I have this service in spring boot app:
#Service
public class FileStorageService {
private final Path root = Paths.get("uploads");
public void init() {
try {
Files.createDirectories(root);
} catch (IOException e) {
throw new RuntimeException("Could not initialize folder for upload!");
}
}
public ResponseEntity<?> save(MultipartFile[] files) {
if (files.length == 0) {
//return ResponseEntity.badRequest();
}
Arrays.asList(files).stream().forEach(file -> {
try {
// Get the file and save it somewhere
byte[] bytes = file.getBytes();
Path path = Paths.get(root + File.separator + file.getOriginalFilename());
Files.write(path, bytes);
} catch (IOException e) {
e.printStackTrace();
}
});
return new ResponseEntity<>(null, HttpStatus.OK);
}
}
There is some folder for saving files "uploads". And i am now wondering what is
spring.servlet.multipart.location=
setting used for. Is this just folder where raw data from request are stored before they are saved in "uploads" folder?
As mentioned in this document for spring.servlet.multipart.location,
location specifies the directory where uploaded files will be stored.
When not specified, a temporary directory will be used.
I dig around some internal code and found out that, it depends on the spring.servlet.multipart.file-size-threshold configuration. As mentioned in the doc
specifies the size threshold after which files will be written to disk. The default is 0.
If you mention any non-zero value then it will attempt to store that much in memory first before, writing to disk.
For your case, if you mention any location in spring.servlet.multipart.location the file will directly stored over that location only.
You may also want to check spring.servlet.multipart.max-file-size and spring.servlet.multipart.max-request-size property.
Related
I am still searching around this subject, but I cannot find a simple solution, and I don't sure it doesn't exist.
Part 1
I have a service on my application that's generating an excel doc, by the dynamic DB data.
public static void
notiSubscribersToExcel(List<NotificationsSubscriber>
data) {
//generating the file dynamically from DB's data
String prefix = "./src/main/resources/static";
String directoryName = prefix + "/documents/";
String fileName = directoryName + "subscribers_list.xlsx";
File directory = new File(directoryName);
if (! directory.exists()){
directory.mkdir();
// If you require it to make the entire directory path including parents,
// use directory.mkdirs(); here instead.
}
try (OutputStream fileOut = new FileOutputStream(fileName)) {
wb.write(fileOut);
fileOut.close();
wb.close();
} catch (FileNotFoundException e) {
// TODO Auto-generated catch block
e.printStackTrace();
} catch (IOException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
Part 2
I want to access it from the browser, so when I call it will get downloaded.
I know that for the static content, all I need to do is to call to the file, from the browser like that:
http://localhost:8080/documents/myfile.xlsx
After I would be able to do it, all I need is to create link to this url from my client app.
The problem -
Currently if I call to the file as above, it will download only the file which have been there in the compiling stage, but if I am generating a new files after the app is running the content won't be available.
It seems that the content is (as it's called) "static" and cannot be changed after startup.
So my question is
is there is a way to define a folder on the app structure that will be dynamic? I just want to access the new generated file.
BTW I found this answer and others which doing configuration methods, or web services, but I don't want all this. And I have tried some of them, but the result is the same.
FYI I don't bundle my client app with the server app, I run them from different hosts
The problem is to download the file with the dynamic content from a Spring app.
This can be solved with Spring BOOT. Here is the solution as shown in this illustration - when i click Download report, my app generates a dynamic Excel report and its downloaded to the browser:
From a JS, make a get request to a Spring Controller:
function DownloadReport(e){
//Post the values to the controller
window.location="../report" ;
}
Here is the Spring Controller GET Method with /report:
#RequestMapping(value = ["/report"], method = [RequestMethod.GET])
#ResponseBody
fun report(request: HttpServletRequest, response: HttpServletResponse) {
// Call exportExcel to generate an EXCEL doc with data using jxl.Workbook
val excelData = excel.exportExcel(myList)
try {
// Download the report.
val reportName = "ExcelReport.xls"
response.contentType = "application/vnd.ms-excel"
response.setHeader("Content-disposition", "attachment; filename=$reportName")
org.apache.commons.io.IOUtils.copy(excelData, response.outputStream)
response.flushBuffer()
} catch (e: Exception) {
e.printStackTrace()
}
}
This code is implemented in Kotlin - but you can implement it as easily in Java too.
