}
这伴随着一个简单的web.xml文件
<web-app xmlns="http://xmlns.jcp.org/xml/ns/javaee"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://xmlns.jcp.org/xml/ns/javaee
http://xmlns.jcp.org/xml/ns/javaee/web-app_3_1.xsd"
version="3.1">
<servlet>
<servlet-name>servlet</servlet-name>
<servlet-class>in.vshukla.MyServlet</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>servlet</servlet-name>
<url-pattern>/*</url-pattern>
</servlet-mapping>
</web-app>
<?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 https://maven.apache.org/xsd/maven-4.0.0.xsd">
<parent>
<groupId>in.vshukla</groupId>
<artifactId>my</artifactId>
<version>0.1-SNAPSHOT</version>
</parent>
<modelVersion>4.0.0</modelVersion>
<artifactId>my-api</artifactId>
<packaging>war</packaging>
<properties>
<javax.version>3.1.0</javax.version>
</properties>
<dependencies>
<dependency>
<groupId>in.vshukla</groupId>
<artifactId>my-data</artifactId>
<version>${project.parent.version}</version>
</dependency>
<dependency>
<groupId>javax.servlet</groupId>
<artifactId>javax.servlet-api</artifactId>
<version>${javax.version}</version>
</dependency>
</dependencies>
</project>
servlet
in.vshukla.MyServlet
servlet
/*
我已经交叉检查了生成的war文件。我在war文件上附加tree命令的输出
.
|-- META-INF
`-- WEB-INF
|-- classes
| |-- in
| | `-- vshukla
| | `-- MyServlet.class
| `-- spring
| `-- spring_config.xml
|-- lib
| |-- javax.servlet-api-3.1.0.jar
| `-- my-data-0.1-SNAPSHOT.jar
`-- web.xml
7 directories, 5 files
.
|--META-INF
`--WEB-INF
|--班级
||--in
|| `--vshukla
||`--MyServlet.class
|”“春天
|`--spring_config.xml
|--解放党
||--javax.servlet-api-3.1.0.jar
|`--my-data-0.1-SNAPSHOT.jar
`--web.xml
7个目录,5个文件
我正在使用maven构建这个。
下面是pom.xml文件
<web-app xmlns="http://xmlns.jcp.org/xml/ns/javaee"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://xmlns.jcp.org/xml/ns/javaee
http://xmlns.jcp.org/xml/ns/javaee/web-app_3_1.xsd"
version="3.1">
<servlet>
<servlet-name>servlet</servlet-name>
<servlet-class>in.vshukla.MyServlet</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>servlet</servlet-name>
<url-pattern>/*</url-pattern>
</servlet-mapping>
</web-app>
<?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 https://maven.apache.org/xsd/maven-4.0.0.xsd">
<parent>
<groupId>in.vshukla</groupId>
<artifactId>my</artifactId>
<version>0.1-SNAPSHOT</version>
</parent>
<modelVersion>4.0.0</modelVersion>
<artifactId>my-api</artifactId>
<packaging>war</packaging>
<properties>
<javax.version>3.1.0</javax.version>
</properties>
<dependencies>
<dependency>
<groupId>in.vshukla</groupId>
<artifactId>my-data</artifactId>
<version>${project.parent.version}</version>
</dependency>
<dependency>
<groupId>javax.servlet</groupId>
<artifactId>javax.servlet-api</artifactId>
<version>${javax.version}</version>
</dependency>
</dependencies>
</project>
in.vshukla
我的
0.1-1快照
4.0.0
我的api
战争
3.1.0
in.vshukla
我的数据
${project.parent.version}
javax.servlet
javax.servlet-api
${javax.version}
即使MyServlet类正在实现一个javaxservlet,我也会遇到这个错误
我做错了什么?您通常不希望在.war中包含javax.servlet-api-3.1.0.jar。它将由框架提供。如果两个类由不同的类加载器加载,即使它们具有相同的包和名称,它们也将被视为不同的类。请粘贴导入部分好吗?如果改为使用公共类MyServlet Extendes HttpServlet呢{@Sarker:添加了导入部分。如果我扩展了HttpServlet,则会出现相同的错误。只需从
.war
文件中删除javax.servlet-api-3.1.0.jar
,正如IanLovejoy在回答中指定的那样,您的问题就会消失。您可能想更强调地说明它,例如,不要包含javax.servlet-api-3.1.0.jar
永远不要将它包含在.war
文件中。@Andreas哈哈,是的,好建议,我想我有时会礼貌地掩盖我的观点。也许是一个中间立场——“你不应该……”好的。我如何防止它被包括在战争中?@venky这取决于它最初是如何进入那里的。你还没有共享你的构建脚本。如果使用Maven,请参阅以下问题:@venky然后添加提供的
。