I would like to determine real file extension for security reason.
How can I do that?
Supposing you really mean to get the true content type of a file (ie it's MIME type) you should refer to this excellent answer.
You can get the true content type of a file in Java using the following code:
File file = new File("filename.asgdsag");
InputStream is = new BufferedInputStream(new FileInputStream(file));
String mimeType = URLConnection.guessContentTypeFromStream(is);
There are a number of ways that you can do this, some more complicated (and more reliable) than others. The page I linked to discusses quite a few of these approaches.
Not sure exactly what you mean, but however you do this it is only going to work for the specific set of file formats which are known to you
you could exclude executables (are you talking windows here?) - there's some file header information here http://support.microsoft.com/kb/65122 - you could scan and block files which look like they have an exe header - is this getting close to what you mean when you say 'real file extension'?
Related
following the question I asked before How to have my java project to use some files without using their absolute path? I found the solution but another problem popped up in creating text files that I want to write into.here's my code:
private String pathProvider() throws Exception {
//finding the location where the jar file has been located
String jarPath=URLDecoder.decode(getClass().getProtectionDomain().getCodeSource().getLocation().getPath(), "UTF-8");
//creating the full and final path
String completePath=jarPath.substring(0,jarPath.lastIndexOf("/"))+File.separator+"Records.txt";
return completePath;
}
public void writeRecord() {
try(Formatter writer=new Formatter(new FileWriter(new File(pathProvider()),true))) {
writer.format("%s %s %s %s %s %s %s %s %n", whichIsChecked(),nameInput.getText(),lastNameInput.getText()
,idInput.getText(),fieldOfStudyInput.getText(),date.getSelectedItem().toString()
,month.getSelectedItem().toString(),year.getSelectedItem().toString());
successful();
} catch (Exception e) {
failure();
}
}
this works and creates the text file wherever the jar file is running from but my problem is that when the information is been written to the file, the numbers,symbols, and English characters are remained but other characters which are in Persian are turned into question marks. like: ????? 111 ????? ????.although running the app in eclipse doesn't make this problem,running the jar does.
Note:I found the code ,inside pathProvider method, in some person's question.
Your pasted code and the linked question are complete red herrings - they have nothing whatsoever to do with the error you ran into. Also, that protection domain stuff is a hack and you've been told before not to write data files next to your jar files, it's not how OSes (are supposed to) work. Use user.home for this.
There is nothing in this method that explains the question marks - the string, as returned, has plenty of issues (see above), but NOT that it will result in question marks in the output.
Files are fundamentally bytes. Strings are fundamentally characters. Therefore, when you write code that writes a string to a file, some code somewhere is converting chars to bytes.
Make sure the place where that happens includes a charset encoding.
Use the new API (I think you've also been told to do this, by me, in an earlier question of yours) which defaults to UTF-8. Alternatively, specify UTF-8 when you write. Note that the usage of UTF-8 here is about the file name, not the contents of it (as in, if you put persian symbols in the file name, it's not about persian symbols in the contents of the file / in the contents you want to write).
Because you didn't paste the code, I can't give you specific details as there are hundreds of ways to do this, and I do not know which one you used.
To write to a file given a String representing its path:
Path p = Paths.get(completePath);
Files.write("Hello, World!", p);
is all you need. This will write as UTF_8, which can handle persian symbols (because the Files API defaults to UTF-8 if you specify no encoding, unlike e.g. new File, FileOutputStream, FileWriter, etc).
If you're using outdated APIs: new BufferedWriter(new OutputStreamWriter(new FileOutputStream(thePath), StandardCharsets.UTF-8) - but note that this is a resource leak bug unless you add the appropriate try-with-resources.
If you're using FileWriter: FileWriter is broken, never use this class. Use something else.
If you're converting the string on its own, it's str.getBytes(StandardCharsets.UTF_8), not str.getBytes().
I am using getClassLoader().getResources to find the path for Jsoup to parse.
String path = JsoupDemo1.class.getClassLoader().getResource("student.xml").getPath();
Document document = Jsoup.parse(new File(path), "utf-8");
Elements names = document.getElementsByTag("name");
System.out.println(names.size());
My student.xml has been placed under the src folder in my module "day11_xml" and this code snippet comes from the class JsoupDemo1 in the package cn.itcast.xml.jsoup under the same module of "day11_xml". The error messages reads as follows:
java.io.FileNotFoundException:/Users/dingshun/Downloads/New%20Java%20Projects/demo/out/production/day11_xml/student.xml (No such file or directory)
I don't get it, as I can find the exact file in the given path. I'm confused, but could you guys help me out? Also, I'm new to both Java programming and this forum and if this question sounds silly or my question format is not right, please let me know.
What you're doing looks good. Maybe use the stream version JSoup.parse.
URL url = JsoupDemo1.class.getClassLoader().getResource("student.xml");
InputStream stream = JsoupDemo1.class.getClassLoader().getResourceAsStream("student.xml");
document = Jsoup.parse(stream, "utf-8", url.toURI()toString());
The documentation linked seems to imply it will work with html not xml, so maybe you need to use the other argument which provides a parser?
