I followed this tutorial for uploading a file in my JSF2 application.
The application works fine but I am unhappy with one aspect.
While rebuilding the request, the File sent via request is saved somewhere on the disk.
Even though the file is saved I need to rename the file with a name which is available after entering the Managed Bean containing the action method.
Therefore I decided to create a new file with de desired name, copy the already saved file, and then delete the unneeded one.
private File uploadFile;
//...
try {
BufferedWriter bw = new BufferedWriter(new FileWriter(newFile));
BufferedReader br = new BufferedReader(new FileReader(uploadFile));
String line = "";
while ((line = br.readLine()) != null){
bw.write(line);
}
} catch (Exception e){}
The new file appears in the desired location but this error is thrown when I'm trying to open the file: "Invalid or unsupported PNG file"
These are my questions:
Is there a better way to solve this problem?
Is this solution the best way to upload a picture? Is there a reason to save the file before the business logic when there may be need to resize the picture or the desired name is not available yet.
LE:
I know abot this tutorial as well but I'm trying to do this mojarra only.
There is a rename method built into java.io.File object already, I'd be surprised if it didn't work for your situation.
public boolean renameTo(File dest)
Renames the file denoted by this abstract pathname.
Many aspects of the behavior of this method are inherently platform-dependent:
The rename operation might not be able to move a file from one filesystem to
another, it might not be atomic, and it might not succeed if a file with the
destination abstract pathname already exists. The return value should always
be checked to make sure that the rename operation was successful.
You can also check if a file exists before saving it, and you can use the ImageIO class to do validations on the uploaded file before performing the initial save.
Don't use Reader and Writer when you deal with binary files like images. Use streams: FileInputStream and FileOutputStream. And the best variant is to use #Perception solution with renameTo method.
Readers read file as if it consists of characters (e.g. txt, properties, yaml files). Image files are not characters, they are binary and you must use streams for that.
Related
I am working on a Springboot application that has to return a zip file to a frontend when the user downloads some report. I want to create a zip file without writing the zip file or the original files to disk.
The directory I want to zip contains other directories, that contain the actual files. For example, dir1 has subDir1 and subDir2 inside, subDir1 will have two file subDir1File1.pdf and subDir1File2.pdf. subDir2 will also have files inside.
I can do this easily by creating the physical files on the disk. However, I feel it will be more elegant to return these files without writing to disk.
You would use ByteArrayOutputStream if the scope was to write to memory. In essence, the zip file would be entirely contained in memory, so be sure that you don't risk to have too many requests at once and that the file size is reasonable in size! Otherwise this approach can seriously backfire!
You can use following snippet :
public static byte[] zip(final String str) throws IOException {
if (StringUtils.isEmpty(str)) {
throw new IllegalArgumentException("Cannot zip null or empty string");
}
ByteArrayOutputStream bos = new ByteArrayOutputStream();
try (GZIPOutputStream gos = new GZIPOutputStream(bos)) {
gos.write(str.getBytes(StandardCharsets.UTF_8));
}
return bos.toByteArray();
}
But as stated in another answer, make sure you are not risking your program too much by loading everything into your java memory.
Please note that you should stream whenever possible. In your case, you could write your data to https://docs.oracle.com/javase/8/docs/api/index.html?java/util/zip/ZipOutputStream.html.
The only downside of this appproach is: the client won't be able to show a download status bar, because the server will not be able to send the "Content-length" header. That's because the size of a ZIP file can only be known after it has been generated, but the server needs to send the headers first. So - no temporary zip file - no file size beforehand.
You are also talking about subdirectories. This is just a naming issue when dealing with a ZIP stream. Each zip item needs to be named like this: "directory/directory2/file.txt". This will produce subdirectories when unzipping.
I am trying to open a file for reading or create the file if it was not there.
I use this code:
String location = "/test1/test2/test3/";
new File(location).mkdirs();
location += "fileName.properties";
Path confDir = Paths.get(location);
InputStream in = Files.newInputStream(confDir, StandardOpenOption.CREATE);
in.close();
And I get java.nio.file.NoSuchFileException
Considering that I am using StandardOpenOption.CREATE option, the file should be created if it is not there.
Any idea why I am getting this exception?
It seems that you want one of two quite separate things to happen:
If the file exists, read it; or
If the file does not exist, create it.
The two things are mutually exclusive but you seem to have confusingly merged them. If the file did not exist and you've just created it, there's no point in reading it. So keep the two things separate:
Path confDir = Paths.get("/test1/test2/test3");
Files.createDirectories(confDir);
Path confFile = confDir.resolve("filename.properties");
if (Files.exists(confFile))
try (InputStream in = Files.newInputStream(confFile)) {
// Use the InputStream...
}
else
Files.createFile(confFile);
Notice also that it's better to use "try-with-resources" instead of manually closing the InputStream.
