Web 服务无法在 Tomcat 和 Eclipse 上正确执行

Question

我正在尝试创建一个 Web Service 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/ 我已按照 post 建议的相同步骤进行操作，并且我运行使用 apache Tomcat v7 对项目进行了调整。但是，输出没有按照教程所说的那样显示。当我运行我的项目时，我收到一个错误 HTTP Status 404 - /RestJerseyExample/ ; The resource is not available。我对此很陌生，我不明白我可能哪里出错了。我正在研究 Windows 7。非常感谢建议！

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>

此外，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>

Answer 1

您没有注册您的资源。您应该在 web.xml 中添加 <servlet></servlet> 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>

然后尝试以下 URL： http://localhost:8080/crunchify/

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