I need to get properties file from war.
String fileName = loadFileName();
try {
FileInputStream fis = new FileInputStream(new File(Thread.currentThread().getContextClassLoader().getResource(fileName).toURI()));
property.load(fis);
}
.........
private String loadFileName(){
if(currentLocale.equals(new Locale("en"))){
return "file:///locale/output/language.properties";
} else {
return "file:///locale/output/language_ru.properties";
}
}
This files locate in main/resources/locale/output folder. Unfortunately,this approach is not working, all property values is null. How to load files from war correctly? Server: JBoss AS 7
A WAR is an archive file. You can't address "files" within the archive as files. Just open the resource via #getResourceAsStream(String name). name is probably /locale/output/language.properties here.
Of course you shouldn't build i18n yourself.
Related
I have deployed a spring-boot application JAR file. Now, I want to upload the image from android and store it in the myfolder of resource directory. But unable to get the path of resource directory.
Error is:
java.io.FileNotFoundException: src/main/resources/static/myfolder/myimage.png
(No such file or directory)
This is the code for storing the file in the resource folder
private final String RESOURCE_PATH = "src/main/resources";
String filepath = "/myfolder/";
public String saveFile(byte[] bytes, String filepath, String filename) throws MalformedURLException, IOException {
File file = new File(RESOURCE_PATH + filepath + filename);
OutputStream out = new FileOutputStream(file);
try {
out.write(bytes);
} catch (Exception e) {
e.printStackTrace();
} finally {
out.close();
}
return file.getName();
}
UPDATED:
This is what I have tried
private final String RESOURCE_PATH = "config/";
controller class:
String filepath = "myfolder/";
String filename = "newfile.png"
public String saveFile(byte[] bytes, String filepath, String filename) throws MalformedURLException, IOException {
//reading old file
System.out.println(Files.readAllBytes(Paths.get("config","myfolder","oldfile.png"))); //gives noSuchFileException
//writing new file
File file = new File(RESOURCE_PATH + filepath + filename);
OutputStream out = new FileOutputStream(file); //FileNotFoundException
try {
out.write(bytes);
} catch (Exception e) {
e.printStackTrace();
} finally {
out.close();
}
return file.getName();
}
Project structure:
+springdemo-0.0.1-application.zip
+config
+myfolder
-oldfile.png
-application.properties
+lib
+springdemo-0.0.1.jar
+start.sh
-springdemo-0.0.1.jar //running this jar file
Usually when you deploy an application (or start it using Java), you start a JAR file. You don't have a resource folder. You can have one and access it, too, but it certainly won't be src/main/resources.
When you build your final artifact (your application), it creates a JAR (or EAR or WAR) file and your resources, which you had in your src/main/resources-folder, are copied over to the output directory and included in the final artifact. That folder simply does not exist when the application is run (assuming you are trying to run it standalone).
During the build process target/ is created and contains the classes, resources, test-resources and the likes (assuming you are building with Maven; it is a little different if you build using Gradle or Ant or by hand).
What you can do is create a folder e.g. docs next to your final artifact, give it the appropriate permissions (chmod/chown) and have your application output files into that folder. This folder is then expected to exist on the target machine running your artifact, too, so if it doesn't, it would mean the folder does not exist or the application lacks the proper permissions to read from / write to that folder.
If you need more details, don't hesitate to ask.
Update:
To access a resource, which is bundled and hence inside your artifact (e.g. final.jar), you should be able to retrieve it by using e.g. the following:
testText = new String(ControllerClass.class.getResourceAsStream("/test.txt").readAllBytes());
This is assuming your test.txt file is right under src/main/resources and was bundled to be directly in the root of your JAR-file (or target folder where your application is run from). ControllerClass is the controller, which is accessing the file. readAllBytes just does exactly this: read all the bytes from a text file. For accessing images inside your artifact, you might want to use ImageIO.
IF you however want to access an external file, which is not bundled and hence not inside your artifact, you may use File image = new File(...) where ... would be something like "docs/image.png". This would require you to create a folder called docs next to your JAR-artifact and put a file image.png inside of it.
You of course also may work with streams and there are various helpful libraries for working with input- and output streams.
The following was meant for AWT, but it works in case you really want to access the bytes of your image: ImageIO. In a controller you usually wouldn't want to do that, but rather have your users access (and thus download) it from a given available folder.
I hope this helps :).
I need to get this file from the resources folder in the File object, not in InputSream.
I am using below code, working file on eclipse but FoleNotFoundException on the server. )Using AWS EC2)
Code:
URL res = ResidentHelperService.class.getClassLoader().getResource("key.pem");
System.out.println("resource path2 :" + res);
File privateKeyFile = Paths.get(res.toURI()).toFile();
After printing path looks like:
:jar:file:/home/centos/myproject/microservices/user-service/target/user-service-0.0.1-SNAPSHOT.jar!/BOOT-INF/lib/project-common-utility-0.0.1-SNAPSHOT.jar!/key.pem
I have added dependency on the common jar to user service pom.
