【问题标题】:Web Services not executing properly on Tomcat and EclipseWeb 服务无法在 Tomcat 和 Eclipse 上正确执行
【发布时间】:2015-08-27 08:24:06
【问题描述】:

我正在尝试创建一个Web Service,它将温度从C 转换为F。我遵循了一个基本教程: http://crunchify.com/how-to-build-restful-service-with-java-using-jax-rs-and-jersey/ 我遵循了帖子建议的相同步骤,并且正在使用apache Tomcat v7 运行该项目。但是,输出并没有按照教程所说的方式显示。当我运行我的项目时,我收到一个错误,上面写着HTTP Status 404 - /RestJerseyExample/ ; The resource is not available。我对此很陌生,我不明白我哪里可能出错了。我正在处理Windows 7。建议非常感谢!

这是我的web.xml

<?xml version="1.0" encoding="UTF-8"?><web-app xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"xmlns="http://java.sun.com/xml/ns/javaee" xmlns:web="http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd"
xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_3_0.xsd"
version="3.0">
<display-name>CrunchifyRESTJerseyExample</display-name>
<welcome-file-list>
    <welcome-file>index.html</welcome-file>
    <welcome-file>index.htm</welcome-file>
    <welcome-file>index.jsp</welcome-file>
    <welcome-file>default.html</welcome-file>
    <welcome-file>default.htm</welcome-file>
    <welcome-file>default.jsp</welcome-file>
</welcome-file-list>

<servlet>
    <servlet-name>Jersey Web Application</servlet-name>
    <servlet-class>com.sun.jersey.spi.container.servlet.ServletContainer</servlet-class>
    <load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
    <servlet-name>Jersey Web Application</servlet-name>
    <url-pattern>/crunchify/*</url-pattern>
</servlet-mapping>

另外,这是我的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>RESTJerseyExample</groupId>
<artifactId>RESTJerseyExample</artifactId>
<version>0.0.1-SNAPSHOT</version>
<packaging>war</packaging>
<build>
    <sourceDirectory>src</sourceDirectory>
    <plugins>
    <plugin>
        <artifactId>maven-compiler-plugin</artifactId>
        <version>3.1</version>
        <configuration>
        <source>1.8</source>
        <target>1.8</target>
        </configuration>
    </plugin>
    <plugin>
        <artifactId>maven-war-plugin</artifactId>
        <version>2.4</version>
        <configuration>
        <warSourceDirectory>WebContent</warSourceDirectory>
        <failOnMissingWebXml>false</failOnMissingWebXml>
        </configuration>
    </plugin>
    </plugins>
</build>
<dependencies>
    <dependency>
        <groupId>asm</groupId>
        <artifactId>asm</artifactId>
        <version>3.3.1</version>
    </dependency>
    <dependency>
        <groupId>com.sun.jersey</groupId>
        <artifactId>jersey-bundle</artifactId>
        <version>1.19</version>
    </dependency>
    <dependency>
        <groupId>org.json</groupId>
        <artifactId>json</artifactId>
        <version>20140107</version>
    </dependency>
    <dependency>
        <groupId>com.sun.jersey</groupId>
        <artifactId>jersey-server</artifactId>
        <version>1.19</version>
    </dependency>
    <dependency>
        <groupId>com.sun.jersey</groupId>
        <artifactId>jersey-core</artifactId>
        <version>1.19</version>
    </dependency>
</dependencies>
</project>

【问题讨论】:

  • 请提供web.xml等配置和资源文件
  • 请检查我的编辑
  • 您是否尝试通过“/RestJerseyExample/”上下文访问您的服务器?我认为应该是“/CrunchifyRESTJerseyExample/crunchify/{your_resource_context}”
  • 教程中提到了第一个。然而,我的班级名称是第二个..

标签: eclipse web-services tomcat7 restful-url


【解决方案1】:

您没有注册您的资源。您应该在web.xml 中添加您的&lt;servlet&gt;&lt;/servlet&gt; 配置:

<servlet>
    <servlet-name>Jersey Web Application</servlet-name>
    <servlet-class>com.sun.jersey.spi.container.servlet.ServletContainer</servlet-class>
    <init-param>
        <param-name>com.sun.jersey.config.property.packages</param-name>
        <param-value>com.crunchify.restjersey</param-value>
    </init-param>
    <load-on-startup>1</load-on-startup>
</servlet>

并尝试以下 URL: http://localhost:8080/crunchify/

【讨论】:

  • 刚刚尝试了你的建议,我仍然收到错误 404
  • 你想访问什么资源?
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