从 URL 获取文件名

这应该可以解决问题(我把错误处理留给你) :

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);
}

我想到了这个:

String url = "http://www.example.com/some/path/to/a/file.xml";
String file = url.substring(url.lastIndexOf('/')+1, url.lastIndexOf('.'));

String fileName = url.substring( url.lastIndexOf('/')+1, url.length() );


String fileNameWithoutExtn = fileName.substring(0, fileName.lastIndexOf('.'));

* ;

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");


}

我希望这能帮到你。

从 String 创建 URL 对象。首先，当您有一个 URL 对象时，有一些方法可以轻松地提取出您需要的任何信息片段。

我强烈推荐 Javaalmanac 网站，它有大量的例子，但是已经被移动了。你可能会对 http://exampledepot.8waytrips.com/egs/java.io/File2Uri.html感兴趣:

// 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
}

Andy 的答案使用 split ()重做了一遍:

Url u= ...;
String[] pathparts= u.getPath().split("\\/");
String filename= pathparts[pathparts.length-1].split("\\.", 1)[0];

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;
}

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;
}

这样吧:

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);

不用重造轮子，用 Apache (法语)怎么样:

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
}


}

在 Android 中，这是最简单的方法。我知道它不会在 Java 中工作，但它可能有助于 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;
}

Urls 最终可以有参数，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;
}

为了返回文件名 没有延期和 没有参数，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 扩展名不带参数的文件名，使用以下命令:

/** 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 "";
}
}

保持简单:

/**
* 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];
}

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 "";
}
}

获取 文件 名称与分机，没有延期，只有分机，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);

记录结果:

File Name(13656): test.png
File Name Without Extension(13656): test
File Extension(13656): .png

希望对你有帮助。

如果您不需要去除文件扩展名，这里有一种方法可以做到这一点，而不需要求助于容易出错的 String 操作，也不需要使用外部库。使用 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()

String fileName = url.substring(url.lastIndexOf('/') + 1);

如果只想从 java.net.URL 获取文件名(不包括任何查询参数) ，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"

将被转换为此输出 String:

image.png

我发现有些 url 在直接传递给 FilenameUtils.getName时会返回不想要的结果，这需要包装以避免被利用。

比如说,

System.out.println(FilenameUtils.getName("http://www.google.com/.."));

报税表

..

我怀疑没人会同意。

下面的函数似乎工作得很好，并显示了其中的一些测试用例，当无法确定文件名时，它返回 null。

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;
}
}

下面的示例中包含了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;
}
}
}

除了所有的高级方法，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);
}
}

一句话:

new File(uri.getPath).getName

完整的代码(在一个标量 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，可以去掉查询参数和协议方案:

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

Urllib中的 Url对象允许访问路径的未转义文件名:

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());

有一些方法:

Java7文件 I/O:

String fileName = Paths.get(strUrl).getFileName().toString();

Apache Commons:

String fileName = FilenameUtils.getName(strUrl);

使用新泽西州:

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);

我和你有同样的问题，我用这个解决了它:

var URL = window.location.pathname; // Gets page name
var page = URL.substring(URL.lastIndexOf('/') + 1);
console.info(page)

如果使用 春天，则有一个 帮手来处理 URI。 解决办法如下:

List<String> pathSegments = UriComponentsBuilder.fromUriString(url).build().getPathSegments();
String filename = pathSegments.get(pathSegments.size()-1);

返回 new File (Uri.parse (url) . getPath ()) . getName ()

字符串 fileNameWithExtension = url.substring (url.lastIndexOf (’/’) + 1) ;

字符串 fileNameWithExtension.substring (0，fileNameWithExtension.lastIndexOf (’.’)) ;