using java 8, tomcat 8
Hi, i am loading a file using properties, but i have a check before loading which returns the same properties object if its already been loaded (not null). which is a normal case scenario but i want to know if there is any way that if any change occur in target file, and some trigger should be called and refreshes all the properties objects. here is my code.
public static String loadConnectionFile(String keyname) {
String message = "";
getMessageFromConnectionFile();
if (propertiesForConnection.containsKey(keyname))
message = propertiesForConnection.getProperty(keyname);
return message;
}
public static synchronized void getMessageFromConnectionFile() {
if (propertiesForConnection == null) {
FileInputStream fileInput = null;
try {
File file = new File(Constants.GET_CONNECTION_FILE_PATH);
fileInput = new FileInputStream(file);
Reader reader = new InputStreamReader(fileInput, "UTF-8");
propertiesForConnection = new Properties();
propertiesForConnection.load(reader);
} catch (Exception e) {
Utilities.printErrorLog(Utilities.convertStackTraceToString(e), logger);
} finally {
try {
fileInput.close();
} catch (Exception e) {
Utilities.printErrorLog(Utilities.convertStackTraceToString(e), logger);
}
}
}
}
the loadConnectionFile method executes first and calls getMessageFromConnectionFile which has check implemented for "null", now if we remove that check it will definitely load updated file every time but it will slower the performance. i want an alternate way.
hope i explained my question.
thanks in advance.
Java has a file watcher service. It is an API. You can "listen" for changes in files and directories. So you can listen for changes to your properties file, or the directory in which your properties file is located. The Java Tutorials on Oracle's OTN Web site has a section on the watcher service.
Good Luck,
Avi.
I've exported my Java console application to a Jar file, but when I run the jar and call code that parses in a JSON file I get a java.lang.IllegalArgumentException
Does anyone know why the exception is being thrown when I run the program as a JAR? The parsing works fine when the application is run from Eclipse.
This is the exact error that is output when I execute the jar file and call the code that parses the JSON file:
Exception in thread "main" java.lang.IllegalArgumentException: URI is not hierar
chical
at java.io.File.<init>(Unknown Source)
at gmit.GameParser.parse(GameParser.java:44)
at gmit.Main.main(Main.java:28)
This is how the parsing is being done in my GameParser class:
public class GameParser {
private static final String GAME_FILE = "/resources/game.json";
private URL sourceURL = getClass().getResource(GAME_FILE);
private int locationId;
private List<Location> locations;
private List<Item> items;
private List<Character> characters;
public void parse() throws IOException, URISyntaxException {
ObjectMapper mapper = new ObjectMapper();
mapper.configure(DeserializationFeature.FAIL_ON_UNKNOWN_PROPERTIES, false);
try {
// read from file, convert it to Location class
Location loc = new Location();
loc = mapper.readValue(new File(sourceURL.toURI()), Location.class);
Item item = mapper.readValue(new File(sourceURL.toURI()), Item.class);
GameCharacter character = mapper.readValue(new File(sourceURL.toURI()), GameCharacter.class);
// display to console
System.out.println(loc.toString());
System.out.println(item.toString());
System.out.println(character.toString());
} catch (JsonGenerationException e) {
e.printStackTrace();
} catch (JsonMappingException e) {
e.printStackTrace();
} catch (IOException e) {
e.printStackTrace();
}
}
}
This is the folder structure of my project:
The call getClass().getResource(GAME_FILE); will return a URL relative to this class. If you are executing your program from a JAR file, it will return a URL pointing to a JAR file.
Files in java can only represent direct filesystem files, not the ones in zip/jar archives.
To fix it:
Try to use getClass().getResourceAsStream() and use that instead of Files or
extract the files into some directory and use File in the same way as you are trying now.