Actually, it turned out that Jsoup could not find my file because the path name "New%20Java%20Projects" has spaces between them. When I reload the file in a folder which has no spaces in its name, it works out just fine. So it can parse xml using parse(File in, String charsetName) method. It seems it cannot parse path name which has spaces in it.
I want to get mimetype of a file can anyone please help me
I want MIME Type like this...
File file=new File("example.jpeg");
String MimeTypeOfFile=/*files mimetype*/;
Thank You in Advance
You can use the Apache Tika Library: It detects and extracts metadata and text from over a thousand different file types
http://tika.apache.org/0.7/detection.html
It has various methods like extension checking or reading file data to detect mime-type. It would be easy and efficient rather than writing yourself.
Example :
System.out.println(new Tika().detect(new File(PATH_TO_FILE)));
I recently found out about java.util.Properties, which allows me to write and read from a config without writing my own function for it.
I was excited since it is so easy to use, but later noticed a flaw when I stored the modified config file.
Here is my code, quite simple for now:
FileWriter writer = null;
Properties configFile = new Properties();
configFile.load(ReadFileTest.class.getClassLoader().getResourceAsStream("config.txt"));
String screenwidth = configFile.getProperty("screenwidth");
String screenheight = configFile.getProperty("screenheight");
System.out.println(screenwidth);
System.out.println(screenheight);
configFile.setProperty("screenwidth", "1024");
configFile.setProperty("screenheight", "600");
try {
writer = new FileWriter("config.txt" );
configFile.store(writer, null);
} catch (IOException e) {
e.printStackTrace();
}
writer.flush();
writer.close();
The problem I noticed was that the config file I try to edit is stored like this:
foo: bar
bar: foo
foobar: barfoo
However, the output after properties.store(writer, null) is this:
foo=bar
bar=foo
foobar=barfoo
The config file I edit is not for my program, it is for an other application that needs the config file to be in the format shown above with : as divider or else it will reset the configuration to default.
Does anybody know how to easily change this?
I searched through the first 5 Google pages now but found noone with a similar problem.
I also checked the Javadoc and found no function that allows me to change it without writing a class for myself.
I would like to use Properties for now since it is there and quite easy to use.
I also got the idea of just replacing all = with : after I saved the file but maybe someone got a better suggestion?
Don't use a tool that isn't designed for the task - don't use Properties here. Instead, I'd just write your own - should be easy enough.
You can still use a Properties instance as your "store", but don't use it for serializing the properties to text. Instead, just use a FileWriter, iterate through the properties, and write the lines yourself - as key + ": " + value.
New idea here
Your comment about converting the = to : got me thinking: Properties.store() writes to a Stream. You could use an in-memory ByteArrayOutputStream, convert as appropriate in memory before you write to a file, then write the file. Likewise for Properties.load(). Or you could insert FilterXXXs instead. (I'd probably do it in memory).
I was looking into how hard it would be to subclass. It's nearly impossible. :-(
If you look at the source code for Properties, (I'm looking at Java 6) store() calls store0(). Now, unfortunately, store0 is private, not protected, and the "=" is given as a magic constant, not something read from a property. And it calls another private method called saveConvert() that also has a lot of magic constants.
Overall, I rate this code as D- quality. It breaks almost all the rules of good code and good style.
But, it's open source, so, theoretically, you could copy and paste (and improve!) a bunch of code into your own BetterProperties class.
I'm doing a file upload, and I want to get the Mime type from the uploaded file.
I was trying to use the request.getContentType(), but when I call:
String contentType = req.getContentType();
It will return:
multipart/form-data; boundary=---------------------------310662768914663
How can I get the correct value?
Thanks in advance
It sounds like as if you're homegrowing a multipart/form-data parser. I wouldn't recommend to do that. Rather use a decent one like Apache Commons FileUpload. For uploaded files, it offers a FileItem#getContentType() to extract the client-specified content type, if any.
String contentType = item.getContentType();
If it returns null (just because the client didn't specify it), then you can take benefit of ServletContext#getMimeType() based on the file name.
String filename = FilenameUtils.getName(item.getName());
String contentType = getServletContext().getMimeType(filename);
This will be resolved based on <mime-mapping> entries in servletcontainer's default web.xml (in case of for example Tomcat, it's present in /conf/web.xml) and also on the web.xml of your webapp, if any, which can expand/override the servletcontainer's default mappings.
You however need to keep in mind that the value of the multipart content type is fully controlled by the client and also that the client-provided file extension does not necessarily need to represent the actual file content. For instance, the client could just edit the file extension. Be careful when using this information in business logic.
Related:
How to upload files in JSP/Servlet?
How to check whether an uploaded file is an image?
just use:
public String ServletContext.getMimeType(String file)
You could use MimetypesFileTypeMap
String contentType = new MimetypesFileTypeMap().getContentType(fileName)); // gets mime type
However, you would encounter the overhead of editing the mime.types file, if the file type is not already listed. (Sorry, I take that back, as you could add instances to the map programmatically and that would be the first place that it checks)