Question

我跟随this guide建立我的hello world网络服务并陷入困境，这是我的代码：

MyResource.java:

package com.example;

import javax.ws.rs.GET;
import javax.ws.rs.Path;
import javax.ws.rs.Produces;

@Path("myresource")
public class MyResource {
    @GET
    @Produces("text/plain")
    public String getIt() {
        return "Got it!";
    }
}

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/maven-v4_0_0.xsd">
<modelVersion>4.0.0</modelVersion>

<groupId>ts</groupId>
<artifactId>mtest</artifactId>
<packaging>war</packaging>
<version>1.0-SNAPSHOT</version>
<name>mtest Maven Webapp</name>
<url>http://maven.apache.org</url>

<dependencies>
    <dependency>
        <groupId>org.glassfish.jersey.containers</groupId>
        <artifactId>jersey-container-jetty-http</artifactId>
        <version>2.19</version>
    </dependency>

    <dependency>
        <groupId>org.glassfish.jersey.containers</groupId>
        <artifactId>jersey-container-jetty-servlet</artifactId>
        <version>2.19</version>
    </dependency>
</dependencies>
<build>
    <finalName>mtest</finalName>
    <plugins>
        <plugin>
            <groupId>org.eclipse.jetty</groupId>
            <artifactId>jetty-maven-plugin</artifactId>
            <version>9.3.0.M1</version>
        </plugin>
    </plugins>
</build>

web.xml:

<web-app version="3.0"
     xmlns="http://java.sun.com/xml/ns/javaee"
     xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance">

<!-- Servlet declaration can be omitted in which case
     it would be automatically added by Jersey -->
<servlet>
    <servlet-name>javax.ws.rs.core.Application</servlet-name>
</servlet>

<servlet-mapping>
    <servlet-name>javax.ws.rs.core.Application</servlet-name>
    <url-pattern>/myresource/*</url-pattern>
</servlet-mapping>

现在我的问题是：

启动Jetty之后，我可以从http://localhost:8080/获取我的jsp的hello世界，但http://localhost:8080/myresource返回404
Intellij抱怨Element web-app must be declared

Answer 1

在web.xml中添加servlet声明，如下所示

<servlet>
    <servlet-name>javax.ws.rs.core.Application</servlet-name>
    <servlet-class>org.glassfish.jersey.servlet.ServletContainer</servlet-class>
    <init-param>
        <param-name>jersey.config.server.provider.packages</param-name>
        <param-value>com.example</param-value>
    </init-param>
    <load-on-startup>1</load-on-startup>
</servlet>

应该有用。有时这条路需要这样一个条。

@Path("/myresource")

看看，如果你把

<url-pattern>/myresource/*</url-pattern>

你必须点击

http://localhost:8080/myresource/myresource

一个是您的网络配置，另一个是您的资源路径。

但是，如果您在tomcat中尝试此操作，则还需要包含artifactId，如下所示：

http://localhost:8080/mtest/myresource/myresource

我希望这对你有帮助。

基于Jersey和Jetty的RESTful Hello世界

1 个答案: