I'm using PathMatcher and SimpleFileVisitor to iterate over directory and find all the files that start only with a certain prefix. However, I can't get any files although there are some files that match my preference.
Example of file required:
Prefix_some_text.csv
Here is the Main code that invokes the call for SimpleFileVisitor class, and it uses regex pattern with the prefix and suppose to find all files starting with the certain pattern:
String directoryAsString = "C:/Users";
String pattern = "Prefix";
SearchFileByWildcard sfbw = new SearchFileByWildcard();
try {
List<String> actual = sfbw.searchWithWc(Paths.get(directoryAsString),pattern);
} catch (IOException e) {
e.printStackTrace();
}
The implementation of SearchFileByWildcard class that uses SimpleFileVisitor :
static class SearchFileByWildcard {
List<String> matchesList = new ArrayList<String>();
List<String> searchWithWc(Path rootDir, String pattern) throws IOException {
matchesList.clear();
FileVisitor<Path> matcherVisitor = new SimpleFileVisitor<Path>() {
#Override
public FileVisitResult visitFile(Path file, BasicFileAttributes attribs) throws IOException {
FileSystem fs = FileSystems.getDefault();
PathMatcher matcher = fs.getPathMatcher("regex:" + pattern);
Path name = file.getFileName(); //takes the filename from the full path
if (matcher.matches(name)) {
matchesList.add(name.toString());
}
return FileVisitResult.CONTINUE;
}
};
Files.walkFileTree(rootDir, matcherVisitor);
return matchesList;
}
}
I'm debating whether to use glob instead of regex? Or maybe something with my regex is flawed.
It seems like the pattern is wrong. It matches only files named "Prefix". Try to change it in String pattern = "Prefix.*";.
Otherwise you can scan for files which name starts by the string "Prefix".
String name = file.getFileName().toString();
if (name.startsWith(pattern)) {
matchesList.add(name);
}
I am looking for a way to get a list of all resource names from a given classpath directory, something like a method List<String> getResourceNames (String directoryName).
For example, given a classpath directory x/y/z containing files a.html, b.html, c.html and a subdirectory d, getResourceNames("x/y/z") should return a List<String> containing the following strings:['a.html', 'b.html', 'c.html', 'd'].
It should work both for resources in filesystem and jars.
I know that I can write a quick snippet with Files, JarFiles and URLs, but I do not want to reinvent the wheel. My question is, given existing publicly available libraries, what is the quickest way to implement getResourceNames? Spring and Apache Commons stacks are both feasible.
Custom Scanner
Implement your own scanner. For example:
(limitations of this solution are mentioned in the comments)
private List<String> getResourceFiles(String path) throws IOException {
List<String> filenames = new ArrayList<>();
try (
InputStream in = getResourceAsStream(path);
BufferedReader br = new BufferedReader(new InputStreamReader(in))) {
String resource;
while ((resource = br.readLine()) != null) {
filenames.add(resource);
}
}
return filenames;
}
private InputStream getResourceAsStream(String resource) {
final InputStream in
= getContextClassLoader().getResourceAsStream(resource);
return in == null ? getClass().getResourceAsStream(resource) : in;
}
private ClassLoader getContextClassLoader() {
return Thread.currentThread().getContextClassLoader();
}
Spring Framework
Use PathMatchingResourcePatternResolver from Spring Framework.
Ronmamo Reflections
The other techniques might be slow at runtime for huge CLASSPATH values. A faster solution is to use ronmamo's Reflections API, which precompiles the search at compile time.