Please help me to get the file from resources of a common project.
If you have your file in resources folder, the easiest way to access it from the code is probably to use org.springframework.util.ResourceUtils class that Spring provides:
try {
final File file = ResourceUtils.getFile("classpath:key.pem");
....
} catch (FileNotFoundException e) {
e.printStackTrace();
}
Perhaps this way can help you with your issue.
I need to catch some directory within the application. For that I have a small demonstration:
String pkgName = TestClass.class.getPackage().getName();
String relPath = pkgName.replace(".", "/");
URL resource = ClassLoader.getSystemClassLoader().getResource(relPath);
File file = new File(resource.getPath());
System.out.println("Dir exists:" + file.exists());
While running application from IDE I receive my goal and I can find my directory. But running application as JAR file, does not return a valid "file" (from Javas perspective) and my sout gives me back File exists:false. Is there some way to get this file? In this case, the file is a directory.
Java ClassPath is an abstraction that differs from a filesystem abstraction.
A classpath element may exist in two physical ways:
exploded with classpath pointing to the root directory
packed in a JAR archive
Unfortunatelly, the file.getPath does return a File object if classpath is pointing to file system but it does not if you refer to a JAR file.
In 99% of all cases you should read the contents of a resource using InputStream.
Here is a snippet, that uses IOUtils from apache commons-io to load the whole file content into a String.
public static String readResource(final String classpathResource) {
try {
final InputStream is = TestClass.class.getResourceAsStream(classpathResource);
// TODO verify is != null
final String content = IOUtils.toString(
is, StandardCharsets.UTF_8);
return content;
} catch (final IOException e) {
throw new UncheckedIOException(e);
}
}
I have create folder (i.e uploads ) in web application. I want to create one more folder inside "uploads" folder at runtime depends one the username of user. for this i have write below code. This code is creating folder and file but the location is different that i expected.
the location that i am getting is in eclipse location not web application location
D:\PAST\RequiredPlugins\JUNO\eclipse\uploads\datto\adhar.PNG
then i am getting error in FileOutStream that "system can't find the location specified."
public String getFolderName(String folderName, MultipartFile uploadPhoto)
throws ShareMeException {
File uploadfFile = null;
try {
File file = new File("uploads\\" + folderName);
if (!file.exists()) {
file.mkdir();
}
uploadfFile = new File(file.getAbsoluteFile()
+ "\\"+uploadPhoto.getOriginalFilename());
if (uploadfFile.exists()) {
throw new ShareMeException(
"file already exist please rename it");
} else {
uploadfFile.createNewFile();
FileOutputStream fout = new FileOutputStream(uploadfFile);
fout.write(uploadPhoto.getBytes());
fout.flush();
fout.close();
}
} catch (IOException e) {
throw new ShareMeException(e.getMessage());
}
return uploadfFile.getAbsolutePath();
}
i want to save uploaded file in web app "uploads" folder
Your filename is not absolute: uploads\folderName is resolved against the current directory, which the Eclipse launcher sets to JUNO\eclipse.
You should introduce an application variable like APP_HOME and resolve any data directory (including upload) against this variable.
Also, I suggest not to name anything (neither files nor directories) on your filesystem after user-entered input: you are asking for troubles (unicode characters in the user name) and especially security holes (even in combination with the unicode thing). If you really want to use the filesystem, keep the filename anonymous (1.data, 2.data, ...) and keep metadata inside some database.
You can do something on below lines in your webapp:-
String folderPath= request.getServletContext().getRealPath("/");
File file = new File (folderPath+"upload");
file.mkdir();
I have a simple maven project that reads properties from src/main/resources, and runs perfectly fine in eclipse, but when I export this application as a runnable jar although it tells me Exporting resources/test.properties while building the jar, it breaks unless I include the test.properties file in the location I am invoking my shell script from. Why does this happen? How does it work fine in eclipse? When I look into the jar file contents, it is exactly in the same folder - resources, but command line is just not working. Any advise?
Here is the code that reads the properties from the file -
public class Test {
private static Properties properties = new Properties();
static{
InputStream is = null;
try{
is = Test.class.getClassLoader().getResourceAsStream("test.properties");
if(is!=null)
properties.load(is);
}catch(Exception e){
throw new TestException("Unable to load the properties file :" + e.getMessage(), e);
}finally {
IOUtils.closeQuietly(is);
}
}
public static String getProperty(String key){
return properties.getProperty(key);
}
public static String getProperty(String key, String defaultValue){
return properties.getProperty(key, defaultValue);
}
To get any property, I invoke the getProperty method.
Oh, right. I think Eclipse builds the .jar incorrectly. In maven all files in src/main/resources/* are packaged into /*, but you are saying that your jar has the file as /resources/test.properties, and the getResourceAsStream("test.properties") expects the file in the root folder, not inside /resources. This is wrong. Use maven to build .jar.