无法初始化类com.amazonaws.services.sqs.AmazonSQSClient

Question

首先,我是JAVA AWS Eclipse Maven Tomcat的新手...我在尝试下面的代码时遇到以下错误.错误是"HTTP状态500 - java.lang.NoClassDefFoundError:无法初始化类com.amazonaws.services.sqs. AmazonSQSClient" ...

package sms.pii.webservice;

import com.amazonaws.auth.BasicAWSCredentials;
import com.amazonaws.services.sqs.AmazonSQS;
import com.amazonaws.services.sqs.AmazonSQSClient;
import com.amazonaws.services.sqs.model.*;

public class AWSSimpleQueueServiceUtil {

public BasicAWSCredentials credentials;
public AmazonSQS sqs;


public AWSSimpleQueueServiceUtil(){
    try{
        String accessKey= "xxxxxx";
        String secretKey= "xxxxxxxx";
        this.credentials = new BasicAWSCredentials(accessKey,secretKey);
        this.sqs = new AmazonSQSClient(this.credentials);
        //this.sqs.setEndpoint("https://sqs.ap-southeast-1.amazonaws.com");

    }
    catch(Exception e){
        System.out.println("exception while creating awss3client : " + e);
    }
}

public String createNewQueue(String queueName){
    CreateQueueRequest createQueueRequest = new CreateQueueRequest(queueName);
    String queueUrl = this.sqs.createQueue(createQueueRequest).getQueueUrl();
    return queueUrl;
}

public String getQueueUrlByName(String queueName){
    GetQueueUrlRequest getQueueUrlRequest = new GetQueueUrlRequest(queueName);
    return this.sqs.getQueueUrl(getQueueUrlRequest).getQueueUrl();
}

public ListQueuesResult listAllQueues(){
   return this.sqs.listQueues();
}

}

package sms.pii.webservice;

import javax.ws.rs.GET;
import javax.ws.rs.Path;
import javax.ws.rs.PathParam;
import javax.ws.rs.Produces;
import javax.ws.rs.core.MediaType;

import sms.pii.webservice.AWSSimpleQueueServiceUtil; 

@Path("/Queue")
public class TestSQS {

@GET
@Path("/Name/{name}")
@Produces(MediaType.APPLICATION_JSON)
public Student produceJSON( @PathParam("name") String name ) {
    Student st = new Student(name, "kumar",55,21);
    return st;
}

@GET
@Path("/createQueue/{name}")
@Produces(MediaType.TEXT_PLAIN)
public String createQueue(@PathParam("name") String queueName){
    AWSSimpleQueueServiceUtil test = new AWSSimpleQueueServiceUtil();
    return test.createNewQueue(queueName);
}

@GET
@Path("/getQueueUrl/{name}")
@Produces(MediaType.TEXT_PLAIN)
public String getQueueUrl(@PathParam("name") String queueName){
    AWSSimpleQueueServiceUtil test = new AWSSimpleQueueServiceUtil();
    return test.getQueueUrlByName(queueName);
}
}

的pom.xml

<dependencies>
<dependency>
    <groupId>com.sun.jersey</groupId>
    <artifactId>jersey-server</artifactId>
    <version>1.9</version>
</dependency>
<dependency>
    <groupId>com.sun.jersey</groupId>
    <artifactId>jersey-json</artifactId>
    <version>1.9</version>
</dependency>
<dependency>
    <groupId>com.amazonaws</groupId>
    <artifactId>aws-java-sdk</artifactId>
    <version>1.8.9.1</version>
</dependency>
<dependency>
    <groupId>commons-logging</groupId>
    <artifactId>commons-logging</artifactId>
    <version>1.1.1</version>
</dependency>

Answer 1

java.lang.NoClassDefFoundError只是意味着:

"嘿伙计,当你(自动)在Eclipse(和/或Maven)中构建项目(编译时)时,你的IDE能够找到这个类com.amazonaws.services.sqs.AmazonSQSClient.但是当你想要运行时在服务器上(运行时),我找不到它了."

所以你在运行时缺少一个以前编译过的类.

现在请这样做:

A-清洁阶段

1. 在eclipse中去菜单栏 - >清理 - >清理所有项目.
2. 如果eclipse还不支持maven(例如没有m2e):打开命令行(Windows或Linux或其他)并移动到包含eclipse项目的目录并输入"mvn clean".
3. 如果eclipse确实支持m2e,那么直接在项目的eclipse上右键单击 - > maven - > update project.

B-配置阶段:

4. 在你的eclipse项目中,右键单击 - >部署程序集.您将看到一种包含"source"和"Deploy Path"列的表.如果没有源"Maven Dependency"的行,请点击按钮添加 - > Java Build path Entries - > next button - >"Maven Dependency".

5. 添加"Maven依赖关系"后,请确保其部署路径值为"WEB-INF/lib".

C-部署和运行时

6. 右键单击您的项目 - > maven install

7. 右键单击您的项目 - >运行为(或调试为) - >选择您的tomcat而不是启动它.那时你的项目必须已经配置好了.

确保你已经安装了eclipse插件m2e.它将使您在eclipse/maven中的开发生活更轻松.