我总是收到以下错误。有人可以帮我吗?
Exception in thread "main" java.lang.NoClassDefFoundError: org/apache/spark/Logging
at java.lang.ClassLoader.defineClass1(Native Method)
at java.lang.ClassLoader.defineClass(ClassLoader.java:763)
at java.security.SecureClassLoader.defineClass(SecureClassLoader.java:142)
at java.net.URLClassLoader.defineClass(URLClassLoader.java:467)
at java.net.URLClassLoader.access$100(URLClassLoader.java:73)
at java.net.URLClassLoader$1.run(URLClassLoader.java:368)
at java.net.URLClassLoader$1.run(URLClassLoader.java:362)
at java.security.AccessController.doPrivileged(Native Method)
at java.net.URLClassLoader.findClass(URLClassLoader.java:361)
at java.lang.ClassLoader.loadClass(ClassLoader.java:424)
at sun.misc.Launcher$AppClassLoader.loadClass(Launcher.java:331)
at java.lang.ClassLoader.loadClass(ClassLoader.java:357)
at com.datastax.spark.connector.japi.DStreamJavaFunctions.<init>(DStreamJavaFunctions.java:24)
at com.datastax.spark.connector.japi.CassandraStreamingJavaUtil.javaFunctions(CassandraStreamingJavaUtil.java:55)
at SparkStream.main(SparkStream.java:51)
at sun.reflect.NativeMethodAccessorImpl.invoke0(Native Method)
at sun.reflect.NativeMethodAccessorImpl.invoke(NativeMethodAccessorImpl.java:62)
at sun.reflect.DelegatingMethodAccessorImpl.invoke(DelegatingMethodAccessorImpl.java:43)
at java.lang.reflect.Method.invoke(Method.java:498)
at com.intellij.rt.execution.application.AppMain.main(AppMain.java:147)
Caused by: java.lang.ClassNotFoundException: org.apache.spark.Logging
at java.net.URLClassLoader.findClass(URLClassLoader.java:381)
at java.lang.ClassLoader.loadClass(ClassLoader.java:424)
at sun.misc.Launcher$AppClassLoader.loadClass(Launcher.java:331)
at java.lang.ClassLoader.loadClass(ClassLoader.java:357)
... 20 more
当我编译下面的代码时。我在网上搜索过,但没有找到解决方案。添加 saveToCassandra 时出现错误。
import com.datastax.spark.connector.japi.CassandraStreamingJavaUtil;
import org.apache.spark.SparkConf;
import org.apache.spark.api.java.JavaSparkContext;
import org.apache.spark.streaming.Duration;
import org.apache.spark.streaming.api.java.JavaDStream;
import org.apache.spark.streaming.api.java.JavaPairInputDStream;
import org.apache.spark.streaming.api.java.JavaStreamingContext;
import org.apache.spark.streaming.kafka.KafkaUtils;
import java.io.Serializable;
import java.util.Collections;
import java.util.HashMap;
import java.util.Map;
import java.util.Set;
import static com.datastax.spark.connector.japi.CassandraJavaUtil.mapToRow;
/**
* Created by jonas on 10/10/16.
