jarファイルからファイルを読み取る方法
JARファイルにファイルがあります。 1.txt、 例えば。
どうすればアクセスできますか？私のソースコードは：
Double result=0.0;
File file = new File("1.txt")); //how get this file from a jar file
BufferedReader input = new BufferedReader(new FileReader(file));
String line;
while ((line = input.readLine()) != null) {
if(me==Integer.parseInt(line.split(":")[0])){
result= parseDouble(line.split(":")[1]);
}
}
input.close();
return result;
このファイルはファイルシステム上に独立して存在しないため、Fileは使用できません。代わりに、次のようにgetResourceAsStream（）が必要です。
...
InputStream in = getClass().getResourceAsStream("/1.txt");
BufferedReader input = new BufferedReader(new InputStreamReader(in));
...
Jarがクラスパスにある場合：
InputStream is = YourClass.class.getResourceAsStream("1.txt");
クラスパスにない場合は、次の方法でアクセスできます。
URL url = new URL("jar:file:/absolute/location/of/yourJar.jar!/1.txt");
InputStream is = url.openStream();
この答え に似たものが必要です。
その特別な方法で、アーカイブからファイルを引き出す必要があります。
BufferedReader input = new BufferedReader(new InputStreamReader(
this.getClass().getClassLoader().getResourceAsStream("1.txt")));
JarファイルはZipファイルです。...
Jarファイルを読み取るには、試してください
ZipFile file = new ZipFile("whatever.jar");
if (file != null) {
ZipEntries entries = file.entries(); //get entries from the Zip file...
if (entries != null) {
while (entries.hasMoreElements()) {
ZipEntry entry = entries.nextElement();
//use the entry to see if it's the file '1.txt'
//Read from the byte using file.getInputStream(entry)
}
}
}
お役に立てれば。
private String loadResourceFileIntoString(String path) {
//path = "/resources/html/custom.css" for example
BufferedReader buffer = new BufferedReader(new InputStreamReader(getClass().getResourceAsStream(path)));
return buffer.lines().collect(Collectors.joining(System.getProperty("line.separator")));
}
私はこれまでに同じ問題に何度か遭遇しました。 JDK 7では、誰かがクラスパスファイルシステムを書くことを望んでいましたが、残念ながらまだありませんでした。
Springには、クラスパスリソースを非常にうまくロードできるResourceクラスがあります。
答えはとても良かったのですが、クラスパスリソースであるファイルとディレクトリで動作する例を示すことで、議論に追加できると思いました。
この問題を解決するために小さなプロトタイプを作成しました。プロトタイプはすべてのEdgeケースを処理するわけではありませんが、jarファイル内のディレクトリ内のリソースの検索を処理します。
私はかなり以前からStack Overflowを使用しています。質問に答えたのを覚えているのはこれが初めてなので、長く行っても許してくれます（それが私の性質です）。
package com.foo;
import Java.io.File;
import Java.io.FileReader;
import Java.io.InputStreamReader;
import Java.io.Reader;
import Java.net.URI;
import Java.net.URL;
import Java.util.Enumeration;
import Java.util.Zip.ZipEntry;
import Java.util.Zip.ZipFile;
/**
* Prototype resource reader.
* This prototype is devoid of error checking.
*
*
* I have two prototype jar files that I have setup.
* <pre>
* <dependency>
* <groupId>invoke</groupId>
* <artifactId>invoke</artifactId>
* <version>1.0-SNAPSHOT</version>
* </dependency>
*
* <dependency>
* <groupId>node</groupId>
* <artifactId>node</artifactId>
* <version>1.0-SNAPSHOT</version>
* </dependency>
* </pre>
* The jar files each have a file under /org/node/ called resource.txt.
* <br />
* This is just a prototype of what a handler would look like with classpath://
* I also have a resource.foo.txt in my local resources for this project.
* <br />
*/
public class ClasspathReader {
public static void main(String[] args) throws Exception {
/* This project includes two jar files that each have a resource located
in /org/node/ called resource.txt.
*/
/*
Name space is just a device I am using to see if a file in a dir
starts with a name space. Think of namespace like a file extension
but it is the start of the file not the end.
*/
String namespace = "resource";
//someResource is classpath.
String someResource = args.length > 0 ? args[0] :
//"classpath:///org/node/resource.txt"; It works with files
"classpath:///org/node/"; //It also works with directories
URI someResourceURI = URI.create(someResource);
System.out.println("URI of resource = " + someResourceURI);
someResource = someResourceURI.getPath();
System.out.println("PATH of resource =" + someResource);
boolean isDir = !someResource.endsWith(".txt");
/** Classpath resource can never really start with a starting slash.
* Logically they do, but in reality you have to strip it.
* This is a known behavior of classpath resources.
* It works with a slash unless the resource is in a jar file.
* Bottom line, by stripping it, it always works.
*/
if (someResource.startsWith("/")) {
someResource = someResource.substring(1);
}
/* Use the ClassLoader to lookup all resources that have this name.