This problem happen when you have two files with the same name,i mean in your project you have folder whith name "Images" and in your desktop you have other folder his name "images" automatically JVM choose desktop folder ,so if you want to confirm try to print your URI.Use this example to show your URI before creating your file
try {
URL location = this.getClass().getResource("/WavFile");
System.out.println(location.toURI());
File file = new File(location.toURI());
if (!file.exists()) {
System.out.println(file.mkdirs());
System.out.println(file.getAbsoluteFile());
}else
{
System.out.println(file.getPath());
}
} catch (Exception e) {
e.printStackTrace();
}
protected void executeInternal(JobExecutionContext context) throws JobExecutionException
{
System.out.println("Sending Birthday Wishes... ");
try
{
for(int i=0;i<maillist.length;i++)
{
Email email = new Email();
email.setFrom("spv_it#yahoo.com");
email.setSubject("Happy IndependenceDay");
email.setTo(maillist[i]);
email.setText("<font color=blue><h4>Dear Users,<br><br><br>Wish you a Happy Independence Day!<br><br><br>Regards,<br>Penna Cement Industries Limited</h4></font>");
byte[] data = null;
ClassPathResource img = new ClassPathResource("newLogo.gif");
InputStream inputStream = img.getInputStream();
data = new byte[inputStream.available()];
while((inputStream.read(data)!=-1));
Attachment attachment = new Attachment(data, "HappyBirthDay","image/gif", true);
email.addAttachment(attachment);
emailService.sendEmail(email);
}
}
catch (MessagingException e)
{
e.printStackTrace();
}
catch (Exception e)
{
e.printStackTrace();
}
}
This is the error I'm getting:
java.io.FileNotFoundException: class path resource [newLogo.gif] cannot be opened because it does not exist
at org.springframework.core.io.ClassPathResource.getInputStream(ClassPathResource.java:135)
at com.mail.schedular.BirthdayWisherJob.executeInternal(BirthdayWisherJob.java:55)
at org.springframework.scheduling.quartz.QuartzJobBean.execute(QuartzJobBean.java:66)
at org.quartz.core.JobRunShell.run(JobRunShell.java:223)
at org.quartz.simpl.SimpleThreadPool$WorkerThread.run(SimpleThreadPool.java:549)
The best practise is to read/write or to provide reference of any file is by mentioning the ABSOLUTE PATH of that file.
To your question, It shows the FileNotFoundException because, JVM failed to locate the file in your current directory which is by default your source path. So provide the absolute path in ClassPathResource or copy that image file to your current directory. It will solve your problem.
I think you need to put your file inside inside the src folder , if it's there then check whether it's under some directory which is inside the src directory.
Then give the correct location like given details below
src[dir]----->newLogo.gif
ClassPathResource img = new ClassPathResource("newLogo.gif");
or,
src[dir]----->images[dir]---->newLogo.gif
ClassPathResource img = new ClassPathResource("/images/newLogo.gif");
You got this error since the job is running in a separate quartz thread, I suggest that you locate your file newLogo.gif outside the jar and use the following to load it.
Thread.currentThread().getContextClassLoader().getResource("classpath:image/newLogo.gif");
I am having a problem writing to a .xml file inside of my jar. When I use the following code inside of my Netbeans IDE, no error occurs and it writes to the file just fine.
public void saveSettings(){
Properties prop = new Properties();
FileOutputStream out;
try {
File file = new File(Duct.class.getResource("/Settings.xml").toURI());
out = new FileOutputStream(file);
prop.setProperty("LAST_FILE", getLastFile());
try {
prop.storeToXML(out,null);
} catch (Exception e) {
JOptionPane.showMessageDialog(null, e.toString());
}
try {
out.close();
} catch (Exception e) {
JOptionPane.showMessageDialog(null, e.toString());
}
} catch (Exception e) {
JOptionPane.showMessageDialog(null, e.toString());
}
}
However, when I execute the jar I get an error saying:
IllegalArguementException: uri is not hierachal
Does anyone have an idea of why it's working when i run it in Netbeans, but not working when i execute the jar. Also does anyone have a solution to the problem?
The default class loader expects the classpath to be static (so it can cache heavily), so this approach will not work.
You can put Settings.xml in the file system if you can get a suitable location to put it. This is most likely vendor and platform specific, but can be done.
Add the location of the Settings.xml to the classpath.
I was also struggling with this exception. But finally found out the solution.
When you use .toURI() it returns some thing like
D:/folderName/folderName/Settings.xml
and hence you get the exception "URI is not hierarchical"
To avoid this call the method getPath() on the URI returned, which returns something like
/D:/folderName/folderName/Settings.xml
which is now hierarchical.
In your case, the 5th line in your code should be
File file = new File(Duct.class.getResource("/Settings.xml").toURI().getPath());