Соединение без ассоциации в HQL

Я сталкивался с этой же проблемой несколько раз раньше. В JDK 7 я надеялся, что кто-то напишет файловую систему classpath, но, увы, пока нет.

Spring имеет класс Resource, который позволяет вам очень приятно загружать ресурсы classpath.

Ответы были очень хорошими, но я подумал, что могу добавить к обсуждению пример, который работает с файлами и каталогами, которые являются ресурсами classpath.

Я написал небольшой прототип, чтобы решить эту проблему. Прототип не обрабатывает каждый крайний случай, но он обрабатывает поиск ресурсов в каталогах, которые находятся в jar-файлах.

Я использовал Stack Overflow довольно давно. Это первый раз, когда я помню, как отвечал на вопрос, так что прости меня, если я пойду долго (это моя природа).

   

    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.
    * 
    *             
    *                  invoke
    *                  invoke
    *                  1.0-SNAPSHOT
    *              
    *
    *              
    *                   node
    *                   node
    *                   1.0-SNAPSHOT
    *              
    * 
    * The jar files each have a file under /org/node/ called resource.txt.
    * 

    * 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.
    * 

    */
    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();
                    }

                }

            }
        }

    }


   

Здесь вы можете увидеть более полный пример с примером вывода.