在 Java 中，给定 ajava.net.URL或 aString的形式http://www.example.com/some/path/to/a/file.xml，获取文件名减去扩展名的最简单方法是什么？所以，在这个例子中，我正在寻找返回的东西"file"。
我可以想到几种方法来做到这一点，但我正在寻找易于阅读且简短的东西。
Sietse
问问题
226981 次
27 回答
217
与其重新发明轮子，不如使用 Apache commons-io：
import org.apache.commons.io.FilenameUtils;
public class FilenameUtilTest {
public static void main(String[] args) throws Exception {
URL url = new URL("http://www.example.com/some/path/to/a/file.xml?foo=bar#test");
System.out.println(FilenameUtils.getBaseName(url.getPath())); // -> file
System.out.println(FilenameUtils.getExtension(url.getPath())); // -> xml
System.out.println(FilenameUtils.getName(url.getPath())); // -> file.xml
}
}
于 2013-06-18T11:30:54.790 回答
200
String fileName = url.substring( url.lastIndexOf('/')+1, url.length() );
String fileNameWithoutExtn = fileName.substring(0, fileName.lastIndexOf('.'));
于 2009-03-03T09:37:19.407 回答
38
如果您不需要摆脱文件扩展名，这里有一种方法可以做到这一点，而无需诉诸容易出错的字符串操作，也无需使用外部库。适用于 Java 1.7+：
import java.net.URI
import java.nio.file.Paths
String url = "http://example.org/file?p=foo&q=bar"
String filename = Paths.get(new URI(url).getPath()).getFileName().toString()
于 2015-11-23T12:23:36.773 回答
26
这应该削减它（我将把错误处理留给你）：
int slashIndex = url.lastIndexOf('/');
int dotIndex = url.lastIndexOf('.', slashIndex);
String filenameWithoutExtension;
if (dotIndex == -1) {
filenameWithoutExtension = url.substring(slashIndex + 1);
} else {
filenameWithoutExtension = url.substring(slashIndex + 1, dotIndex);
}
于 2009-03-03T09:33:12.130 回答
24
一个班轮：
new File(uri.getPath).getName
完整代码（在 scala REPL 中）：
import java.io.File
import java.net.URI
val uri = new URI("http://example.org/file.txt?whatever")
new File(uri.getPath).getName
res18: String = file.txt
注意：URI#gePath已经足够智能剥离查询参数和协议的方案。例子：
new URI("http://example.org/hey/file.txt?whatever").getPath
res20: String = /hey/file.txt
new URI("hdfs:///hey/file.txt").getPath
res21: String = /hey/file.txt
new URI("file:///hey/file.txt").getPath
res22: String = /hey/file.txt
于 2017-11-08T15:13:18.070 回答
14
public static String getFileName(URL extUrl) {
//URL: "http://photosaaaaa.net/photos-ak-snc1/v315/224/13/659629384/s659629384_752969_4472.jpg"
String filename = "";
//PATH: /photos-ak-snc1/v315/224/13/659629384/s659629384_752969_4472.jpg
String path = extUrl.getPath();
//Checks for both forward and/or backslash
//NOTE:**While backslashes are not supported in URL's
//most browsers will autoreplace them with forward slashes
//So technically if you're parsing an html page you could run into
//a backslash , so i'm accounting for them here;
String[] pathContents = path.split("[\\\\/]");
if(pathContents != null){
int pathContentsLength = pathContents.length;
System.out.println("Path Contents Length: " + pathContentsLength);
for (int i = 0; i < pathContents.length; i++) {
System.out.println("Path " + i + ": " + pathContents[i]);
}
//lastPart: s659629384_752969_4472.jpg
String lastPart = pathContents[pathContentsLength-1];
String[] lastPartContents = lastPart.split("\\.");
if(lastPartContents != null && lastPartContents.length > 1){
int lastPartContentLength = lastPartContents.length;
System.out.println("Last Part Length: " + lastPartContentLength);
//filenames can contain . , so we assume everything before
//the last . is the name, everything after the last . is the
//extension
String name = "";
for (int i = 0; i < lastPartContentLength; i++) {
System.out.println("Last Part " + i + ": "+ lastPartContents[i]);
if(i < (lastPartContents.length -1)){
name += lastPartContents[i] ;
if(i < (lastPartContentLength -2)){
name += ".";
}
}
}
String extension = lastPartContents[lastPartContentLength -1];
filename = name + "." +extension;
System.out.println("Name: " + name);
System.out.println("Extension: " + extension);
System.out.println("Filename: " + filename);
