尝试在 tomcat 中显示某些内容时出现错误 404

Question

我在这个视频中做了所有事情：但是当我输入时： http://localhost:8080/spring-sample-1.0-SNAPSHOT/hello

出现此错误：

HTTP Status 404 – Not Found
Type Status Report

Message The requested resource [/spring-sample-1.0-SNAPSHOT/hello] is not available

Description The origin server did not find a current representation for the target resource or is not willing to disclose that one exists.

我唯一改变的是： tomcat/bin/setclasspath.bat ，我在那里添加了一行：

set JRE_HOME=C:\Program Files\Java\jre1.8.0_271

因为没有那个服务器不会启动

好的，所以我的应用程序非常简单，我在 java 15 中创建了 mvn 项目，然后是两个类：

配置：

import org.springframework.context.annotation.ComponentScan;
import org.springframework.context.annotation.Configuration;
import org.springframework.web.servlet.config.annotation.EnableWebMvc;
import org.springframework.web.servlet.support.AbstractAnnotationConfigDispatcherServletInitializer;

@Configuration
@ComponentScan({"app"})
@EnableWebMvc
public class Config extends AbstractAnnotationConfigDispatcherServletInitializer {
    @Override
    protected Class<?>[] getRootConfigClasses() {
        return new Class[0];
    }

    @Override
    protected Class<?>[] getServletConfigClasses() {
        return new Class[0];
    }

    @Override
    protected String[] getServletMappings() {
        return new String[0];
    }
}

你好：

import org.springframework.web.bind.annotation.GetMapping;
import org.springframework.web.bind.annotation.RestController;

@RestController
public class Hello {

    @GetMapping("/hello")
    public String get(){
        return "Bycza zagroda!";
    }
}

pom.xml：

<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>pl.bykowski</groupId>
    <artifactId>spring-sample</artifactId>
    <version>1.0-SNAPSHOT</version>
    <packaging>war</packaging>

    <properties>
        <maven.compiler.source>15</maven.compiler.source>
        <maven.compiler.target>15</maven.compiler.target>
    </properties>
    <dependencies>
        <dependency>
            <groupId>org.springframework</groupId>
            <artifactId>spring-webmvc</artifactId>
            <version>5.1.5.RELEASE</version>
        </dependency>
        <dependency>
            <groupId>com.fasterxml.jackson.core</groupId>
            <artifactId>jackson-core</artifactId>
            <version>2.9.8</version>
        </dependency>
        <dependency>
            <groupId>com.fasterxml.jackson.core</groupId>
            <artifactId>jackson-databind</artifactId>
            <version>2.9.8</version>
        </dependency>
        <dependency>
            <groupId>javax.servlet</groupId>
            <artifactId>javax.servlet-api</artifactId>
            <version>4.0.1</version>
            <scope>provided</scope>
        </dependency>
    </dependencies>
    <build>
        <plugins>
            <plugin>
                <groupId>org.apache.maven.plugins</groupId>
                <artifactId>maven-war-plugin</artifactId>
                <version>3.2.3</version>
                <configuration>
                    <failOnMissingWebXml>false</failOnMissingWebXml>
                </configuration>
            </plugin>
        </plugins>
    </build>
</project>

然后我将所有内容打包在war文件中并将其添加到tomcat管理器中： WAR file to deploy

之后我在 tomcat 模块中单击/spring-sample-1.0-SNAPSHOT

然后在最后输入你好

任何想法出了什么问题？ ：/

Answer 1

首先，我建议为操作系统设置 JAVA_HOME 或 JRE_HOME 环境变量。 有关更多详细信息： https://confluence.atlassian.com/doc/setting-the-java_home-variable-in-windows-8895.html无论如何更改tomcat的文件不是一个好主意：）

要解决无法访问 url 的问题，请检查：

对于已安装的 tomcat：

• as rule, name of war file is context name and part of the url(spring-sample-1.0-SNAPSHOT), but that name can be configured in context.xml (Documentation: https://tomcat.apache.org/tomcat- 8.0-doc/config/context.html )
• 打开管理器 window 如您在视频中看到的 (http://localhost:8080/manager/html) 并找到您的 web 应用程序的链接

对于嵌入式 tomcat：

• 上下文路径应该类似于 pom.xml 中的 artifactId

Answer 2

@saver
部署时从 tomcat 记录：

21-Dec-2020 16:49:22.227 INFO [http-nio-8000-exec-17] org.apache.catalina.startup.HostConfig.deployWAR Deploying web application archive [C:\Users\Damian\Desktop\JAVA\apache-tomcat-8.5.61\webapps\spring-sample3-1.0-SNAPSHOT.war]
21-Dec-2020 16:49:24.113 INFO [http-nio-8000-exec-17] org.apache.jasper.servlet.TldScanner.scanJars At least one JAR was scanned for TLDs yet contained no TLDs. Enable debug logging for this logger for a complete list of JARs that were scanned but no TLDs were found in them. Skipping unneeded JARs during scanning can improve startup time and JSP compilation time.
21-Dec-2020 16:49:24.138 INFO [http-nio-8000-exec-17] org.apache.catalina.startup.HostConfig.deployWAR Deployment of web application archive [C:\Users\Damian\Desktop\JAVA\apache-tomcat-8.5.61\webapps\spring-sample3-1.0-SNAPSHOT.war] has finished in [1,907] ms

那么，我应该使用哪些版本的 JRE 和 JDK？

Answer 3

@daniep kajoi you should set path on java 15 for tomcat, or change maven.compiler.source attribute in pom.xml on 1.8 version - one of two options. 我在你的日志中看到你的路径是'spring-sample3-1.0-SNAPSHOT.war'

21-Dec-2020 16:49:22.227 INFO [http-nio-8000-exec-17] org.apache.catalina.startup.HostConfig.deployWAR Deploying web application archive [C:\Users\Damian\Desktop\JAVA\apache-tomcat-8.5.61\webapps\spring-sample3-1.0-SNAPSHOT.war]
21-Dec-2020 16:49:24.113 INFO [http-nio-8000-exec-17] org.apache.jasper.servlet.TldScanner.scanJars At least one JAR was scanned for TLDs yet contained no TLDs. Enable debug logging for this logger for a complete list of JARs that were scanned but no TLDs were found in them. Skipping unneeded JARs during scanning can improve startup time and JSP compilation time.
21-Dec-2020 16:49:24.138 INFO [http-nio-8000-exec-17] org.apache.catalina.startup.HostConfig.deployWAR Deployment of web application archive [C:\Users\Damian\Desktop\JAVA\apache-tomcat-8.5.61\webapps\spring-sample3-1.0-SNAPSHOT.war] has finished in [1,907] ms

尝试打开 url: http://localhost:8080/spring-sample3-1.0-SNAPSHOT.war/hello

Answer 4

我发现了问题：在配置 class 中，您为 servlet 映射和 servlet 配置 class 提供了不正确的值。 请更改配置 class 如下：

@Configuration
@ComponentScan({"app"})
@EnableWebMvc
public class Config extends AbstractAnnotationConfigDispatcherServletInitializer {
    @Override
    protected Class<?>[] getRootConfigClasses() {
        return new Class[0];
    }

    @Override
    protected Class<?>[] getServletConfigClasses() {
        return new Class[] {Config.class};
    }

    @Override
    protected String[] getServletMappings() {
        return new String[]{"/"};
    }
}