كيف تقرأ ملف من ملف جرة؟

Question

لدي ملف في ملف جرة. إنه 1.txt, ، علي سبيل المثال.

كيف يمكنني الوصول إليه؟ رمز المصدر الخاص بي هو:

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;

Answer 1

لا يمكنك استخدام الملف ، لأن هذا الملف غير موجود بشكل مستقل على نظام الملفات. بدلاً من ذلك ، تحتاج إلى getResourCeasStream () ، مثل ذلك:

...
InputStream in = getClass().getResourceAsStream("/1.txt");
BufferedReader input = new BufferedReader(new InputStreamReader(in));
...

Answer 2

إذا كانت جرة الخاصة بك على classpath:

InputStream is = YourClass.class.getResourceAsStream("1.txt");

إذا لم يكن على classpath ، فيمكنك الوصول إليه عبر:

URL url = new URL("jar:file:/absolute/location/of/yourJar.jar!/1.txt");
InputStream is = url.openStream();

Answer 3

شيء مشابه ل هذا الجواب هو ما تحتاجه.

تحتاج إلى سحب الملف من الأرشيف بهذه الطريقة الخاصة.

BufferedReader input = new BufferedReader(new InputStreamReader(
         this.getClass().getClassLoader().getResourceAsStream("1.txt")));

Answer 4

ملف جرة هو ملف مضغوط .....

حتى قراءة ملف جرة ، حاول

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)
      }
    }
}

أتمنى أن يساعدك هذا.

Answer 5

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")));
}

Answer 6

لقد واجهت هذه المشكلة نفسها عدة مرات من قبل. كنت آمل في JDK 7 أن يكتب شخص ما نظام ملفات classpath ، ولكن للأسف لم يكن بعد.

يحتوي Spring على فئة الموارد التي تتيح لك تحميل موارد ClassPath بشكل جيد للغاية.

كانت الإجابات جيدة جدًا ، لكنني اعتقدت أنه يمكنني إضافة إلى المناقشة مع إظهار مثال يعمل مع الملفات والأدلة التي هي موارد ClassPath.

كتبت نموذجًا أوليًا لحل هذه المشكلة بالذات. لا يتعامل النموذج الأولي مع كل حالة حافة ، ولكنه يتعامل مع البحث عن الموارد في الدلائل الموجودة في ملفات JAR.

لقد استخدمت مكدس فائض لبعض الوقت. هذه هي المرة الأولى التي أتذكر فيها الإجابة على سؤال ، لذا سامحني إذا ذهبت إلى فترة طويلة (إنها طبيعتي).

   

    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();
                    }

                }

            }
        }

    }


   

يمكنك رؤية مثال أكمل هنا مع عينة إخراج.

Answer 7

لقد عمل هذا بالنسبة لي لنسخ ملف txt من ملف JAR إلى ملف txt آخر

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();
    }
        }
            }

Answer 8

InputStream inputStreamLastName = this.getClass().getClassLoader().getResourceAsStream("path of file");


    try {

        br  = new BufferedReader(new InputStreamReader(inputStreamLastName, "UTF-8"));
        String sCurrentLine;
        ArrayList<String> lastNamesList = new ArrayList<String>();
        while ((sCurrentLine = br.readLine()) != null) {

            if(sCurrentLine.length()>=min && sCurrentLine.length()<=max){
                lastNamesList.add(sCurrentLine);
            }

        }