}
}
return filename;
}
于 2009-12-06T20:42:35.970 回答
12
获取带扩展名的文件名，不带扩展名，只有扩展名只有 3 行：
String urlStr = "http://www.example.com/yourpath/foler/test.png";
String fileName = urlStr.substring(urlStr.lastIndexOf('/')+1, urlStr.length());
String fileNameWithoutExtension = fileName.substring(0, fileName.lastIndexOf('.'));
String fileExtension = urlStr.substring(urlStr.lastIndexOf("."));
Log.i("File Name", fileName);
Log.i("File Name Without Extension", fileNameWithoutExtension);
Log.i("File Extension", fileExtension);
日志结果：
File Name(13656): test.png
File Name Without Extension(13656): test
File Extension(13656): .png
希望它会帮助你。
于 2015-07-17T05:38:20.780 回答
9
我想出了这个：
String url = "http://www.example.com/some/path/to/a/file.xml";
String file = url.substring(url.lastIndexOf('/')+1, url.lastIndexOf('.'));
于 2009-03-03T09:34:36.777 回答
9
String fileName = url.substring(url.lastIndexOf('/') + 1);
于 2016-02-16T10:18:13.697 回答
9
有一些方法：
Java 7 文件 I/O：
String fileName = Paths.get(strUrl).getFileName().toString();
阿帕奇共享：
String fileName = FilenameUtils.getName(strUrl);
使用泽西岛：
UriBuilder buildURI = UriBuilder.fromUri(strUrl);
URI uri = buildURI.build();
String fileName = Paths.get(uri.getPath()).getFileName();
子串：
String fileName = strUrl.substring(strUrl.lastIndexOf('/') + 1);
于 2018-07-07T07:57:05.633 回答
8
把事情简单化 ：
/**
* This function will take an URL as input and return the file name.
* <p>Examples :</p>
* <ul>
* <li>http://example.com/a/b/c/test.txt -> test.txt</li>
* <li>http://example.com/ -> an empty string </li>
* <li>http://example.com/test.txt?param=value -> test.txt</li>
* <li>http://example.com/test.txt#anchor -> test.txt</li>
* </ul>
*
* @param url The input URL
* @return The URL file name
*/
public static String getFileNameFromUrl(URL url) {
String urlString = url.getFile();
return urlString.substring(urlString.lastIndexOf('/') + 1).split("\\?")[0].split("#")[0];
}
于 2014-11-07T21:50:04.843 回答
5
这是在 Android 中执行此操作的最简单方法。我知道它不适用于 Java，但它可以帮助 Android 应用程序开发人员。
import android.webkit.URLUtil;
public String getFileNameFromURL(String url) {
String fileNameWithExtension = null;
String fileNameWithoutExtension = null;
if (URLUtil.isValidUrl(url)) {
fileNameWithExtension = URLUtil.guessFileName(url, null, null);
if (fileNameWithExtension != null && !fileNameWithExtension.isEmpty()) {
String[] f = fileNameWithExtension.split(".");
if (f != null & f.length > 1) {
fileNameWithoutExtension = f[0];
}
}
}
return fileNameWithoutExtension;
}
于 2013-10-11T08:53:14.800 回答
3
从字符串创建一个 URL 对象。当您第一次拥有一个 URL 对象时，有一些方法可以轻松提取您需要的任何信息片段。
我强烈推荐 Javaalmanac 网站，该网站有大量示例，但后来已经移动。您可能会发现http://exampledepot.8waytrips.com/egs/java.io/File2Uri.html很有趣：
// Create a file object
File file = new File("filename");
// Convert the file object to a URL
URL url = null;
try {
// The file need not exist. It is made into an absolute path
// by prefixing the current working directory
url = file.toURL(); // file:/d:/almanac1.4/java.io/filename
} catch (MalformedURLException e) {
}
// Convert the URL to a file object
file = new File(url.getFile()); // d:/almanac1.4/java.io/filename
// Read the file contents using the URL
try {
// Open an input stream
InputStream is = url.openStream();
// Read from is
is.close();
} catch (IOException e) {
// Could not open the file
}
于 2009-03-03T10:18:02.460 回答
2
如果您只想从 java.net.URL 获取文件名（不包括任何查询参数），您可以使用以下函数：
public static String getFilenameFromURL(URL url) {
return new File(url.getPath().toString()).getName();
}
例如，这个输入 URL：
"http://example.com/image.png?version=2&modificationDate=1449846324000"
将被翻译成这个输出字符串：
image.png
于 2016-08-25T09:59:57.750 回答
2
我发现一些 url 在直接传递时会FilenameUtils.getName返回不需要的结果，因此需要将其包装起来以避免被利用。
例如，
System.out.println(FilenameUtils.getName("http://www.google.com/.."));
返回
..