*/
public class SparkStream implements Serializable{
public static void main(String[] args) throws Exception{
SparkConf conf = new SparkConf(true)
.setAppName("TwitterToCassandra")
.setMaster("local[*]")
.set("spark.cassandra.connection.host", "127.0.0.1")
.set("spark.cassandra.connection.port", "9042");
;
JavaSparkContext sc = new JavaSparkContext(conf);
JavaStreamingContext ssc = new JavaStreamingContext(sc, new Duration(5000));
Map<String, String> kafkaParams = new HashMap<>();
kafkaParams.put("bootstrap.servers", "localhost:9092");
Set<String> topics = Collections.singleton("Test");
JavaPairInputDStream<String, String> directKafkaStream = KafkaUtils.createDirectStream(
ssc,
String.class,
String.class,
kafka.serializer.StringDecoder.class,
kafka.serializer.StringDecoder.class,
kafkaParams,
topics
);
JavaDStream<Tweet> createTweet = directKafkaStream.map(s -> createTweet(s._2));
CassandraStreamingJavaUtil.javaFunctions(createTweet)
.writerBuilder("mykeyspace", "rawtweet", mapToRow(Tweet.class))
.saveToCassandra();
ssc.start();
ssc.awaitTermination();
}
public static Tweet createTweet(String rawKafka){
String[] splitted = rawKafka.split("\\|");
Tweet t = new Tweet(splitted[0], splitted[1], splitted[2], splitted[3]);
return t;
}
}
我的 pom 如下。
<?xml version="1.0" encoding="UTF-8"?>
<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>com.company</groupId>
<artifactId>Sentiment</artifactId>
<version>1.0-SNAPSHOT</version>
<build>
<plugins>
<plugin>
<groupId>org.apache.maven.plugins</groupId>
<artifactId>maven-compiler-plugin</artifactId>
<configuration>
<source>1.8</source>
<target>1.8</target>
</configuration>
</plugin>
</plugins>
</build>
<repositories>
<repository>
<id>twitter4j.org</id>
<name>twitter4j.org Repository</name>
<url>http://twitter4j.org/maven2</url>
<releases>
<enabled>true</enabled>
</releases>
<snapshots>
<enabled>true</enabled>
</snapshots>
</repository>
</repositories>
<dependencies>
<dependency>
<groupId>org.apache.spark</groupId>
<artifactId>spark-streaming_2.11</artifactId>
<version>2.0.1</version>
</dependency>
<dependency>
<groupId>org.apache.spark</groupId>
<artifactId>spark-core_2.10</artifactId>
<version>2.0.0</version>
</dependency>
<dependency>
<groupId>org.apache.spark</groupId>
<artifactId>spark-sql_2.10</artifactId>
<version>2.0.0</version>
</dependency>
<dependency>
<groupId>org.apache.spark</groupId>
<artifactId>spark-streaming-kafka-0-8_2.11</artifactId>
<version>2.0.1</version>
</dependency>
<!-- https://mvnrepository.com/artifact/org.scala-lang/scala-library -->
<dependency>
<groupId>org.scala-lang</groupId>
<artifactId>scala-library</artifactId>
<version>2.11.8</version>
</dependency>
<!-- https://mvnrepository.com/artifact/com.datastax.spark/spark-cassandra-connector_2.10 -->
<dependency>
<groupId>com.datastax.spark</groupId>
<artifactId>spark-cassandra-connector_2.10</artifactId>
<version>1.6.2</version>
</dependency>
<dependency>
<groupId>org.apache.kafka</groupId>
<artifactId>kafka_2.10</artifactId>
<version>0.9.0.0</version>
</dependency>
<dependency>
<groupId>org.twitter4j</groupId>
<artifactId>twitter4j-core</artifactId>
<version>[4.0,)</version>
</dependency>
<dependency>
<groupId>org.twitter4j</groupId>
<artifactId>twitter4j-stream</artifactId>
<version>4.0.4</version>
</dependency>
<dependency>
<groupId>org.twitter4j</groupId>
<artifactId>twitter4j-async</artifactId>
<version>4.0.4</version>
</dependency>
</dependencies>
</project>
最佳答案
org.apache.spark.Logging 在 Spark 1.5.2 或更低版本中可用。它不在 2.0.0 中。请更改版本如下
