通过使用jar:
URI scheme,虽然可能不明智,但有可能读取基本上已重命名为.zip文件(.ear,.war,.jar等)的存档格式。
例如,当uri
变量评估为单个顶级归档文件时,例如以下代码,效果很好。当uri
等于jar:file:///Users/justingarrick/Desktop/test/my_war.war!/
时
private FileSystem createZipFileSystem(Path path) throws IOException {
URI uri = URI.create("jar:" + path.toUri().toString());
FileSystem fs;
try {
fs = FileSystems.getFileSystem(uri);
} catch (FileSystemNotFoundException e) {
fs = FileSystems.newFileSystem(uri, new HashMap<>());
}
return fs;
}
但是,当URI包含嵌套的归档文件(例如URI)时,
getFileSystem
和newFileSystem
调用会失败,并显示IllegalArgumentException
。当uri
等于jar:jar:file:///Users/justingarrick/Desktop/test/my_war.war!/some_jar.jar!/
(.war内的.jar)时。嵌套存档文件是否有有效的
java.net.URI
方案? 最佳答案
如上述乔纳斯·柏林(Jonas Berlin)的评论所述,答案是否。从java.net.JarURLConnection source:
/* get the specs for a given url out of the cache, and compute and
* cache them if they're not there.
*/
private void parseSpecs(URL url) throws MalformedURLException {
String spec = url.getFile();
int separator = spec.indexOf("!/");
/*
* REMIND: we don't handle nested JAR URLs
*/
if (separator == -1) {
throw new MalformedURLException("no !/ found in url spec:" + spec);
}
jarFileURL = new URL(spec.substring(0, separator++));
entryName = null;
/* if ! is the last letter of the innerURL, entryName is null */
if (++separator != spec.length()) {
entryName = spec.substring(separator, spec.length());
entryName = ParseUtil.decode (entryName);
}
}
关于java - 嵌套归档文件是否存在有效的java.net.URI?,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/26935050/