我怀疑有人愿意允许。
以下函数似乎工作正常，并显示了其中一些测试用例，并且null在无法确定文件名时返回。
public static String getFilenameFromUrl(String url)
{
if (url == null)
return null;
try
{
// Add a protocol if none found
if (! url.contains("//"))
url = "http://" + url;
URL uri = new URL(url);
String result = FilenameUtils.getName(uri.getPath());
if (result == null || result.isEmpty())
return null;
if (result.contains(".."))
return null;
return result;
}
catch (MalformedURLException e)
{
return null;
}
}
以下示例中包含一些简单的测试用例：
import java.util.Objects;
import java.net.URL;
import org.apache.commons.io.FilenameUtils;
class Main {
public static void main(String[] args) {
validateFilename(null, null);
validateFilename("", null);
validateFilename("www.google.com/../me/you?trex=5#sdf", "you");
validateFilename("www.google.com/../me/you?trex=5 is the num#sdf", "you");
validateFilename("http://www.google.com/test.png?test", "test.png");
validateFilename("http://www.google.com", null);
validateFilename("http://www.google.com#test", null);
validateFilename("http://www.google.com////", null);
validateFilename("www.google.com/..", null);
validateFilename("http://www.google.com/..", null);
validateFilename("http://www.google.com/test", "test");
validateFilename("https://www.google.com/../../test.png", "test.png");
validateFilename("file://www.google.com/test.png", "test.png");
validateFilename("file://www.google.com/../me/you?trex=5", "you");
validateFilename("file://www.google.com/../me/you?trex", "you");
}
private static void validateFilename(String url, String expectedFilename){
String actualFilename = getFilenameFromUrl(url);
System.out.println("");
System.out.println("url:" + url);
System.out.println("filename:" + expectedFilename);
if (! Objects.equals(actualFilename, expectedFilename))
throw new RuntimeException("Problem, actual=" + actualFilename + " and expected=" + expectedFilename + " are not equal");
}
public static String getFilenameFromUrl(String url)
{
if (url == null)
return null;
try
{
// Add a protocol if none found
if (! url.contains("//"))
url = "http://" + url;
URL uri = new URL(url);
String result = FilenameUtils.getName(uri.getPath());
if (result == null || result.isEmpty())
return null;
if (result.contains(".."))
return null;
return result;
}
catch (MalformedURLException e)
{
return null;
}
}
}
于 2016-12-20T14:07:49.750 回答
1
Urls最后可以有参数，这个
/**
* Getting file name from url without extension
* @param url string
* @return file name
*/
public static String getFileName(String url) {
String fileName;
int slashIndex = url.lastIndexOf("/");
int qIndex = url.lastIndexOf("?");
if (qIndex > slashIndex) {//if has parameters
fileName = url.substring(slashIndex + 1, qIndex);
} else {
fileName = url.substring(slashIndex + 1);
}
if (fileName.contains(".")) {
fileName = fileName.substring(0, fileName.lastIndexOf("."));
}
return fileName;
}
于 2013-10-31T20:39:17.327 回答
1
urllib中的Url对象允许您访问路径的非转义文件名。这里有些例子：
String raw = "http://www.example.com/some/path/to/a/file.xml";
assertEquals("file.xml", Url.parse(raw).path().filename());
raw = "http://www.example.com/files/r%C3%A9sum%C3%A9.pdf";
assertEquals("résumé.pdf", Url.parse(raw).path().filename());
于 2017-11-12T20:16:17.353 回答
0
使用 split() 重做安迪的回答：
Url u= ...;
String[] pathparts= u.getPath().split("\\/");
String filename= pathparts[pathparts.length-1].split("\\.", 1)[0];
于 2009-03-03T10:27:17.600 回答
0
public String getFileNameWithoutExtension(URL url) {
String path = url.getPath();
if (StringUtils.isBlank(path)) {
return null;
}
if (StringUtils.endsWith(path, "/")) {
//is a directory ..