Here is the code
Source: forums.devx.com/showthread.php?t=153784
import java.io.File;
import java.io.IOException;
import java.util.ArrayList;
import java.util.Collection;
import java.util.Enumeration;
import java.util.regex.Pattern;
import java.util.zip.ZipEntry;
import java.util.zip.ZipException;
import java.util.zip.ZipFile;
/**
* list resources available from the classpath # *
*/
public class ResourceList{
/**
* for all elements of java.class.path get a Collection of resources Pattern
* pattern = Pattern.compile(".*"); gets all resources
*
* #param pattern
* the pattern to match
* #return the resources in the order they are found
*/
public static Collection<String> getResources(
final Pattern pattern){
final ArrayList<String> retval = new ArrayList<String>();
final String classPath = System.getProperty("java.class.path", ".");
final String[] classPathElements = classPath.split(System.getProperty("path.separator"));
for(final String element : classPathElements){
retval.addAll(getResources(element, pattern));
}
return retval;
}
private static Collection<String> getResources(
final String element,
final Pattern pattern){
final ArrayList<String> retval = new ArrayList<String>();
final File file = new File(element);
if(file.isDirectory()){
retval.addAll(getResourcesFromDirectory(file, pattern));
} else{
retval.addAll(getResourcesFromJarFile(file, pattern));
}
return retval;
}
private static Collection<String> getResourcesFromJarFile(
final File file,
final Pattern pattern){
final ArrayList<String> retval = new ArrayList<String>();
ZipFile zf;
try{
zf = new ZipFile(file);
} catch(final ZipException e){
throw new Error(e);
} catch(final IOException e){
throw new Error(e);
}
final Enumeration e = zf.entries();
while(e.hasMoreElements()){
final ZipEntry ze = (ZipEntry) e.nextElement();
final String fileName = ze.getName();
final boolean accept = pattern.matcher(fileName).matches();
if(accept){
retval.add(fileName);
}
}
try{
zf.close();
} catch(final IOException e1){
throw new Error(e1);
}
return retval;
}
private static Collection<String> getResourcesFromDirectory(
final File directory,
final Pattern pattern){
final ArrayList<String> retval = new ArrayList<String>();
final File[] fileList = directory.listFiles();
for(final File file : fileList){
if(file.isDirectory()){
retval.addAll(getResourcesFromDirectory(file, pattern));
} else{
try{
final String fileName = file.getCanonicalPath();
final boolean accept = pattern.matcher(fileName).matches();
if(accept){
retval.add(fileName);
}
} catch(final IOException e){
throw new Error(e);
}
}
}
return retval;
}
/**
* list the resources that match args[0]
*
* #param args
* args[0] is the pattern to match, or list all resources if
* there are no args
*/
public static void main(final String[] args){
Pattern pattern;
if(args.length < 1){
pattern = Pattern.compile(".*");
} else{
pattern = Pattern.compile(args[0]);
}
final Collection<String> list = ResourceList.getResources(pattern);
for(final String name : list){
System.out.println(name);
}
}
}
If you are using Spring Have a look at PathMatchingResourcePatternResolver
Using Reflections
Get everything on the classpath:
Reflections reflections = new Reflections(null, new ResourcesScanner());
Set<String> resourceList = reflections.getResources(x -> true);
Another example - get all files with extension .csv from some.package:
Reflections reflections = new Reflections("some.package", new ResourcesScanner());
Set<String> resourceList = reflections.getResources(Pattern.compile(".*\\.csv"));
So in terms of the PathMatchingResourcePatternResolver this is what is needed in the code:
#Autowired
ResourcePatternResolver resourceResolver;
public void getResources() {
resourceResolver.getResources("classpath:config/*.xml");
}
If you use apache commonsIO you can use for the filesystem (optionally with extension filter):
Collection<File> files = FileUtils.listFiles(new File("directory/"), null, false);
and for resources/classpath:
List<String> files = IOUtils.readLines(MyClass.class.getClassLoader().getResourceAsStream("directory/"), Charsets.UTF_8);
If you don't know if "directoy/" is in the filesystem or in resources you may add a
if (new File("directory/").isDirectory())
or
if (MyClass.class.getClassLoader().getResource("directory/") != null)
before the calls and use both in combination...
The most robust mechanism for listing all resources in the classpath is currently to use this pattern with ClassGraph, because it handles the widest possible array of classpath specification mechanisms, including the new JPMS module system. (I am the author of ClassGraph.)