<dependency>
<groupId>org.apache.spark</groupId>
<artifactId>spark-streaming_2.11</artifactId>
<version>1.5.2</version>
</dependency>
<dependency>
<groupId>org.apache.spark</groupId>
<artifactId>spark-core_2.10</artifactId>
<version>1.5.2</version>
</dependency>
<dependency>
<groupId>org.apache.spark</groupId>
<artifactId>spark-sql_2.10</artifactId>
<version>1.5.2</version>
</dependency>
<dependency>
<groupId>org.apache.spark</groupId>
<artifactId>spark-streaming-kafka-0-8_2.11</artifactId>
<version>1.6.2</version>
</dependency>
关于java.lang.NoClassDefFoundError : org/apache/spark/Logging 错误,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/40287289/
大约一年前,我决定确保每个包含非唯一文本的Flash通知都将从模块中的方法中获取文本。我这样做的最初原因是为了避免一遍又一遍地输入相同的字符串。如果我想更改措辞,我可以在一个地方轻松完成,而且一遍又一遍地重复同一件事而出现拼写错误的可能性也会降低。我最终得到的是这样的:moduleMessagesdefformat_error_messages(errors)errors.map{|attribute,message|"Error:#{attribute.to_s.titleize}#{message}."}enddeferror_message_could_not_find(obje
我主要使用Ruby来执行此操作,但到目前为止我的攻击计划如下:使用gemsrdf、rdf-rdfa和rdf-microdata或mida来解析给定任何URI的数据。我认为最好映射到像schema.org这样的统一模式,例如使用这个yaml文件,它试图描述数据词汇表和opengraph到schema.org之间的转换:#SchemaXtoschema.orgconversion#data-vocabularyDV:name:namestreet-address:streetAddressregion:addressRegionlocality:addressLocalityphoto:i
我真的很习惯使用Ruby编写以下代码:my_hash={}my_hash['test']=1Java中对应的数据结构是什么? 最佳答案 HashMapmap=newHashMap();map.put("test",1);我假设? 关于java-等价于Java中的RubyHash,我们在StackOverflow上找到一个类似的问题: https://stackoverflow.com/questions/22737685/
我遵循MichaelHartl的“RubyonRails教程:学习Web开发”,并创建了检查用户名和电子邮件长度有效性的测试(名称最多50个字符,电子邮件最多255个字符)。test/helpers/application_helper_test.rb的内容是:require'test_helper'classApplicationHelperTest在运行bundleexecraketest时,所有测试都通过了,但我看到以下消息在最后被标记为错误:ERROR["test_full_title_helper",ApplicationHelperTest,1.820016791]test
我是rails的新手,想在form字段上应用验证。myviewsnew.html.erb.....模拟.rbclassSimulation{:in=>1..25,:message=>'Therowmustbebetween1and25'}end模拟Controller.rbclassSimulationsController我想检查模型类中row字段的整数范围,如果不在范围内则返回错误信息。我可以检查上面代码的范围,但无法返回错误消息提前致谢 最佳答案 关键是您使用的是模型表单,一种显示ActiveRecord模型实例属性的表单。c
我正在尝试编写一个将文件上传到AWS并公开该文件的Ruby脚本。我做了以下事情:s3=Aws::S3::Resource.new(credentials:Aws::Credentials.new(KEY,SECRET),region:'us-west-2')obj=s3.bucket('stg-db').object('key')obj.upload_file(filename)这似乎工作正常,除了该文件不是公开可用的,而且我无法获得它的公共(public)URL。但是当我登录到S3时,我可以正常查看我的文件。为了使其公开可用,我将最后一行更改为obj.upload_file(file
我克隆了一个rails仓库,我现在正尝试捆绑安装背景:OSXElCapitanruby2.2.3p173(2015-08-18修订版51636)[x86_64-darwin15]rails-v在您的Gemfile中列出的或native可用的任何gem源中找不到gem'pg(>=0)ruby'。运行bundleinstall以安装缺少的gem。bundleinstallFetchinggemmetadatafromhttps://rubygems.org/............Fetchingversionmetadatafromhttps://rubygems.org/...Fe
在Cooper的书BeginningRuby中,第166页有一个我无法重现的示例。classSongincludeComparableattr_accessor:lengthdef(other)@lengthother.lengthenddefinitialize(song_name,length)@song_name=song_name@length=lengthendenda=Song.new('Rockaroundtheclock',143)b=Song.new('BohemianRhapsody',544)c=Song.new('MinuteWaltz',60)a.betwee
我是Google云的新手,我正在尝试对其进行首次部署。我的第一个部署是RubyonRails项目。我基本上是在关注thisguideinthegoogleclouddocumentation.唯一的区别是我使用的是我自己的项目,而不是他们提供的“helloworld”项目。这是我的app.yaml文件runtime:customvm:trueentrypoint:bundleexecrackup-p8080-Eproductionconfig.ruresources:cpu:0.5memory_gb:1.3disk_size_gb:10当我转到我的项目目录并运行gcloudprevie
我有两个Rails模型,即Invoice和Invoice_details。一个Invoice_details属于Invoice,一个Invoice有多个Invoice_details。我无法使用accepts_nested_attributes_forinInvoice通过Invoice模型保存Invoice_details。我收到以下错误:(0.2ms)BEGIN(0.2ms)ROLLBACKCompleted422UnprocessableEntityin25ms(ActiveRecord:4.0ms)ActiveRecord::RecordInvalid(Validationfa