I am implementing a Branch Predictor for one of my classes and I am trying to read files from my src folder in Eclipse but for some reason it is not able to open the files. I have done this before with the exact same process so I'm not sure what is different.
traceFile is set from the command line and if I print "input", it will print out the correct file path and I have confirmed it is there manually.
ClassLoader loader = BiModalPredictor.class.getClassLoader();
File input = new File(loader.getResource(traceFile).getFile());
Scanner fin = new Scanner(input);
Is there any insight as to why this might be happening? I've tried restarting Eclipse, refreshing the files, and I've also tested it on another program which worked. No idea why it can't find this file.
Resources on the classpath, i.e. available through the classloaders getResource method, will not be files on the file system when your application is deployed as a jar file, or deployed in general. Do not use File with such resources, instead use getResourceAsStream to access the resource content.
Besides, your code is wrong. getResource() returns a URL. If you want a File object from a URL, you should use new File(uri), where the URI is obtained by calling url.toURI().
File input = new File(loader.getResource(traceFile).toURI());
Related
i have following line
File file = ResourceUtils.getFile("classpath:calculation.csv");
and i also tried
File file = ResourceUtils.getFile("classpath:/calculation.csv");
but both will throw an error
java.io.FileNotFoundException: class path resource [calculation.csv] cannot be resolved to absolute file path because it does not exist
but i do have calculation.csv in by resources folder..
why is this?
I need to read file from resources folder, and it should also work in server enviroment
EDIT:
ClassLoader classLoader = getClass().getClassLoader();
File file = new File(classLoader.getResource("calculation.csv").getFile());
works just as fine, so not at all..
EDIT2:
tried with folder.. i have both calculation.csv and csv/calculation.csv in my resources folder now..
none of the above work, with /csv/ added.
what kind of path does this thing want?!
EDIT3:
aaand
File file = new ClassPathResource("calculation.csv").getFile();
is also no go, what even is this..
Loading file (as FILE) wont work. You must use it as resource. Files inside JAR will not work as file anyway. This is also what your "check" code shows.
classLoader.getResource("calculation.csv") works, because you are using classloader to get resource, not filesystem to get file (which is what File api does). It could work, if you would deal with non packed application. Once you pack your app into JAR, file path will be like your/path/to/jar.jar!someResource - note ! mark (and that is what you would see as well). So basicly it will return File instance, you that you wont be able to use anyway, as file system has no access to it.
You could alternatively try to extract it first with ResourceUtiuls#extractJarFileURL(URL jarUrl) and then use extracted file.
I think, that in most cases Class#getResourceAsStream is the way to go and I think that it should fit your needs as well to read content of resource.
If I do this
fis = new FileInputStream(new File(".").getAbsolutePath() + "/sudoinput.txt");
Its trying to write to this location on the server. I am not sure if this is a writable
place.
FILE NAME (fos)::::::::::::::::::/opt/tomcat/temp/./sudoinput.txt
FILE NAME (fis)::::::::::::::::::/opt/tomcat/temp/./sudoinput.txt
I wanted to write to
webapps/sudoku/WEB-INF/classes
which is basically
C:\Users...\git\sudo-project\sudo\src\main\resources
On Eclipse Windows 7 I get this
error
src\main\resources\sudoinput.txt (The system cannot find the path specified)
if I give
fis = new FileInputStream("src/main/resources/sudoinput.txt");
I have tried this too:
fis = new FileInputStream("src\\main\\resources\\sudoinput.txt");
but doesn't work.
how should I create a fileinputstream to be able to write to src/main/resources ?
please note that I am using eclipse windows to do dev and will be uploading the .war file on to a unix server if this changes the way in which the paths need to be specified.
The src/main/resources folder is a folder that is supposed to contain resources for your application. As you noted, maven packages these files to the root of your file so that you can access them in your library.
Have a look at the Maven documentation about the standard directory layout.
In certain cases, it is possible to write to the context but it is not a good idea to try it. Depending on how your webapp is deployed, you might not be able to write into the directory. Consider the case when you deploy a .war archive. This would mean that you try to write into the war archive and this won't be possible.
A better idea would be to use a temporary file. In that way you can be sure this will work, regardless of the way your web application is deployed.