List<String> resourceNames;
try (ScanResult scanResult = new ClassGraph().acceptPaths("x/y/z").scan()) {
resourceNames = scanResult.getAllResources().getNames();
}
The Spring framework's PathMatchingResourcePatternResolver is really awesome for these things:
private Resource[] getXMLResources() throws IOException
{
ClassLoader classLoader = MethodHandles.lookup().getClass().getClassLoader();
PathMatchingResourcePatternResolver resolver = new PathMatchingResourcePatternResolver(classLoader);
return resolver.getResources("classpath:x/y/z/*.xml");
}
Maven dependency:
<dependency>
<groupId>org.springframework</groupId>
<artifactId>spring-core</artifactId>
<version>LATEST</version>
</dependency>
This should work (if spring is not an option):
public static List<String> getFilenamesForDirnameFromCP(String directoryName) throws URISyntaxException, UnsupportedEncodingException, IOException {
List<String> filenames = new ArrayList<>();
URL url = Thread.currentThread().getContextClassLoader().getResource(directoryName);
if (url != null) {
if (url.getProtocol().equals("file")) {
File file = Paths.get(url.toURI()).toFile();
if (file != null) {
File[] files = file.listFiles();
if (files != null) {
for (File filename : files) {
filenames.add(filename.toString());
}
}
}
} else if (url.getProtocol().equals("jar")) {
String dirname = directoryName + "/";
String path = url.getPath();
String jarPath = path.substring(5, path.indexOf("!"));
try (JarFile jar = new JarFile(URLDecoder.decode(jarPath, StandardCharsets.UTF_8.name()))) {
Enumeration<JarEntry> entries = jar.entries();
while (entries.hasMoreElements()) {
JarEntry entry = entries.nextElement();
String name = entry.getName();
if (name.startsWith(dirname) && !dirname.equals(name)) {
URL resource = Thread.currentThread().getContextClassLoader().getResource(name);
filenames.add(resource.toString());
}
}
}
}
}
return filenames;
}
My way, no Spring, used during a unit test:
URI uri = TestClass.class.getResource("/resources").toURI();
Path myPath = Paths.get(uri);
Stream<Path> walk = Files.walk(myPath, 1);
for (Iterator<Path> it = walk.iterator(); it.hasNext(); ) {
Path filename = it.next();
System.out.println(filename);
}
With Spring it's easy. Be it a file, or folder, or even multiple files, there are chances, you can do it via injection.
This example demonstrates the injection of multiple files located in x/y/z folder.
import org.springframework.beans.factory.annotation.Value;
import org.springframework.core.io.Resource;
import org.springframework.stereotype.Service;
#Service
public class StackoverflowService {
#Value("classpath:x/y/z/*")
private Resource[] resources;
public List<String> getResourceNames() {
return Arrays.stream(resources)
.map(Resource::getFilename)
.collect(Collectors.toList());
}
}
It does work for resources in the filesystem as well as in JARs.
Used a combination of Rob's response.
final String resourceDir = "resourceDirectory/";
List<String> files = IOUtils.readLines(Thread.currentThread().getClass().getClassLoader().getResourceAsStream(resourceDir), Charsets.UTF_8);
for (String f : files) {
String data = IOUtils.toString(Thread.currentThread().getClass().getClassLoader().getResourceAsStream(resourceDir + f));
// ... process data
}
I think you can leverage the [Zip File System Provider][1] to achieve this. When using FileSystems.newFileSystem it looks like you can treat the objects in that ZIP as a "regular" file.
In the linked documentation above:
Specify the configuration options for the zip file system in the java.util.Map object passed to the FileSystems.newFileSystem method. See the [Zip File System Properties][2] topic for information about the provider-specific configuration properties for the zip file system.
Once you have an instance of a zip file system, you can invoke the methods of the [java.nio.file.FileSystem][3] and [java.nio.file.Path][4] classes to perform operations such as copying, moving, and renaming files, as well as modifying file attributes.
The documentation for the jdk.zipfs module in [Java 11 states][5]:
The zip file system provider treats a zip or JAR file as a file system and provides the ability to manipulate the contents of the file. The zip file system provider can be created by [FileSystems.newFileSystem][6] if installed.
Here is a contrived example I did using your example resources. Note that a .zip is a .jar, but you could adapt your code to instead use classpath resources:
Setup
cd /tmp
mkdir -p x/y/z
touch x/y/z/{a,b,c}.html
echo 'hello world' > x/y/z/d
zip -r example.zip x
Java
import java.io.IOException;
import java.net.URI;
import java.nio.file.FileSystem;
import java.nio.file.FileSystems;
import java.nio.file.Files;
import java.util.Collections;
import java.util.stream.Collectors;
public class MkobitZipRead {
public static void main(String[] args) throws IOException {
final URI uri = URI.create("jar:file:/tmp/example.zip");
try (
final FileSystem zipfs = FileSystems.newFileSystem(uri, Collections.emptyMap());
) {
Files.walk(zipfs.getPath("/")).forEach(path -> System.out.println("Files in zip:" + path));
System.out.println("-----");
final String manifest = Files.readAllLines(
zipfs.getPath("x", "y", "z").resolve("d")
).stream().collect(Collectors.joining(System.lineSeparator()));
System.out.println(manifest);
}
}
}
Output
Files in zip:/
Files in zip:/x/
Files in zip:/x/y/
Files in zip:/x/y/z/
Files in zip:/x/y/z/c.html
Files in zip:/x/y/z/b.html
Files in zip:/x/y/z/a.html
Files in zip:/x/y/z/d
-----
hello world
Neither of answers worked for me even though I had my resources put in resources folders and followed the above answers. What did make a trick was:
#Value("file:*/**/resources/**/schema/*.json")
private Resource[] resources;
Expanding on Luke Hutchinsons answer above, using his ClassGraph library, I was able to easily get a list of all files in a Resource folder with almost no effort at all.
