164

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。

4

27 回答 27

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&amp;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