Look for all resources that match the location we are looking for. */
Enumeration resources = null;
/* Check the context classloader first. Always use this if available. */
try {
resources =
Thread.currentThread().getContextClassLoader().getResources(someResource);
} catch (Exception ex) {
ex.printStackTrace();
}
if (resources == null || !resources.hasMoreElements()) {
resources = ClasspathReader.class.getClassLoader().getResources(someResource);
}
//Now iterate over the URLs of the resources from the classpath
while (resources.hasMoreElements()) {
URL resource = resources.nextElement();
/* if the resource is a file, it just means that we can use normal mechanism
to scan the directory.
*/
if (resource.getProtocol().equals("file")) {
//if it is a file then we can handle it the normal way.
handleFile(resource, namespace);
continue;
}
System.out.println("Resource " + resource);
/*
Split up the string that looks like this:
jar:file:/Users/rick/.m2/repository/invoke/invoke/1.0-SNAPSHOT/invoke-1.0-SNAPSHOT.jar!/org/node/
into
this /Users/rick/.m2/repository/invoke/invoke/1.0-SNAPSHOT/invoke-1.0-SNAPSHOT.jar
and this
/org/node/
*/
String[] split = resource.toString().split(":");
String[] split2 = split[2].split("!");
String zipFileName = split2[0];
String sresource = split2[1];
System.out.printf("After split Zip file name = %s," +
" \nresource in Zip %s \n", zipFileName, sresource);
/* Open up the Zip file. */
ZipFile zipFile = new ZipFile(zipFileName);
/* Iterate through the entries. */
Enumeration entries = zipFile.entries();
while (entries.hasMoreElements()) {
ZipEntry entry = entries.nextElement();
/* If it is a directory, then skip it. */
if (entry.isDirectory()) {
continue;
}
String entryName = entry.getName();
System.out.printf("Zip entry name %s \n", entryName);
/* If it does not start with our someResource String
then it is not our resource so continue.
*/
if (!entryName.startsWith(someResource)) {
continue;
}
/* the fileName part from the entry name.
* where /foo/bar/foo/bee/bar.txt, bar.txt is the file
*/
String fileName = entryName.substring(entryName.lastIndexOf("/") + 1);
System.out.printf("fileName %s \n", fileName);
/* See if the file starts with our namespace and ends with our extension.
*/
if (fileName.startsWith(namespace) && fileName.endsWith(".txt")) {
/* If you found the file, print out
the contents fo the file to System.out.*/
try (Reader reader = new InputStreamReader(zipFile.getInputStream(entry))) {
StringBuilder builder = new StringBuilder();
int ch = 0;
while ((ch = reader.read()) != -1) {
builder.append((char) ch);
}
System.out.printf("Zip fileName = %s\n\n####\n contents of file %s\n###\n", entryName, builder);
} catch (Exception ex) {
ex.printStackTrace();
}
}
//use the entry to see if it's the file '1.txt'
//Read from the byte using file.getInputStream(entry)
}
}
}
/**
* The file was on the file system not a Zip file,
* this is here for completeness for this example.
* otherwise.
*
* @param resource
* @param namespace
* @throws Exception
*/
private static void handleFile(URL resource, String namespace) throws Exception {
System.out.println("Handle this resource as a file " + resource);
URI uri = resource.toURI();
File file = new File(uri.getPath());
if (file.isDirectory()) {
for (File childFile : file.listFiles()) {
if (childFile.isDirectory()) {
continue;
}
String fileName = childFile.getName();
if (fileName.startsWith(namespace) && fileName.endsWith("txt")) {
try (FileReader reader = new FileReader(childFile)) {
StringBuilder builder = new StringBuilder();
int ch = 0;
while ((ch = reader.read()) != -1) {
builder.append((char) ch);
}
System.out.printf("fileName = %s\n\n####\n contents of file %s\n###\n", childFile, builder);
} catch (Exception ex) {
ex.printStackTrace();
}
}
}
} else {
String fileName = file.getName();
if (fileName.startsWith(namespace) && fileName.endsWith("txt")) {
try (FileReader reader = new FileReader(file)) {
StringBuilder builder = new StringBuilder();
int ch = 0;
while ((ch = reader.read()) != -1) {
builder.append((char) ch);
}
System.out.printf("fileName = %s\n\n####\n contents of file %s\n###\n", fileName, builder);
} catch (Exception ex) {
ex.printStackTrace();
}
}
}
}
}
これは私にjarファイルから別のtxtファイルにtxtファイルをコピーするために働いた
public static void copyTextMethod() throws Exception{
String inputPath = "path/to/.jar";
String outputPath = "Desktop/CopyText.txt";
File resStreamOut = new File(outputPath);
int readBytes;
JarFile file = new JarFile(inputPath);
FileWriter fw = new FileWriter(resStreamOut);
try{
Enumeration<JarEntry> entries = file.entries();
while (entries.hasMoreElements()){
JarEntry entry = entries.nextElement();
if (entry.getName().equals("readMe/tempReadme.txt")) {
System.out.println(entry +" : Entry");
InputStream is = file.getInputStream(entry);
BufferedWriter output = new BufferedWriter(fw);
while ((readBytes = is.read()) != -1) {
output.write((char) readBytes);
}
System.out.println(outputPath);
output.close();
}
}
} catch(Exception er){
er.printStackTrace();
}
}
}