Accordingly to the JavaDocs you should have used newOutputStream() method instead, and then you will create the file:
OutputStream out = Files.newOutputStream(confDir, StandardOpenOption.CREATE);
out.close();
JavaDocs:
// Opens a file, returning an input stream to read from the file.
static InputStream newInputStream(Path path, OpenOption... options)
// Opens or creates a file, returning an output stream that
// may be used to write bytes to the file.
static OutputStream newOutputStream(Path path, OpenOption... options)
The explanation is that OpenOption constants usage relies on wether you are going to use it within a write(output) stream or a read(input) stream. This explains why OpenOption.CREATE only works deliberatery with the OutputStream but not with InputStream.
NOTE: I agree with #EJP, you should take a look to Oracle's tutorials to create files properly.
I think you intended to create an OutputStream (for writing to) instead of an InputStream (which is for reading)
Another handy way of creating an empty file is using apache-commons FileUtils like this
FileUtils.touch(new File("/test1/test2/test3/fileName.properties"));
I would like to be able to upload files in my JSF2.2 web application, so I started using the new <h:inputFile> component.
My only question is, how can I specify the location, where the files will be saved in the server? I would like to get hold of them as java.io.File instances. This has to be implemented in the backing bean, but I don't clearly understand how.
JSF won't save the file in any predefined location. It will basically just offer you the uploaded file in flavor of a javax.servlet.http.Part instance which is behind the scenes temporarily stored in server's memory and/or temporary disk storage location which you shouldn't worry about.
Important is that you need to read the Part as soon as possible when the bean action (listener) method is invoked. The temporary storage may be cleared out when the HTTP response associated with the HTTP request is completed. In other words, the uploaded file won't necessarily be available in a subsequent request.
So, given a
<h:form enctype="multipart/form-data">
<h:inputFile value="#{bean.uploadedFile}">
<f:ajax listener="#{bean.upload}" />
</h:inputFile>
</h:form>
You have basically 2 options to save it:
1. Read all raw file contents into a byte[]
You can use InputStream#readAllBytes() for this.
private Part uploadedFile; // +getter+setter
private String fileName;
private byte[] fileContents;
public void upload() {
fileName = Paths.get(uploadedFile.getSubmittedFileName()).getFileName().toString(); // MSIE fix.
try (InputStream input = uploadedFile.getInputStream()) {
fileContents = input.readAllBytes();
}
catch (IOException e) {
// Show faces message?
}
}
Note the Path#getFileName(). This is a MSIE fix as to obtaining the submitted file name. This browser incorrectly sends the full file path along the name instead of only the file name.
In case you're not on Java 9 yet and therefore can't use InputStream#readAllBytes(), then head to Convert InputStream to byte array in Java for all other ways to convert InputStream to byte[].
Keep in mind that each byte of an uploaded file costs one byte of server memory. Be careful that your server don't exhaust of memory when users do this too often or can easily abuse your system in this way. If you want to avoid this, better use (temporary) files on local disk file system instead.
2. Or, write it to local disk file system
In order to save it to the desired location, you need to get the content by Part#getInputStream() and then copy it to the Path representing the location.
private Part uploadedFile; // +getter+setter
private File savedFile;
public void upload() {
String fileName = Paths.get(uploadedFile.getSubmittedFileName()).getFileName().toString(); // MSIE fix.
savedFile = new File(uploads, fileName);
try (InputStream input = file.getInputStream()) {
Files.copy(input, savedFile.toPath());
}
catch (IOException e) {
// Show faces message?
}
}
Note the Path#getFileName(). This is a MSIE fix as to obtaining the submitted file name. This browser incorrectly sends the full file path along the name instead of only the file name.
The uploads folder and the filename is fully under your control. E.g. "/path/to/uploads" and Part#getSubmittedFileName() respectively. Keep in mind that any existing file would be overwritten, you might want to use File#createTempFile() to autogenerate a filename. You can find an elaborate example in this answer.
Do not use Part#write() as some prople may suggest. It will basically rename the file in the temporary storage location as identified by #MultipartConfig(location). Also do not use ExternalContext#getRealPath() in order to save the uploaded file in deploy folder. The file will get lost when the WAR is redeployed for the simple reason that the file is not contained in the original WAR. Always save it on an absolute path outside the deploy folder.
For a live demo of upload-and-preview feature, check the demo section of the <o:inputFile> page on OmniFaces showcase.
See also:
Write file into disk using JSF 2.2 inputFile
How to save uploaded file in JSF
Recommended way to save uploaded files in a servlet application
My Question: How do I open a file (in the system default [external] program for the file) without saving the file to disk?
My Situation: I have files in my resources and I want to display those without saving them to disk first. For example, I have an xml file and I want to open it on the user's machine in the default program for reading xml file without saving it to the disk first.
