谁能告诉我如何在没有扩展名的情况下获取文件名? 例子:
fileNameWithExt = "test.xml";
fileNameWithOutExt = "test";
谁能告诉我如何在没有扩展名的情况下获取文件名? 例子:
fileNameWithExt = "test.xml";
fileNameWithOutExt = "test";
当前回答
最简单的方法是使用正则表达式。
fileNameWithOutExt = "test.xml".replaceFirst("[.][^.]+$", "");
上面的表达式将删除最后一个点后面跟一个或多个字符。这是一个基本的单元测试。
public void testRegex() {
assertEquals("test", "test.xml".replaceFirst("[.][^.]+$", ""));
assertEquals("test.2", "test.2.xml".replaceFirst("[.][^.]+$", ""));
}
其他回答
请看下面的测试程序:
public class javatemp {
static String stripExtension (String str) {
// Handle null case specially.
if (str == null) return null;
// Get position of last '.'.
int pos = str.lastIndexOf(".");
// If there wasn't any '.' just return the string as is.
if (pos == -1) return str;
// Otherwise return the string, up to the dot.
return str.substring(0, pos);
}
public static void main(String[] args) {
System.out.println ("test.xml -> " + stripExtension ("test.xml"));
System.out.println ("test.2.xml -> " + stripExtension ("test.2.xml"));
System.out.println ("test -> " + stripExtension ("test"));
System.out.println ("test. -> " + stripExtension ("test."));
}
}
输出:
test.xml -> test
test.2.xml -> test.2
test -> test
test. -> test
如果你不喜欢导入完整的apache.commons,我提取了相同的功能:
public class StringUtils {
public static String getBaseName(String filename) {
return removeExtension(getName(filename));
}
public static int indexOfLastSeparator(String filename) {
if(filename == null) {
return -1;
} else {
int lastUnixPos = filename.lastIndexOf(47);
int lastWindowsPos = filename.lastIndexOf(92);
return Math.max(lastUnixPos, lastWindowsPos);
}
}
public static String getName(String filename) {
if(filename == null) {
return null;
} else {
int index = indexOfLastSeparator(filename);
return filename.substring(index + 1);
}
}
public static String removeExtension(String filename) {
if(filename == null) {
return null;
} else {
int index = indexOfExtension(filename);
return index == -1?filename:filename.substring(0, index);
}
}
public static int indexOfExtension(String filename) {
if(filename == null) {
return -1;
} else {
int extensionPos = filename.lastIndexOf(46);
int lastSeparator = indexOfLastSeparator(filename);
return lastSeparator > extensionPos?-1:extensionPos;
}
}
}
public static String getFileExtension(String fileName) {
if (TextUtils.isEmpty(fileName) || !fileName.contains(".") || fileName.endsWith(".")) return null;
return fileName.substring(fileName.lastIndexOf(".") + 1);
}
public static String getBaseFileName(String fileName) {
if (TextUtils.isEmpty(fileName) || !fileName.contains(".") || fileName.endsWith(".")) return null;
return fileName.substring(0,fileName.lastIndexOf("."));
}
流利的方式:
public static String fileNameWithOutExt (String fileName) {
return Optional.of(fileName.lastIndexOf(".")).filter(i-> i >= 0)
.filter(i-> i > fileName.lastIndexOf(File.separator))
.map(i-> fileName.substring(0, i)).orElse(fileName);
}
简单起见,使用Java的String.replaceAll()方法,如下所示:
String fileNameWithExt = "test.xml";
String fileNameWithoutExt
= fileNameWithExt.replaceAll( "^.*?(([^/\\\\\\.]+))\\.[^\\.]+$", "$1" );
当filenamewitheext包含完全限定路径时,这也可以工作。