URL에서 파일 이름을 얻습니다
https://stackoverflow.com/questions/605696
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Java에서 java.net.URL 또는 a String 형태 http://www.example.com/some/path/to/a/file.xml , 파일 이름을 얻는 가장 쉬운 방법은 확장자를 뺀 것입니까? 그래서이 예에서는 돌아 오는 것을 찾고 있습니다. "file".
나는 이것을 할 수있는 몇 가지 방법을 생각할 수 있지만 읽기 쉬운 무언가를 찾고 있습니다.
해결책
바퀴를 재창조하는 대신 apache를 사용하는 것은 어떻습니까 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
}
}
다른 팁
String fileName = url.substring( url.lastIndexOf('/')+1, url.length() );
String fileNameWithoutExtn = fileName.substring(0, fileName.lastIndexOf('.'));
이것은 그것을 잘라야합니다 (오류 처리를 당신에게 맡길 것입니다) :
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);
}
파일 확장자를 제거 할 필요가 없다면 오류가 발생하기 쉬운 문자열 조작에 의지하지 않고 외부 라이브러리를 사용하지 않고 수행 할 수있는 방법이 있습니다. Java 1.7+와 함께 작동합니다.
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()
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;
}
얻다 파일 확장자 이름, 확장없이, 확장 만 단지 3 줄로 :
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);
로그 결과 :
File Name(13656): test.png
File Name Without Extension(13656): test
File Extension(13656): .png
그것이 당신을 도울 수 있기를 바랍니다.
나는 이것을 생각해 냈다 :
String url = "http://www.example.com/some/path/to/a/file.xml";
String file = url.substring(url.lastIndexOf('/')+1, url.lastIndexOf('.'));
Keep it simple :
/**
* 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];
}
String fileName = url.substring(url.lastIndexOf('/') + 1);
One liner:
new File(uri.getPath).getName
Complete code:
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
Note: URI#gePath is already intelligent enough to strip off query parameters and the protocol's scheme. Examples:
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
Here is the simplest way to do it in Android. I know it will not work in Java but It may help Android application developer.
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;
}
Create an URL object from the String. When first you have an URL object there are methods to easily pull out just about any snippet of information you need.
I can strongly recommend the Javaalmanac web site which has tons of examples, but which has since moved. You might find http://exampledepot.8waytrips.com/egs/java.io/File2Uri.html interesting:
// 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
}
There are some ways:
Java 7 File I/O:
String fileName = Paths.get(strUrl).getFileName().toString();
Apache Commons:
String fileName = FilenameUtils.getName(strUrl);
Using Jersey:
UriBuilder buildURI = UriBuilder.fromUri(strUrl);
URI uri = buildURI.build();
String fileName = Paths.get(uri.getPath()).getFileName();
Substring:
String fileName = strUrl.substring(strUrl.lastIndexOf('/') + 1);
If you want to get only the filename from a java.net.URL (not including any query parameters), you could use the following function:
public static String getFilenameFromURL(URL url) {
return new File(url.getPath().toString()).getName();
}
For example, this input URL:
"http://example.com/image.png?version=2&modificationDate=1449846324000"
Would be translated to this output String:
image.png
I've found that some urls when passed directly to FilenameUtils.getName return unwanted results and this needs to be wrapped up to avoid exploits.
For example,
System.out.println(FilenameUtils.getName("http://www.google.com/.."));
returns
..
which I doubt anyone wants to allow.
The following function seems to work fine, and shows some of these test cases, and it returns null when the filename can't be determined.
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;
}
}
This is wrapped up with some simple tests cases in the following example:
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;
}
}
}
Urls can have parameters in the end, this
/**
* 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;
}
The Url object in urllib allows you to access the path's unescaped filename. Here are some examples:
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());
andy's answer redone using split():
Url u= ...;
String[] pathparts= u.getPath().split("\\/");
String filename= pathparts[pathparts.length-1].split("\\.", 1)[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;
}
How about this:
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);
In order to return filename without extension and without parameters use the following:
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
In order to return filename with extension without params use this:
/** 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 "";
}
}
Beyond the all advanced methods, my simple trick is 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);
}
}
I have the same problem, with yours. I solved it by this:
var URL = window.location.pathname; // Gets page name
var page = URL.substring(URL.lastIndexOf('/') + 1);
console.info(page)
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 "";
}
}
import 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");
}
I hope this will help you.
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