问题描述
我花了一些时间试图找到一种方法来计算JAR中文件夹中的文件数。我汇总了几个代码的示例，这些代码用于实现不同的目的。当我通过Eclipse运行代码时它很好，但是在导出到JAR后它失败并返回0.在这种情况下，我使用的文件夹路径只是rules /。我将不胜感激任何建议或样品。谢谢。
I've spent a bit of time trying to find a way to count the number of files in a folder within a JAR. I put together several examples of code that served different purposes to make this work. It counts just fine when I run the code through Eclipse but after exporting to a JAR it fails and returns 0. In this case, my folder path I use is just "rules/". I would appreciate any recommendations or samples. Thanks.
public static int countFiles(String folderPath) throws IOException { //Counts the number of files in a specified folder
ClassLoader loader = ToolSet.class.getClassLoader();
InputStream is = loader.getResourceAsStream(folderPath);
try {
byte[] c = new byte[1024];
int count = 0;
int readChars = 0;
boolean empty = true;
while ((readChars = is.read(c)) != -1) {
empty = false;
for (int i = 0; i < readChars; ++i) {
if (c[i] == '\n') {
++count;
}
}
}
return (count == 0 && !empty) ? 1 : count;
} finally {
is.close();
}
}
编辑：
以下与我原来的问题不完全匹配，但多亏了MadProgrammer，我能够减少我的代码，甚至无需计算文件。代码打击搜索我的JAR中的每个文件，查找以.rules结尾的文件，打开文件，在文件中搜索与searchBox.getText（）匹配的字符串，附加结果，然后继续下一个 .rulesfile。
The following doesn't exactly match my original question but thanks to MadProgrammer I was able to reduce my code and eliminate the need to even count the files. The code blow searches every file in my JAR looking for those that end with ".rules", opens the file, searches the file for a string that matches "searchBox.getText()", appends results, and continues on to the next ".rules" file.
StringBuilder results = new StringBuilder();
int count = 0;
JarFile jf = null;
try {
String path = ToolSet.class.getProtectionDomain().getCodeSource().getLocation().getPath();
String decodedPath = URLDecoder.decode(path, "UTF-8");
jf = new JarFile(new File(decodedPath));
Enumeration<JarEntry> entries = jf.entries();
while (entries.hasMoreElements()) {
JarEntry entry = entries.nextElement();
if (entry.getName().endsWith(".rules")) {
String name = entry.getName();
InputStream in = ToolSet.class.getResourceAsStream(name);
InputStreamReader isr = new InputStreamReader(in);
BufferedReader bf = new BufferedReader(isr);
String line;
while ((line = bf.readLine()) != null) {
String lowerText = line.toLowerCase();
if(lowerText.indexOf(searchBox.getText().toLowerCase()) > 0) {
results.append(line + "\n");
count++;
}
}
bf.close();
}
}
} catch (IOException ex) {
try {
jf.close();
} catch (Exception e2) {
}
}
if(count>0) {
logBox.setText(results.toString());
} else {
logBox.setText("No matches could be found");
}
推荐答案
Jar文件本质上是一个带有清单的Zip文件。
A Jar file is essentially a Zip file with a manifest.
Jar / Zip文件实际上没有磁盘这样的目录概念。他们只是有一个有名字的条目列表。这些名称可能包含某种类型的路径分隔符，并且某些条目实际上可能被标记为目录（并且往往没有任何与它们相关联的字节，仅用作标记）
Jar/Zip files don't actually have a concept of directories like disks do. They simply have a list of entries that have names. These names may contain some kind path separator and some entries may actually be marked as directories (and tend not to have any bytes associated with them, merely acting as markers)
如果你想找到给定路径中的所有资源，你将不得不打开Jar文件并自己检查它的条目，例如......
If you want to find all the resources within a given path, you're going to have to open the Jar file and inspect it's entries yourself, for example...
JarFile jf = null;
try {
String path = "resources";
jf = new JarFile(new File("dist/ResourceFolderCounter.jar"));
Enumeration<JarEntry> entries = jf.entries();
while (entries.hasMoreElements()) {
JarEntry entry = entries.nextElement();
if (!entry.isDirectory()) {
String name = entry.getName();
name = name.replace(path + "/", "");
if (!name.contains("/")) {
System.out.println(name);
}
}
}
} catch (IOException ex) {
try {
jf.close();
} catch (Exception e) {
}
}
现在，这需要您知道要使用的Jar文件的名称，这可能有问题，因为您可能希望列出来自多个不同Jars的资源...
Now, this requires you to know the name of the Jar file you want to use, this may be problematic, as you may wish to list resources from a number of different Jars...
更好的解决方案是在构建时生成某种资源查找文件，其中包含您可能需要的所有资源名称，甚至可以键入特定名称......
A better solution would be to generate some kind of "resource lookup" file at build time, which contained all the names of the resources that you might need, maybe even keyed to particular names...
这样你可以简单地使用...
This way you could simple use...
BufferedReader reader = null;
try {
reader = new BufferedReader(new InputStreamReader(getClass().getResourceAsInputStream("/resources/MasterResourceList.txt")));
String name = null;
while ((name = br.readLine()) != null) {
URL url = getClass().getResource(name);
}
} finally {
try {
br.close();
} catch (Exception exp) {
}
}
例如......
您甚至可以使用资源数对文件进行播种;）
You could even seed the file with the number of resources ;)
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