Let's say that in your resource folder, you have a folder called MyImages. This is how easy it is to get a URL list of all the files in that folder:
import io.github.classgraph.ClassGraph;
import io.github.classgraph.ResourceList;
import io.github.classgraph.ScanResult;
public static LinkedList<URL> getURLList (String folder) {
LinkedList<URL> urlList = new LinkedList<>();
ScanResult scanResult = new ClassGraph().enableAllInfo().scan();
ResourceList resources = scanResult.getAllResources();
for (URL url : resources.getURLs()) {
if (url.toString().contains(folder)) {
urlList.addLast(url);
}
}
return urlList;
}
Then you simply do this:
LinkedList<URL> myURLFileList = getURLList("MyImages");
The URLs can then be loaded into streams or use Apache's FileUtils to copy the files somewhere else like this:
String outPath = "/My/Output/Path";
for(URL url : myURLFileList) {
FileUtils.copyURLToFile(url, new File(outPath, url.getFile()));
}
I think ClassGraph is a pretty slick library for making tasks like this very simple and easy to comprehend.
Based on #rob 's information above, I created the implementation which I am releasing to the public domain:
private static List<String> getClasspathEntriesByPath(String path) throws IOException {
InputStream is = Main.class.getClassLoader().getResourceAsStream(path);
StringBuilder sb = new StringBuilder();
while (is.available()>0) {
byte[] buffer = new byte[1024];
sb.append(new String(buffer, Charset.defaultCharset()));
}
return Arrays
.asList(sb.toString().split("\n")) // Convert StringBuilder to individual lines
.stream() // Stream the list
.filter(line -> line.trim().length()>0) // Filter out empty lines
.collect(Collectors.toList()); // Collect remaining lines into a List again
}
While I would not have expected getResourcesAsStream to work like that on a directory, it really does and it works well.
I have a Java web application that exports pdf files.
I have to use Crystal Reports 11. I can already export pdfs, the issue is that it only works locally because the .rpt file has a reference to an XML file in my machine.
So, when I want to export a report and the .rpt file can't find the file, even though I'm giving it a new dataset to work with it still throws a not found exception. I tried changing the file's connection programmatically but it always throws an exception related to the connection.
public InputStream export() throws ReportSDKException, IOException, IllegalArgumentException, IllegalAccessException {
ReportClientDocument reportClientDoc = new ReportClientDocument();
reportClientDoc.setLocale(Locale.forLanguageTag(localeTag));
reportClientDoc.open(reportPath, OpenReportOptions._discardSavedData);
DatabaseController databaseController = reportClientDoc.getDatabaseController();
IConnectionInfo oldConn = databaseController.getConnectionInfos(null).get(0);
IConnectionInfo newConn = resolveConnection(reportClientDoc).get(0);
int replaceParams = DBOptions._ignoreCurrentTableQualifiers | DBOptions._doNotVerifyDB;
databaseController.replaceConnection(oldConn, newConn, null,replaceParams);
reportClientDoc.getDatabaseController().setDataSource(this.dataset);
ParameterFieldController parameterController = reportClientDoc.getDataDefController()
.getParameterFieldController();
for (Param<Double> p : doubleParams) {
parameterController.setCurrentValue(p.subReportName, p.fieldName, p.value);
}
for (Param<Object> p : objectParams) {
parameterController.setCurrentValue(p.subReportName, p.fieldName, p.value);
}
return reportClientDoc.getPrintOutputController().export(this.format);
}
private ConnectionInfos resolveConnection(ReportClientDocument reportClientDoc) throws ReportSDKException {
IConnectionInfo oldConnection = new ConnectionInfo();
DatabaseController dbController = reportClientDoc.getDatabaseController();
oldConnection = dbController.getConnectionInfos(null).getConnectionInfo(0);
String xsdPath = Paths.get(this.xsdPath).toAbsolutePath().toString();