What I have been doing: So far I have just saved the file to a temporary location, but I have no way of knowing when they no longer need the file so I don't know when/if to delete it. Here's my SSCCE code for that (well, it's mostly sscce, except for the resource... You'll have to create that on your own):
package main;
import java.io.*;
public class SOQuestion {
public static void main(String[] args) throws IOException {
new SOQuestion().showTemplate();
}
/** Opens the temporary file */
private void showTemplate() throws IOException {
String tempDir = System.getProperty("java.io.tmpdir") + "\\BONotifier\\";
File parentFile = new File(tempDir);
if (!parentFile.exists()) {
parentFile.mkdirs();
}
File outputFile = new File(parentFile, "template.xml");
InputStream inputStream = getClass().getResourceAsStream("/resources/template.xml");
int size = 4096;
try (OutputStream out = new FileOutputStream(outputFile)) {
byte[] buffer = new byte[size];
int length;
while ((length = inputStream.read(buffer)) > 0) {
out.write(buffer, 0, length);
}
inputStream.close();
}
java.awt.Desktop.getDesktop().open(outputFile);
}
}
Because of this line:
String tempDir = System.getProperty("java.io.tmpdir") + "\\BONotifier\\";
I deduce that you're working on Windows. You can easily make this code multiplatform, you know.
The answer to your question is: no. The Desktop class needs to know where the file is in order to invoke the correct program with a parameter. Note that there is no method in that class accepting an InputStream, which could be a solution.
Anyway, I don't see where the problem is: you create a temporary file, then open it in an editor or whatever. That's fine. In Linux, when the application is exited (normally) all its temporary files are deleted. In Windows, the user will need to trigger the temporary files deletion. However, provided you don't have security constraints, I can't understand where the problem is. After all, temporary files are the operating system's concern.
Depending on how portable your application needs to be, there might be no "one fits all" solution to your problem. However, you can help yourself a bit:
At least under Linux, you can use a pipe (|) to direct the output of one program to the input of another. A simple example for that (using the gedit text editor) might be:
echo "hello world" | gedit
This will (for gedit) open up a new editor window and show the contents "hello world" in a new, unsaved document.
The problem with the above is, that this might not be a platform-independent solution. It will work for Linux and probably OS X, but I don't have a Windows installation here to test it.
Also, you'd need to find out the default editor by yourself. This older question and it's linked article give some ideas on how this might work.
I don't understand your question very well. I can see only two possibilities to your question.
Open an existing file, and you wish to operate on its stream but do not want to save any modifications.
Create a file, so that you could use file i/o to operate on the file stream, but you don't wish to save the stream to file.
In either case, your main motivation is to exploit file i/o existingly available to your discretion and programming pleasure, am I correct?
I have feeling that the question is not that simple and this my answer is probably not the answer you seek. However, if my understanding of the question does coincide with your question ...
If you wish to use Stream io, instead of using FileOutputStream or FileInputStream which are consequent to your opening a File object, why not use non-File InputStream or OutputStream? Your file i/o utilities will finally boil down to manipulating i/o streams anyway.
http://docs.oracle.com/javase/7/docs/api/java/io/OutputStream.html
http://docs.oracle.com/javase/7/docs/api/java/io/InputStream.html
No need to involve temp files.
I want to save a Unique ID (which is a String) which gets created when I launch my Java application. Now I want to save this somewhere (I think in some file on the disk) so that when I relaunch my application I should be able to read it and use that ID.
I want to know what is the good way to saving such ID. I am thinking of creating a Properties file and save it then retrieve it from it when I relaunch application. Is there a better or standard way for this?
EDIT :
Additionally what should be the folder location for storing on the disk. Should it be relative to my execution path or some Logged-in user specific path?
1. If its the same Java application that writes or reads this String, then use Serialization, it will be in non-readable form when saved.
2. If reading and writing is from different program, then use Text file.
3. Using Property file will be also a good approach.
If your app/program needs to store more data at some point sqlite3 might be the best option for you. It is easy to implement and use.
Download sqlite3
EDIT: How many IDs will be stored in the app? If there are just a few, a textfile or property file is enough.
EDIT2: Navigate to your Documents folder on your machine and you will see folders of programs/games. Thats where you should place the file/db. However you can also store it in the installation path on your hard drive. Also make sure your user launches the app trough a shortcut, not the actual execution file
Use the FileWriter and File classes from Java.
It should be something like that:
File f = new File(your path here);
if (f.exists()){
BufferedReader br = new BufferedReader(new FileReader(your path here));
String a = br.readLine();
br.close();
}else{
FileWriter fw = new FileWriter(your path here);
fw.write(your ID String);
fw.flush();
fw.close();
I hope this is want u meant.
Best regards
edit: just noticed too late that your edited your post....