我需要在Unix cli中使用此命令在java中的等效行为:
ls /data/archive/users/*/*.xml
哪个输出我:
/data/archive/users/2012/user1.xml
/data/archive/users/2013/user2.xml
Java 6是否有简单的等效实现?
答案 0 :(得分:2)
使用java.util.Scanner
获取用户输入并使用java.io.File.listFiles(FilenameFilter)
方法获取具有特定过滤器的文件夹中的文件列表。
答案 1 :(得分:0)
是的,有,它被称为list
类的File
方法。有关详细信息,请参阅Javadoc。
答案 2 :(得分:0)
我忘了这来自哪里,但这应该是一个好的开始。还有更多可通过Google获得。
public class RegexFilenameFilter implements FilenameFilter {
/**
* Only file name that match this regex are accepted by this filter
*/
String regex = null; // setting the filter regex to null causes any name to be accepted (same as ".*")
public RegexFilenameFilter() {
}
public RegexFilenameFilter(String filter) {
setWildcard(filter);
}
/**
* Set the filter from a wildcard expression as known from the windows command line
* ("?" = "any character", "*" = zero or more occurances of any character")
*
* @param sWild the wildcard pattern
*
* @return this
*/
public RegexFilenameFilter setWildcard(String sWild) {
regex = wildcardToRegex(sWild);
// throw PatternSyntaxException if the pattern is not valid
// this should never happen if wildcardToRegex works as intended,
// so thiw method does not declare PatternSyntaxException to be thrown
Pattern.compile(regex);
return this;
}
/**
* Set the regular expression of the filter
*
* @param regex the regular expression of the filter
*
* @return this
*/
public RegexFilenameFilter setRegex(String regex) throws java.util.regex.PatternSyntaxException {
this.regex = regex;
// throw PatternSyntaxException if the pattern is not valid
Pattern.compile(regex);
return this;
}
/**
* Tests if a specified file should be included in a file list.
*
* @param dir the directory in which the file was found.
*
* @param name the name of the file.
*
* @return true if and only if the name should be included in the file list; false otherwise.
*/
public boolean accept(File dir, String name) {
boolean bAccept = false;
if (regex == null) {
bAccept = true;
} else {
bAccept = name.toLowerCase().matches(regex);
}
return bAccept;
}
/**
* Converts a windows wildcard pattern to a regex pattern
*
* @param wild - Wildcard patter containing * and ?
*
* @return - a regex pattern that is equivalent to the windows wildcard pattern
*/
private static String wildcardToRegex(String wild) {
if (wild == null) {
return null;
}
StringBuilder buffer = new StringBuilder();
char[] chars = wild.toLowerCase().toCharArray();
for (int i = 0; i < chars.length; ++i) {
if (chars[i] == '*') {
buffer.append(".*");
} else if (chars[i] == '?') {
buffer.append('.');
} else if (chars[i] == ';') {
buffer.append('|');
} else if ("+()^$.{}[]|\\".indexOf(chars[i]) != -1) {
buffer.append('\\').append(chars[i]); // prefix all metacharacters with backslash
} else {
buffer.append(chars[i]);
}
}
return buffer.toString();
}
}
答案 3 :(得分:0)
这是我使用的代码,它适用于相对路径和绝对路径:
DirectoryScanner scanner = new DirectoryScanner();
if (!inputPath.startsWith("/") || inputPath.startsWith(".")) {
scanner.setBasedir(".");
}
scanner.setIncludes(new String[]{inputPath});
scanner.setCaseSensitive(false);
scanner.scan();
String[] foundFiles = scanner.getIncludedFiles();
(来自org.apache.tools.ant的DirectoryScanner)