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[英]How to access the excels sheets while running the Jar file inside it and copy outside jar?
[英]How to copy file inside jar to outside the jar?
我想從jar中復制一個文件。 我要復制的文件將被復制到工作目錄之外。 我做了一些測試,我嘗試的所有方法都以0字節文件結束。
編輯 :我希望通過程序復制文件,而不是手動。
首先,我想說之前發布的一些答案是完全正確的,但我想給我的,因為有時我們不能在GPL下使用開源庫,或者因為我們懶得下載jar XD或者什么你的理由是一個獨立的解決方案。
下面的函數復制Jar文件旁邊的資源:
/**
* Export a resource embedded into a Jar file to the local file path.
*
* @param resourceName ie.: "/SmartLibrary.dll"
* @return The path to the exported resource
* @throws Exception
*/
static public String ExportResource(String resourceName) throws Exception {
InputStream stream = null;
OutputStream resStreamOut = null;
String jarFolder;
try {
stream = ExecutingClass.class.getResourceAsStream(resourceName);//note that each / is a directory down in the "jar tree" been the jar the root of the tree
if(stream == null) {
throw new Exception("Cannot get resource \"" + resourceName + "\" from Jar file.");
}
int readBytes;
byte[] buffer = new byte[4096];
jarFolder = new File(ExecutingClass.class.getProtectionDomain().getCodeSource().getLocation().toURI().getPath()).getParentFile().getPath().replace('\\', '/');
resStreamOut = new FileOutputStream(jarFolder + resourceName);
while ((readBytes = stream.read(buffer)) > 0) {
resStreamOut.write(buffer, 0, readBytes);
}
} catch (Exception ex) {
throw ex;
} finally {
stream.close();
resStreamOut.close();
}
return jarFolder + resourceName;
}
只需將ExecutingClass更改為您的類的名稱,並將其命名為:
String fullPath = ExportResource("/myresource.ext");
正如GOXR3PLUS所回答並由安迪·托馬斯注意到的,您可以通過以下方式實現此目的:
Files.copy( InputStream in, Path target, CopyOption... options)
有關詳細信息,請參閱GOXR3PLUS答案
鑒於您對0字節文件的評論,我不得不假設您正在嘗試以編程方式執行此操作,並且,鑒於您的標記,您正在使用Java進行此操作。 如果這是真的,那么只需使用Class.getResource()來獲取指向JAR中文件的URL,然后使用Apache Commons IO FileUtils.copyURLToFile()將其復制到文件系統。 例如:
URL inputUrl = getClass().getResource("/absolute/path/of/source/in/jar/file");
File dest = new File("/path/to/destination/file");
FileUtils.copyURLToFile(inputUrl, dest);
最有可能的是,您現在擁有的任何代碼的問題是您(正確地)使用緩沖輸出流來寫入文件但是(錯誤地)無法關閉它。
哦,你應該修改你的問題澄清你到底如何想這樣做(編程,不是語言,...)
Java 8(實際上是自1.7以來的FileSystem)附帶了一些很酷的新類/方法來處理這個問題。 有人已經提到JAR基本上是ZIP文件,你可以使用
final URI jarFileUril = URI.create("jar:file:" + file.toURI().getPath());
final FileSystem fs = FileSystems.newFileSystem(jarFileUri, env);
(見Zip文件 )
然后你可以使用一種方便的方法,如:
fs.getPath("filename");
然后你可以使用Files類
try (final Stream<Path> sources = Files.walk(from)) {
sources.forEach(src -> {
final Path dest = to.resolve(from.relativize(src).toString());
try {
if (Files.isDirectory(from)) {
if (Files.notExists(to)) {
log.trace("Creating directory {}", to);
Files.createDirectories(to);
}
} else {
log.trace("Extracting file {} to {}", from, to);
Files.copy(from, to, StandardCopyOption.REPLACE_EXISTING);
}
} catch (IOException e) {
throw new RuntimeException("Failed to unzip file.", e);
}
});
}
注意:我嘗試解壓縮JAR文件以進行測試
使用Java 7+更快的方式,以及獲取當前目錄的代碼:
/**
* Copy a file from source to destination.
*
* @param source
* the source
* @param destination
* the destination
* @return True if succeeded , False if not
*/
public static boolean copy(InputStream source , String destination) {
boolean succeess = true;
System.out.println("Copying ->" + source + "\n\tto ->" + destination);
try {
Files.copy(source, Paths.get(destination), StandardCopyOption.REPLACE_EXISTING);
} catch (IOException ex) {
logger.log(Level.WARNING, "", ex);
succeess = false;
}
return succeess;
}
測試它( icon.png是應用程序包圖像中的圖像):
copy(getClass().getResourceAsStream("/image/icon.png"),getBasePathForClass(Main.class)+"icon.png");
關於代碼行( getBasePathForClass(Main.class)
): - >檢查我在這里添加的答案:) - > 用Java獲取當前工作目錄
使用JarInputStream類:
// assuming you already have an InputStream to the jar file..
JarInputStream jis = new JarInputStream( is );
// get the first entry
JarEntry entry = jis.getNextEntry();
// we will loop through all the entries in the jar file
while ( entry != null ) {
// test the entry.getName() against whatever you are looking for, etc
if ( matches ) {
// read from the JarInputStream until the read method returns -1
// ...
// do what ever you want with the read output
// ...
// if you only care about one file, break here
}
// get the next entry
entry = jis.getNextEntry();
}
jis.close();
另請參見: JarEntry
強大的解決方案:
public static void copyResource(String res, String dest, Class c) throws IOException {
InputStream src = c.getResourceAsStream(res);
Files.copy(src, Paths.get(dest), StandardCopyOption.REPLACE_EXISTING);
}
你可以像這樣使用它:
File tempFileGdalZip = File.createTempFile("temp_gdal", ".zip");
copyResource("/gdal.zip", tempFileGdalZip.getAbsolutePath(), this.getClass());
要將文件從jar復制到外部,需要使用以下方法:
getResourceAsStream()
獲取一個InputStream
到jar文件中的文件 FileOutputStream
打開目標文件 示例代碼,其中還包含一個不替換現有值的變量:
public File saveResource(String name) throws IOException {
return saveResource(name, true);
}
public File saveResource(String name, boolean replace) throws IOException {
return saveResource(new File("."), name, replace)
}
public File saveResource(File outputDirectory, String name) throws IOException {
return saveResource(outputDirectory, name, true);
}
public File saveResource(File outputDirectory, String name, boolean replace)
throws IOException {
File out = new File(outputDirectory, name);
if (!replace && out.exists())
return out;
// Step 1:
InputStream resource = this.getClass().getResourceAsStream(name);
if (resource == null)
throw new FileNotFoundException(name + " (resource not found)");
// Step 2 and automatic step 4
try(InputStream in = resource;
OutputStream writer = new BufferedOutputStream(
new FileOutputStream(out))) {
// Step 3
byte[] buffer = new byte[1024 * 4];
int length;
while((length = in.read(buffer)) >= 0) {
writer.write(buffer, 0, length);
}
}
return out;
}
jar只是一個zip文件。 解壓縮(使用您熟悉的任何方法)並正常復制文件。
${JAVA_HOME}/bin/jar -cvf /path/to.jar
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