Java IO文件前缀字符串太短-但不是 | IO文件前缀字符串太短

本文介绍了Java IO文件前缀字符串太短-但不是的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！

问题描述

尝试从一个目录创建简单的.tar.gz文件.有我的代码:

Trying to create simple .tar.gz file from one directory. There is my code:

File destinationFile = new File("/var/www/swOfflineFeeds/Companies/2/")
File sourceFile = new File("/var/www/swOfflineFeeds/Companies/2/64cacf30-b294-49f4-b166-032a808d73cd/")
println("destinationFile exists: " + destinationFile.exists()) //prints true
println("sourceFile exists: " + sourceFile.exists()) //prints true

Archiver arch = ArchiverFactory.createArchiver(ArchiveFormat.TAR, CompressionType.GZIP)
File archiveFile = arch.create("64cacf30-b294-49f4-b166-032a808d73cd", destinationFile, sourceFile)

我收到错误消息:

| Error 2018-02-15 12:47:08,925 [http-bio-8183-exec-1] ERROR errors.GrailsExceptionResolver  - IllegalArgumentException occurred when processing request: [GET] /socialwall/test/index
Prefix string too short. Stacktrace follows:
Message: Prefix string too short
    Line | Method
->> 1978 | createTempFile in java.io.File
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
|     51 | create         in org.rauschig.jarchivelib.ArchiverCompressorDecorator
|     19 | index . . . .  in com.manas.socialwall.TestController$$EQjisHuy
|    198 | doFilter       in grails.plugin.cache.web.filter.PageFragmentCachingFilter
|     63 | doFilter . . . in grails.plugin.cache.web.filter.AbstractFilter
|   1145 | runWorker      in java.util.concurrent.ThreadPoolExecutor
|    615 | run . . . . .  in java.util.concurrent.ThreadPoolExecutor$Worker
^    745 | run            in java.lang.Thread

您可以看到文件名正确.我用谷歌搜索，有人提到Java IO File类中的错误.这是真的 ?如何避免这个问题?

As you can see filename is correct. I googled and some people mention bug in Java IO File class. Is this true ? How to avoid this problem?

推荐答案

检查使用的库代码，我们看到创建方法看起来像:

Checking the library code used we see that the create method look like :

public File create(String archive, File destination, File... sources) throws IOException {
     ...
     File temp = File.createTempFile(destination.getName(), archiver.getFilenameExtension(), destination);

前缀是第一个参数.如果您检查 File.getName() 做:

The prefix is the first parameter. If you check what File.getName() do :

以您为例.

File destinationFile = new File("/var/www/swOfflineFeeds/Companies/2/");
System.out.println(destinationFile.getName());

收到的前缀太短，无法创建临时文件，它期望前缀至少为3个字符.请参阅 File.createTempFile

The prefix received is too short for the temp file to create, it expect a prefixe of at least 3 character. See File.createTempFile

在您的情况下，您似乎只是提供了一个文件夹，而是提供了文件名(至少3个字符).

In your case, it seems you just provide a folder, provide the file name instead (with at least 3 character).

迈克尔已经创建了一个问题.

这篇关于Java IO文件前缀字符串太短-但不是的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！