In my Spring Camel app, I try to move or delete a file base on the destinationFolder property. If destinationFolder=null, I want the file to be deleted. If destinationFolder!=null, I want the file to be moved to destinationFolder.
String destinationFolder;
//In the Camel routeBuilder:
from("file://C:/folder1?move=" + destinationFolder)
What will happen in destinationFolder is null? Does the file get move to default location?
When I set destinationFolder=null, I see the file is deleted in folder1.
If you set the move option then the file component will move the file, you cannot set it to null and then have it automatic delete the file. By default the file is moved to a folder named .camel.
So either set delete=true or set move to some folder name to move the files.
First, you should know when to use "move", "delete" &"noop" and how it will works in Apache camel
Note :1) If your destination path is not existed then file will delete automatically.
Note :2) If you are not used "noop=true" in Camel URL then file file will delete(if your destination path is null)
Reference : - enter link description here
Basic Test Code:
import org.apache.camel.builder.RouteBuilder;
import org.apache.camel.impl.DefaultCamelContext;
public class SFTPTest {
public static void main(String[] args) throws Exception {
DefaultCamelContext ctx = null;
try{
ctx = new DefaultCamelContext();
ctx.addRoutes(new RouteBuilder() {
#Override
public void configure() throws Exception {
String filepath = "file:///camelexample/?fileName=test.txt&move=null";
from(filepath)
.log("File processed");
}
});
ctx.start();
Thread.sleep(5000);
ctx.stop();
}catch (Exception e){
System.err.println("Exception is : "+e.getLocalizedMessage());
}finally {
try{
ctx.stop();
}catch (Exception e){
System.err.println("Exception is : "+e.getLocalizedMessage());
}
}
}
}
Related
I would like to generate a blog posts overview. For that I want to read the html files from a folder inside the templates folder in the resources folder where Spring Boot stores its templates.
I tried that but it doesnt return an error but also list no files.
What is the way to go here?
Thanks
#Controller
public class Route {
#Autowired
private ResourceLoader resourceLoader;
#RequestMapping("/")
public String home() throws IOException {
final String path = "templates/blog";
final Resource res = resourceLoader.getResource("templates/blog");
try (final BufferedReader reader = new BufferedReader(new InputStreamReader(res.getInputStream()))) {
reader.lines().forEachOrdered(System.out::println);
}
return "blog/a";
}
}
#Controller
public class Route {
#Value("classpath:templates/blog/*")
private Resource[] resources;
#RequestMapping("/")
public String home() throws IOException {
for (final Resource res : resources) {
System.out.println(res.getFilename());
}
return "blog/a";
}
}
did the trick to me.
You should be able to achieve this using NIO2.
In order for NIO2 to work, it requires the concept of FileSystem, and one can be created from the jar URI. Then this file system can be used with Files/Paths.
The code below contains two branches - the first handles loading the files from inside Jar, the second branch - when the code runs from IDE or via "mvn spring-boot:run".
All streams are being used via try-with-resources so they will be auto-closed.
The find function starts from the top of the file system and recursively searches for html files.
public static void readFile(String location) throws URISyntaxException {
URI uri = Objects.requireNonNull(ReadFromJar.class.getClassLoader().getResource(location)).toURI();
if (uri.getScheme().equals("jar")) { //inside jar
try (FileSystem fs = FileSystems.newFileSystem(uri, Collections.emptyMap())) { //build a new FS that represents the jar's contents
Files.find(fs.getPath("/"), 10, (path, fileAttr) -> // control the search depth (e.g. 10)
fileAttr.isRegularFile() //match only files
&& path.toString().contains("blog") //match only files in paths containing "blog"
&& path.getFileName().toString().matches(".*\\.html")) // match only html files
.forEach(ReadFromJar::printFileContent);
} catch (IOException ex) {
ex.printStackTrace();
}
}
else { //from IDE or spring-boot:run
final Path path = Paths.get(uri);
try (DirectoryStream<Path> dirStream = Files.newDirectoryStream(path)) {
dirStream.forEach(ReadFromJar::printFileContent);
} catch (IOException e) {
e.printStackTrace();
}
}
}
private static void printFileContent(final Path file) {
try {
System.out.println("Full path: " + file.toAbsolutePath().toString());
Files.lines(file).forEach(System.out::println);
} catch (IOException e) {
e.printStackTrace();
}
}
I'm trying to create a Java learning Android application.
It should have a code playground for users; which can run simple Java code.
I save users code in a java code and I'm trying to run it.
I'm using JANINO, but the problem is that I can not load external class including users code with it.
Here is my code:
String UsersCodeInput;
public void Run() throws FileNotFoundException, IOException
{
File sourcepath = new File(Environment.getExternalStorageDirectory()+"/MainFolder");
File JavaFile = new File(sourcepath.getPath()+"A.java");
FileOutputStream fos = new FileOutputStream(JavaFile);
fos.write(UsersCodeInput.getBytes());
fos.close();
// now Users Code is Saved.
//trying to run it by means of janino :
ClassLoader cl = new JavaSourceClassLoader(
this.getClass().getClassLoader(), // parentClassLoader
new File[] {sourcepath}, //our sourceFolder
"UTF-8" //Encodeing
);
try
{
// problem is here. next line is not working well; throws classNotFound.
Object o = cl.loadClass(".A").newInstance();
//the class implements runnable.
((Runnable) o).run();
}
catch (ClassNotFoundException e)
{Toast.makeText(this,e.getCause().toString(),Toast.LENGTH_SHORT).show();}
catch (InstantiationException e)
{Toast.makeText(this,e.getCause().toString(),Toast.LENGTH_SHORT).show();}
catch (IllegalAccessException e)
{Toast.makeText(this,e.getCause().toString(),Toast.LENGTH_SHORT).show();}
}
And here is JANINOs tutorial ; Read "source code compiler" part
http://janino-compiler.github.io/janino/
Thanks.
