Question

我有wsdl文件，我用它生成java代码并在eclipse中实现web服务。

现在我的下一个任务是，我必须使用maven或任何东西生成包含所有tomcat jar依赖项的jar / war文件。

因此用户无需安装tomcat，他只需运行jar / war文件即可在某个端口使用这些Web服务。

如何实现呢？

有可能这样做吗？

请帮忙。

更多详情：

我尝试使用tomcat maven插件，按照以下链接提供的说明操作： http://maksim.sorokin.dk/it/2011/01/20/jax-ws-web-services-maven-tomcat/

它创建了war文件，但我无法访问这些soap端点。

<html><head><title>Apache Tomcat/7.0.47 - Error report</title><style><!--H1 {font-family:Tahoma,Arial,sans-serif;color:white;background-color:#525D76;font-size:22px;} H2 {font-family:Tahoma,Arial,sans-serif;color:white;background-color:#525D76;font-size:16px;} H3 {font-family:Tahoma,Arial,sans-serif;color:white;background-color:#525D76;font-size:14px;} BODY {font-family:Tahoma,Arial,sans-serif;color:black;background-color:white;} B {font-family:Tahoma,Arial,sans-serif;color:white;background-color:#525D76;} P {font-family:Tahoma,Arial,sans-serif;background:white;color:black;font-size:12px;}A {color : black;}A.name {color : black;}HR {color : #525D76;}--></style> </head><body><h1>HTTP Status 404 - /Project/services/SubscriberProvisioningService</h1><HR size="1" noshade="noshade"><p><b>type</b> Status report</p><p><b>message</b> <u>/Project/services/SubscriberProvisioningService</u></p><p><b>description</b> <u>The requested resource is not available.</u></p><HR size="1" noshade="noshade"><h3>Apache Tomcat/7.0.47</h3></body></html>

以下是配置完成：

IN POM.XML

  <groupId>org.apache.tomcat.maven</groupId>
                <artifactId>tomcat7-maven-plugin</artifactId>
                <version>2.2</version>
                <executions>
                    <execution>
                        <id>default-cli</id>
                        <goals>
                            <goal>run</goal>
                        </goals>
                        <configuration>
                            <port>8080</port>
                            <path>/</path>
                            <useSeparateTomcatClassLoader>true</useSeparateTomcatClassLoader>
                        </configuration>
                    </execution>
                </executions>

的beans.xml

    <import resource="classpath:META-INF/cxf/cxf.xml" />

    <jaxws:endpoint
      id="SubscriberProvisioningService_PortType"
      serviceName="s:SubscriberProvisioningService"
     enpointName="e:SubscriberProvisioningService"
     implementor="com.www.subscriber.SubscriberProvisioningService_BindingImpl"
      address="http://localhost:8080/Project/services/SubscriberProvisioningService" xmlns:e="http://service.jaxws.cxf.apache.org/endpoint"      xmlns:s="http://service.jaxws.cxf.apache.org/service"/>

使用以下命令运行：

 mvn clean package -DskipTests=true tomcat7:run

[INFO] Packaging webapp
[INFO] Assembling webapp [Project] in     [/home/projects/webservices/Project/target/Project-0.0.1-SNAPSHOT]
[INFO] Processing war project
[INFO] Copying webapp resources  [/home/projects/webservices/Project/src/main/webapp]
[INFO] Webapp assembled in [115 msecs]
[INFO] Building war: /home/projects/webservices/Project/target/Project-0.0.1-SNAPSHOT.war
[INFO]
[INFO] >>> tomcat7-maven-plugin:2.2:run (default-cli) @ Project >>>
[INFO]
[INFO] --- maven-resources-plugin:2.6:resources (default-resources) @ Project ---
[WARNING] Using platform encoding (UTF-8 actually) to copy filtered resources, i.e. build is platform dependent!
[INFO] skip non existing resourceDirectory /home/projects/webservices/Project/src/main/resources
[INFO]
[INFO] --- maven-compiler-plugin:2.5.1:compile (default-compile) @ Project ---
[INFO] Nothing to compile - all classes are up to date
[INFO]
[INFO] <<< tomcat7-maven-plugin:2.2:run (default-cli) @ Project <<<
[INFO]
[INFO] --- tomcat7-maven-plugin:2.2:run (default-cli) @ Project ---
[INFO] Running war on http://localhost:8080/
[INFO] Creating Tomcat server configuration at /home/projects/webservices/Project/target/tomcat
[INFO] create webapp with contextPath:

Answer 1

可能

如果要将tomcat用作Web容器，则要使用嵌入模式。也许这可以帮助： https://devcenter.heroku.com/articles/create-a-java-web-application-using-embedded-tomcat

还有tomcat的替代品，如码头，灰熊和底板。

使用Maven构建Soap Web项目

1 个答案: