使用Apache Camel解壓縮文件？

Question

如何使用Apache Camel從zip文件中提取文件（一個或多個）？可能嗎？

我想這

from("file:/home/myinputzip?noop=true&delay=5000&moveFailed=error") 
    .split(new ZipSplitter()) 
    .streaming().convertBodyTo(String.class)  
    .to("file:/home/myinputzip") 
.end(); 

當我啓動應用程序，該文件是從解壓縮後，但駱駝拋出一個異常，並且文件被移動到文件夾錯誤：

ERROR: org.apache.camel.processor.DefaultErrorHandler - Failed delivery for (MessageId: ID-ubuntu-35217-1377806407437-0-5 on ExchangeId: ID-ubuntu-35217-1377806407437-0-7). Exhausted after delivery attempt: 1 caught: org.apache.camel.component.file.GenericFileOperationFailedException: Cannot write null body to file: /home/myinputzip/aVIII_crrpfp201304.cap org.apache.camel.component.file.GenericFileOperationFailedException: Cannot write null body to file: /home/myinputzip/aVIII_crrpfp201304.cap at org.apache.camel.component.file.FileOperations.storeFile(FileOperations.java:194) at org.apache.camel.component.file.GenericFileProducer.writeFile(GenericFileProducer.java:257) at org.apache.camel.component.file.GenericFileProducer.processExchange(GenericFileProducer.java:159) at org.apache.camel.component.file.GenericFileProducer.process(GenericFileProducer.java:80) at org.apache.camel.util.AsyncProcessorConverterHelper$ProcessorToAsyncProcessorBridge.process(AsyncProcessorConverterHelper.java:61) at org.apache.camel.util.AsyncProcessorHelper.process(AsyncProcessorHelper.java:73) at org.apache.camel.processor.SendProcessor$2.doInAsyncProducer(SendProcessor.java:122) at org.apache.camel.impl.ProducerCache.doInAsyncProducer(ProducerCache.java:298) at org.apache.camel.processor.SendProcessor.process(SendProcessor.java:117)

誰能幫我？

Answer 1

我解決了如下

from("file:/home/myinputzip?noop=true&delay=5000&moveFailed=error") 
    .split(new ZipSplitter()) 
    .streaming().convertBodyTo(String.class) 
     .choice() 
      .when(body().isNotNull()) 
       .to("file:/home/myinputzip") 
     .end() 
.end(); 

Answer 2

我同意你的路線應該工作，但是，你有沒有嘗試過這樣的事情：

from("file:/home/myinputzip?noop=true&delay=5000&moveFailed=error") 
    .unmarshal().zip() 
    .split(body(Iterable.class)) 
    .streaming().convertBodyTo(String.class)  
    .to("file:/home/myinputzip") 
.end(); 

我不得不採取從http://camel.465427.n5.nabble.com/zip-file-best-practices-td5713437.html

我希望這個作品對你這個例子！