I want to create file path in controller
Already created file path and it's working
try {
Files.write(Paths.get("D:\\app\\app\\java.ini"), data, StandardCharsets.UTF_8, StandardOpenOption.CREATE);
} catch (IOException e) {
e.printStackTrace();
}
Now i want to change this D:\\app\\app\\java.ini and i want to create like resources/java.ini
I don't to give any system full path.
Thanx
The sample code below shows the example that you need:
public class MyClass
{
public static void main(String[] args) {
String test = MyClass.class.getProtectionDomain()
.getCodeSource().getLocation().getPath();
System.out.println(test);
}
}
Assuming that the method you are running is called MyClass, you can have this snippet inside your Java method:
try {
String location = MyClass.class.getProtectionDomain()
.getCodeSource().getLocation().getPath() + "resources/java.ini");
Files.write(location, data, StandardCharsets.UTF_8, StandardOpenOption.CREATE);
} catch (IOException e) {
e.printStackTrace();
}
I am facing error MultipartFile error for one day after having upgrade from Spring Boot 1.2.7 to 1.3.1.
What I notice is that the default is now Jetty 9.2 and no more Tomcat 8. Everything was fine until I tried to write an uploaded file using MultipartFile.transferTo(File file) method..
MultipartFile.transferTo() method is calling an implementation of javax.servlet.http.Part Which is implemented this way for tomcat 8
#Override
public void write(String fileName) throws IOException {
File file = new File(fileName);
if (!file.isAbsolute()) {
file = new File(location, fileName);
}
try {
fileItem.write(file);
} catch (Exception e) {
throw new IOException(e);
}
}
and this way for jetty 9.2
public void write(String fileName) throws IOException
{
if (_file == null)
{
_temporary = false;
//part data is only in the ByteArrayOutputStream and never been written to disk
_file = new File (_tmpDir, fileName);
BufferedOutputStream bos = null;
try
{
bos = new BufferedOutputStream(new FileOutputStream(_file));
_bout.writeTo(bos);
bos.flush();
}
finally
{
if (bos != null)
bos.close();
_bout = null;
}
}
else
{
//the part data is already written to a temporary file, just rename it
_temporary = false;
File f = new File(_tmpDir, fileName);
if (_file.renameTo(f))
_file = f;
}
}
What's wrong with the Jetty implementation is that is waiting for file name File.getName() and not the absolute path name File.getPath() which is provided by the call of StandardMultipartHttpServletRequest.transferTo(File file)
#Override
public void transferTo(File dest) throws IOException, IllegalStateException {
this.part.write(dest.getPath());
}
Is this a bug ? Note that this occurs since I have upgraded from spring boot 1.2.7 to 1.3.1. The default was Tomcat and now it is Jetty...
Per the javadoc for javax.servlet.http.Part.write(String filename) the filename parameter is ...
The file is created relative to the location as specified in the
MultipartConfig
In the code you referenced in Jetty 9.2, namely this ...
jetty-9.2.14.v20151106 - MultiPartInputStreamParser.write(String fileName)
You'll see that there's 2 possible code paths it takes, the first is the "in memory" path, and the second is "file on disk" approach.
In both cases, when you specify a filename to Part.write(String) that name is relative to your MultiPartConfig.location (a configuration of which you haven't detailed in your question).
The implementation of MultiPartInputStreamParser has a _tmpDir which is configured from the webapp's MultiPartConfig.location.
If you want this to behave properly, would highly recommend you define a MultiPartConfig.location that is appropriate for your application, instead of relying on the container to pick one.
The Tomcat approach of allowing absolute filenames in Part.write(String) is actually not allowed in the servlet spec (mainly as its a security issue that can be used to cause havoc on a system)
Ok, at the moment if you want to get rid of this error you can switch back to Tomcat instead of Jetty.
Put tomcat into your dependencies:
compile('org.springframework.boot:spring-boot-starter-tomcat')
And declare tomcat as container:
#Bean
public TomcatEmbeddedServletContainerFactory tomcatEmbeddedServletContainerFactory() {
return new TomcatEmbeddedServletContainerFactory();
}
Is there a way using ResourceLoader to get a list of "sub resources" in a directory in the jar?
For example, given sources
src/main/resources/mydir/myfile1.txt
src/main/resources/mydir/myfile2.txt
and using
#Autowired
private ResourceLoader resourceLoader;
I can get to the directory
Resource dir = resourceLoader.getResource("classpath:mydir")
dir.exists() // true
but not the files within the dir. If I could get the file, I could call dir.getFile().listFiles(), but
dir.getFile() // explodes with FileNotFoundException
But I can't find a way to get the "child" resources.
You can use a ResourcePatternResolver to get all the resources that match a particular pattern. For example:
Resource[] resources = resourcePatternResolver.getResources("/mydir/*.txt")
You can have a ResourcePatternResolver injected in the same way as ResourceLoader.
Based on Bohemian's comment and another answer, I used the following to get an input streams of all YAMLs under a directory and sub-directories in resources (Note that the path passed doesn't begin with /):
private static Stream<InputStream> getInputStreamsFromClasspath(
String path,
PathMatchingResourcePatternResolver resolver
) {
try {
return Arrays.stream(resolver.getResources("/" + path + "/**/*.yaml"))
.filter(Resource::exists)
.map(resource -> {
try {
return resource.getInputStream();
} catch (IOException e) {
return null;
}
})
.filter(Objects::nonNull);
} catch (IOException e) {
logger.error("Failed to get definitions from directory {}", path, e);
return Stream.of();
}
}