我正在构建用于使用maven从mysql中选择数据的restful应用程序。我对POM配置非常困惑,如下所示:
<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>com.tutorialacademy.rest</groupId>
<artifactId>helloworld</artifactId>
<packaging>war</packaging>
<version>0.0.1-SNAPSHOT</version>
<name>helloworld Maven Webapp</name>
<url>http://maven.apache.org</url>
<properties>
<project.build.sourceEncoding>UTF-8</project.build.sourceEncoding>
</properties>
<repositories>
<repository>
<id>maven2-repository.java.net</id>
<name>Java.net Repository for Maven</name>
<url>http://download.java.net/maven/2/</url>
<layout>default</layout>
</repository>
</repositories>
<dependencies>
<dependency>
<groupId>junit</groupId>
<artifactId>junit</artifactId>
<version>3.8.1</version>
<scope>test</scope>
</dependency>
<dependency>
<groupId>com.sun.jersey</groupId>
<artifactId>jersey-server</artifactId>
<version>1.9</version>
</dependency>
<dependency>
<groupId>org.apache.httpcomponents</groupId>
<artifactId>httpclient</artifactId>
<version>4.3.3</version>
</dependency>
<dependency>
<groupId>com.googlecode.json-simple</groupId>
<artifactId>json-simple</artifactId>
<version>1.1</version>
</dependency>
<dependency>
<groupId>org.json</groupId>
<artifactId>json</artifactId>
<version>20080701</version>
</dependency>
<dependency>
<groupId>javax.servlet</groupId>
<artifactId>javax.servlet-api</artifactId>
<version>3.0.1</version>
<scope>provided</scope>
</dependency>
<dependency>
<groupId>com.google.code.gson</groupId>
<artifactId>gson</artifactId>
<version>2.6.2</version>
<scope>provided</scope>
</dependency>
</dependencies>
<build>
<sourceDirectory>src/main/java</sourceDirectory>
<plugins>
<plugin>
<artifactId>maven-war-plugin</artifactId>
<version>2.4</version>
<configuration>
<warSourceDirectory>WebContent</warSourceDirectory>
<failOnMissingWebXml>false</failOnMissingWebXml>
</configuration>
</plugin>
<plugin>
<artifactId>maven-compiler-plugin</artifactId>
<version>3.1</version>
<configuration>
<source>1.8</source>
<target>1.8</target>
</configuration>
</plugin>
</plugins>
</build>
</project>
&#13;
我是多个包,如下所示:
问题是我是否需要为每个包添加另一个groupid和artifactid。当我访问在com.tutorialacademy.rest包类下编码为@path的路径http://localhost:8080/helloworld/rest/exportfile/json时,它确实有效:
但是当我访问webService包下的路径http://localhost:8080/helloworld/rest/helloService/hello/时。这给了我错误:
The origin server did not find a current representation for the target resource or is not willing to disclose that one exists.
我的web.xml如下:
<servlet>
<servlet-name>helloworld</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.tutorialacademy.rest</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>helloworld</servlet-name>
<url-pattern>/rest/*</url-pattern>
</servlet-mapping>
答案 0 :(得分:1)
<init-param>
<param-name>com.sun.jersey.config.property.packages</param-name>
<param-value>com.tutorialacademy.rest</param-value>
</init-param>
param-value列出了应该使用@Path
注释的类扫描的包,以便Jersey可以注册这些类。您只列出了您说的有效的一个包。您可以添加多于逗号分隔的包
<param-value>
com.tutorialacademy.rest,
webService
</param-value>
虽然通常的做法是为您的项目提供顶级包。例如
com.foo
com.foo.dao
com.foo.model
com.foo.resources
以递归方式扫描您在param-value中列出的包。所以你可以在这种情况下使用com.foo
,然后所有子包也会被扫描。