我已经编写了一个oozie工作流,它创建HAR归档,然后运行需要从该归档读取数据的MR-job。1.创建归档文件2.作业运行时,mapper在分布式缓存中看到归档文件。3.?我怎样才能读懂这篇文章呢?从这个归档中逐行读取数据的API是什么(我的har是一批多个以新行分隔的文本文件)。注意:当我处理存储在DistirubtedCache中的常用文件(不是HAR存档)时,它工作得很好。我在尝试从HAR读取数据时遇到问题。
下面是一个代码片段:
InputStream inputStream;
String cachedDatafileName = System.getProperty(DIST_CACHE_FILE_NAME);
LOG.info(String.format("Looking for[%s]=[%s] in DistributedCache",DIST_CACHE_FILE_NAME, cachedDatafileName));
URI[] uris = DistributedCache.getCacheArchives(getContext().getConfiguration());
URI uriToCachedDatafile = null;
for(URI uri : uris){
if(uri.toString().endsWith(cachedDatafileName)){
uriToCachedDatafile = uri;
break;
}
}
if(uriToCachedDatafile == null){
throw new RuntimeConfigurationException(String.format("Looking for[%s]=[%s] in DistributedCache failed. There is no such file",
DIST_CACHE_FILE_NAME, cachedDatafileName));
}
Path pathToFile = new Path(uriToCachedDatafile);
LOG.info(String.format("[%s] has been found. Uri is: [%s]. The path is:[%s]",cachedDatafileName, uriToCachedDatafile, pathToFile));
FileSystem fileSystem = pathToFile.getFileSystem(getContext().getConfiguration());
HarFileSystem harFileSystem = new HarFileSystem(fileSystem);
inputStream = harFileSystem.open(pathToFile); //NULL POINTER EXCEPTION IS HERE!
return inputStream;发布于 2013-03-13 14:40:50
protected InputStream getInputStreamToDistCacheFile() throws IOException{
InputStream inputStream;
String cachedDatafileName = System.getProperty(DIST_CACHE_FILE_NAME);
LOG.info(String.format("Looking for[%s]=[%s] in DistributedCache",DIST_CACHE_FILE_NAME, cachedDatafileName));
URI[] uris = DistributedCache.getCacheArchives(getContext().getConfiguration());
URI uriToCachedDatafile = null;
for(URI uri : uris){
if(uri.toString().endsWith(cachedDatafileName)){
uriToCachedDatafile = uri;
break;
}
}
if(uriToCachedDatafile == null){
throw new RuntimeConfigurationException(String.format("Looking for[%s]=[%s] in DistributedCache failed. There is no such file",
DIST_CACHE_FILE_NAME, cachedDatafileName));
}
//Path pathToFile = new Path(uriToCachedDatafile +"/stf/db_bts_stf.txt");
Path pathToFile = new Path("har:///"+"home/ssa/devel/megalabs/kyc-solution/kyc-mrjob/target/test-classes/GSMCellSubscriberHomeIntersectionJobDescriptionClusterMRTest/in/gsm_cell_location_stf.har" +"/stf/db_bts_stf.txt");
//Path pathToFile = new Path(("har://home/ssa/devel/megalabs/kyc-solution/kyc-mrjob/target/test-classes/GSMCellSubscriberHomeIntersectionJobDescriptionClusterMRTest/in/gsm_cell_location_stf.har"));
LOG.info(String.format("[%s] has been found. Uri is: [%s]. The path is:[%s]",cachedDatafileName, uriToCachedDatafile, pathToFile));
FileSystem harFileSystem = pathToFile.getFileSystem(context.getConfiguration());
FSDataInputStream fin = harFileSystem.open(pathToFile);
LOG.info("fin: " + fin);
// FileSystem fileSystem = pathToFile.getFileSystem(getContext().getConfiguration());
// HarFileSystem harFileSystem = new HarFileSystem(fileSystem);
// harFileSystem.exists(new Path("har://home/ssa/devel/mycompany/my-solution/my-mrjob/target/test-classes/HomeJobDescriptionClusterMRTest/in/locations.har"));
// LOG.info("harFileSystem.exists(pathToFile):"+ harFileSystem.exists(pathToFile));
// harFileSystem.initialize(uriToCachedDatafile, context.getConfiguration());
FileStatus[] statuses = harFileSystem.listStatus(new Path("har:///"+"har://home/ssa/devel/mycompany/my-solution/my-mrjob/target/test-classes/HomeJobDescriptionClusterMRTest/in/locations.har"));
for(FileStatus fileStatus : statuses){
LOG.info("fileStatus isDir"+fileStatus.isDirectory() +" len:" + fileStatus.getLen());
}
// String tmpPathToFile = "har:///"+pathToFile.toString(); //+"/stf/db_bts_stf.txt";
// Path tmpPath = new Path(tmpPathToFile);
// LOG.info("KILL ME PATH TO FILE IN ARCHIVE: " +tmpPath);
// inputStream = harFileSystem.open(tmpPath);
// return inputStream;
return fin;
}如你所见,这太可怕了。您已经手动读取了存储在归档中的索引文件,并使用索引文件元数据重建路径。如果您知道存档中存储的文件的确切名称(如我的示例中所示),则可以手动构建路径。
这并不方便,我确实期望像Zip->zipEntry这样的东西,当你可以在不知道它的结构的情况下迭代存档的条目。
https://stackoverflow.com/questions/15202026
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