Eclipse用の開発、Javaの7で私が見つけただけで何かを、org.eclipse.core.internal.resources.OS
public abstract class OS {
private static final String INSTALLED_PLATFORM;
public static final char[] INVALID_RESOURCE_CHARACTERS;
private static final String[] INVALID_RESOURCE_BASENAMES;
private static final String[] INVALID_RESOURCE_FULLNAMES;
static {
//find out the OS being used
//setup the invalid names
INSTALLED_PLATFORM = Platform.getOS();
if (INSTALLED_PLATFORM.equals(Platform.OS_WIN32)) {
//valid names and characters taken from http://msdn.microsoft.com/library/default.asp?url=/library/en-us/fileio/fs/naming_a_file.asp
INVALID_RESOURCE_CHARACTERS = new char[] {'\\', '/', ':', '*', '?', '"', '<', '>', '|'};
INVALID_RESOURCE_BASENAMES = new String[] {"aux", "com1", "com2", "com3", "com4", //$NON-NLS-1$ //$NON-NLS-2$ //$NON-NLS-3$ //$NON-NLS-4$ //$NON-NLS-5$
"com5", "com6", "com7", "com8", "com9", "con", "lpt1", "lpt2", //$NON-NLS-1$ //$NON-NLS-2$ //$NON-NLS-3$ //$NON-NLS-4$ //$NON-NLS-5$ //$NON-NLS-6$ //$NON-NLS-7$ //$NON-NLS-8$
"lpt3", "lpt4", "lpt5", "lpt6", "lpt7", "lpt8", "lpt9", "nul", "prn"}; //$NON-NLS-1$ //$NON-NLS-2$ //$NON-NLS-3$ //$NON-NLS-4$ //$NON-NLS-5$ //$NON-NLS-6$ //$NON-NLS-7$ //$NON-NLS-8$ //$NON-NLS-9$
Arrays.sort(INVALID_RESOURCE_BASENAMES);
//CLOCK$ may be used if an extension is provided
INVALID_RESOURCE_FULLNAMES = new String[] {"clock$"}; //$NON-NLS-1$
} else {
//only front slash and null char are invalid on UNIXes
//taken from http://www.faqs.org/faqs/unix-faq/faq/part2/section-2.html
INVALID_RESOURCE_CHARACTERS = new char[] {'/', '\0',};
INVALID_RESOURCE_BASENAMES = null;
INVALID_RESOURCE_FULLNAMES = null;
}
}
/**
* Returns true if the given name is a valid resource name on this operating system,
* and false otherwise.
*/
public static boolean isNameValid(String name) {
//. and .. have special meaning on all platforms
if (name.equals(".") || name.equals("..")) //$NON-NLS-1$ //$NON-NLS-2$
return false;
if (INSTALLED_PLATFORM.equals(Platform.OS_WIN32)) {
//empty names are not valid
final int length = name.length();
if (length == 0)
return false;
final char lastChar = name.charAt(length-1);
// filenames ending in dot are not valid
if (lastChar == '.')
return false;
// file names ending with whitespace are truncated (bug 118997)
if (Character.isWhitespace(lastChar))
return false;
int dot = name.indexOf('.');
//on windows, filename suffixes are not relevant to name validity
String basename = dot == -1 ? name : name.substring(0, dot);
if (Arrays.binarySearch(INVALID_RESOURCE_BASENAMES, basename.toLowerCase()) >= 0)
return false;
return Arrays.binarySearch(INVALID_RESOURCE_FULLNAMES, name.toLowerCase()) < 0;
}
return true;
}
}
お願い、もう存在しません()回答、これは役に立ちません。ファイルが存在する可能性があるかどうかをチェックしようとしています。 –
@ MasterPeter:あなたの質問を書き換えてみてください。最初は、ファイルが存在するかどうかを知りたいと思っているように見えます。実際に必要なのは、ファイルNAMEが有効かどうかを知ることです。私はタイトルを変更しましたが、質問文に触れませんでした。 – OscarRyz
関連性があるようです:http://eng-przemelek.blogspot.com/2009/07/how-to-create-valid-file-name.html – greenoldman