final String SERVER_NAME = dummyXmlPath + " " + xsdPath;
final String DATABASE_DLL = oldConnection.getAttributes().getStringValue("Database DLL");
final String LOCAL_SCHEMA_FILE = xsdPath;
final String SERVER_TYPE = "XML";
final String PREQESERVERNAME = SERVER_NAME;
final String PREQESERVERTYPE = "XML";
final String LOCAL_XML_FILE = dummyXmlPath;
PropertyBag newAttributes = new PropertyBag();
newAttributes.put("Server Name", SERVER_NAME);
newAttributes.put("Database DLL", DATABASE_DLL);
newAttributes.put("Local Schema File", LOCAL_SCHEMA_FILE);
newAttributes.put("PreQEServerName", PREQESERVERNAME);
newAttributes.put("PreQEServerType", PREQESERVERTYPE);
newAttributes.put("Server Type", SERVER_TYPE);
newAttributes.put("Local XML File", LOCAL_XML_FILE);
IConnectionInfo newConnection = (IConnectionInfo) oldConnection.clone(true);
newConnection.setAttributes(newAttributes);
newConnection.setKind(oldConnection.getKind());
ConnectionInfos connectionInfos = new ConnectionInfos();
connectionInfos.add(newConnection);
return connectionInfos;
}
I was able to fix this by adding the Subreport data to the subreports.
SubreportController subreportController = reportClientDoc.getSubreportController();
for (String string : subreportController.querySubreportNames()) {
subreportController.setDataSource(string, dataset);
}
I am using multi text output formate to create multiple files of a single file i.e each line on new file.
This is my code:
public class MOFExample extends Configured implements Tool {
private static double count = 0;
static class KeyBasedMultipleTextOutputFormat extends
MultipleTextOutputFormat<Text, Text> {
#Override
protected String generateFileNameForKeyValue(Text key, Text value,
String name) {
return count++ + "_";// + name;
}
}
/**
* The main job driver.
*/
public int run(final String[] args) throws Exception {
Path csvInputs = new Path(args[0]);
Path outputDir = new Path(args[1]);
JobConf jobConf = new JobConf(super.getConf());
jobConf.setJarByClass(MOFExample.class);
jobConf.setMapperClass(IdentityMapper.class);
jobConf.setInputFormat(KeyValueTextInputFormat.class);
jobConf.setOutputFormat(KeyBasedMultipleTextOutputFormat.class);
jobConf.setOutputValueClass(Text.class);
jobConf.setOutputKeyClass(Text.class);
FileInputFormat.setInputPaths(jobConf, csvInputs);
FileOutputFormat.setOutputPath(jobConf, outputDir);
//jobConf.setNumMapTasks(4);
jobConf.setNumReduceTasks(4);
return JobClient.runJob(jobConf).isSuccessful() ? 0 : 1;
}
public static void main(final String[] args) throws Exception {
int res = ToolRunner.run(new Configuration(), new MOFExample(), args);
System.exit(res);
}
}
This code runs fine on small text file but when the number of lines of input file are greater than 1900 which is yet not a large file it throws an exception:
Exception in thread "main" java.io.IOException: Job failed!
at org.apache.hadoop.mapred.JobClient.runJob(JobClient.java:836)
at MOFExample.run(MOFExample.java:57)
at org.apache.hadoop.util.ToolRunner.run(ToolRunner.java:70)
at MOFExample.main(MOFExample.java:61)
I also tried this tutorial but this one returns empty output directory without any exception when the input file is large however this one also worked fine with small input file.
Note: I am using Single-Node Cluster
problem in parsing special character attributes using jdom
ex
< tag xml:lang="123" >
this case getAttributes() method return null
is there any solution to fix this.
Works without problems for me:
public class TestJdom
{
public static void main(String[] args) throws JDOMException, IOException {
String xmlString = "<test><tag xml:lang=\"123\"></tag></test>";
SAXBuilder builder = new SAXBuilder();
StringReader stringReader = new StringReader(new String(xmlString
.getBytes()));
Document doc = builder.build(stringReader);
List<?> attrs = doc.getRootElement().getChild("tag").getAttributes();
System.out.println(attrs);
}
}
You probably need to set namespace, check http://cs.au.dk/~amoeller/XML/programming/jdomexample.html