return null;
}
File file = new File(url.getPath());
String fileNameWithExt = file.getName();
int sepPosition = fileNameWithExt.lastIndexOf(".");
String fileNameWithOutExt = null;
if (sepPosition >= 0) {
fileNameWithOutExt = fileNameWithExt.substring(0,sepPosition);
}else{
fileNameWithOutExt = fileNameWithExt;
}
return fileNameWithOutExt;
}
于 2011-06-23T07:56:25.210 回答
0
这个怎么样：
String filenameWithoutExtension = null;
String fullname = new File(
new URI("http://www.xyz.com/some/deep/path/to/abc.png").getPath()).getName();
int lastIndexOfDot = fullname.lastIndexOf('.');
filenameWithoutExtension = fullname.substring(0,
lastIndexOfDot == -1 ? fullname.length() : lastIndexOfDot);
于 2013-06-15T09:15:24.317 回答
0
为了返回不带扩展名且不带参数的文件名，请使用以下命令：
String filenameWithParams = FilenameUtils.getBaseName(urlStr); // may hold params if http://example.com/a?param=yes
return filenameWithParams.split("\\?")[0]; // removing parameters from url if they exist
为了返回不带参数的扩展名文件名，请使用：
/** Parses a URL and extracts the filename from it or returns an empty string (if filename is non existent in the url) <br/>
* This method will work in win/unix formats, will work with mixed case of slashes (forward and backward) <br/>
* This method will remove parameters after the extension
*
* @param urlStr original url string from which we will extract the filename
* @return filename from the url if it exists, or an empty string in all other cases */
private String getFileNameFromUrl(String urlStr) {
String baseName = FilenameUtils.getBaseName(urlStr);
String extension = FilenameUtils.getExtension(urlStr);
try {
extension = extension.split("\\?")[0]; // removing parameters from url if they exist
return baseName.isEmpty() ? "" : baseName + "." + extension;
} catch (NullPointerException npe) {
return "";
}
}
于 2014-08-18T18:52:06.173 回答
0
除了所有高级方法之外，我的简单技巧是StringTokenizer：
import java.util.ArrayList;
import java.util.StringTokenizer;
public class URLName {
public static void main(String args[]){
String url = "http://www.example.com/some/path/to/a/file.xml";
StringTokenizer tokens = new StringTokenizer(url, "/");
ArrayList<String> parts = new ArrayList<>();
while(tokens.hasMoreTokens()){
parts.add(tokens.nextToken());
}
String file = parts.get(parts.size() -1);
int dot = file.indexOf(".");
String fileName = file.substring(0, dot);
System.out.println(fileName);
}
}
于 2017-05-31T12:37:21.357 回答
0
如果您使用的是Spring，则有一个帮助程序来处理 URI。这是解决方案：
List<String> pathSegments = UriComponentsBuilder.fromUriString(url).build().getPathSegments();
String filename = pathSegments.get(pathSegments.size()-1);
于 2019-09-04T12:08:49.630 回答
0
返回新文件(Uri.parse(url).getPath()).getName()
于 2020-03-11T16:35:11.697 回答
-1
create a new file with string image path
String imagePath;
File test = new File(imagePath);
test.getName();
test.getPath();
getExtension(test.getName());
public static String getExtension(String uri) {
if (uri == null) {
return null;
}
int dot = uri.lastIndexOf(".");
if (dot >= 0) {
return uri.substring(dot);
} else {
// No extension.
return "";
}
}
于 2015-02-02T14:40:29.250 回答
-2
我有同样的问题，和你的一样。我通过这个解决了它：
var URL = window.location.pathname; // Gets page name
var page = URL.substring(URL.lastIndexOf('/') + 1);
console.info(page)
于 2019-02-11T02:48:38.133 回答
-3
导入java.io.*；
import java.net.*;
public class ConvertURLToFileName{
public static void main(String[] args)throws IOException{
BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
System.out.print("Please enter the URL : ");
String str = in.readLine();
try{
URL url = new URL(str);
System.out.println("File : "+ url.getFile());
System.out.println("Converting process Successfully");
}
catch (MalformedURLException me){
System.out.println("Converting process error");
}
我希望这能帮到您。
于 2009-03-03T10:13:02.363 回答