Agree with Sandiip Patil. If you didn't have folder inside your resources then path will be /sudoinput.txt or in folder /folder_name/sudoinput.txt. For getting file from resources you should use YourClass.class.getResource("/filename.txt");
For example
Scanner scanner = new Scanner(TestStats.class.getResourceAsStream("/123.txt"));
or
Scanner scanner = new Scanner(new `FileInputStream(TestStats.class.getResource("/123.txt").getPath()));`
Also look at: this
You can keep the file created under resources and call .class.getresource(your_file_name_or_path_separated_with_forward_slash);
See if it works for you.
If you like to create files in webapps/sudoku/WEB-INF/classes which is in the end within the created WAR file which can be achieved by putting the files you want into src/main/resources/
This means in other words you need to create the folder src/main/resources and put the files you like into this directory.
I have an application that creates a temporary mp3-file and puts it in a directory like C:\
File tempfile = File.createTempFile("something", ".mp3", new File("C:\\));
I'm able to read it by just using that same tempfile again.
Everything works fine in the Eclipse IDE.
But when I export my project for as a Runnable jar, my files are still being made correctly (I can play them with some normal music player like iTunes) but I can't seem to read them anymore in my application.
I found out that I need to use something like getClass().getResource("/relative/path/in/jar.mp3") for using resource files that are in the jar. But this doesn't seem to work if I want to select a file from a certain location in my file system like C:\something.mp3
Can somebody help me on this one?
It seems you dont have file name of the temp files . When you was running your program in eclipse that instance was creating a processing files, but after you made a runable you are not able to read those file that instance in eclipse created, You runable file can create its own temp file and can process them,
To make temp files globe put there (path + name ) entries in some db or property file
For example of you will create a temp file from the blow code
File tempfile = File.createTempFile("out", ".txt", new File("D:\\"));
FileWriter fstream = new FileWriter(tempfile);//write in file
out = new BufferedWriter(fstream);
the out will not be out.txt file it will be
out6654748541383250156.txt // it mean a randum number will be append with file
and you code in runable jar is no able to find these temp files
getClass().getResource() only reads resources that are on your classpath. The path that is passed to getResource() is, in fact, a path relative to any paths on your current classpath. This sounds a bit confusing, so I'll give an example:
If your classpath includes a directory C:\development\resources, you would be able to load any file under this directory using getResource(). For example, there is a file C:\development\resources\mp3\song.mp3. You could load this file by calling
getClass().getResource("mp3/song.mp3");
Bottom line: if you want to read files using getResource(), you will need those files to be on your classpath.
For loading from both privileged JARs and the file system, I have had to use two different mechanisms:
getClass().getClassLoader().getResource(path), and if that returns null,
new File(path).toURI().toURL();
You could turn this into a ResourceResolver strategy that uses the classpath method and one or more file methods (perhaps using different base paths).
InputStream inp = new FileInputStream("src/main/resources/ExportHour.xls");
I have a file in the src/main/resources folder of my Java Spring project.
I am attempting to create an inputstream in one of my Controllers, however I always get a file not found exception. When I change the path location to point specifically to the file on my machine, it works fine.
Any way I can make it so the file can be found within the java project?
Try with spring ClassPathResource.
InputStream inp = new ClassPathResource("ExportHour.xls").getInputStream();
That is because the resources folder in maven is put in your jar file directly i.e. the ExportHours.xls file is put inside your jar in the root directory.
It sounds like you could just change the working directory of your process - it's not where you think it is, I suspect. For example, I suggest you write
File file = new File("src/main/resources/ExportHour.xls");
and then log file.getAbsolutePath(), to see what exact file it's using.
However, you should almost certainly not be using a FileInputStream anyway. It would be better to use something like:
InputStream inp = Foo.class.getResourceAsStream("/ExportHour.xls");
... for some class Foo which has a classloader which includes the resources you need.
(Or possibly /resources/ExportHour.xls", depending on your build structure.)
That way even when you've built all of this into a jar file, you'll still be able to open the resource.
Been looking for this for the past 2 hours and can't find anything (I've found solutions to the same problem but with images, not text files).
Pretty much, I made a program that reads a text file. The file is a list of names and IDs. Using Eclipse, I put the file in my src folder and in the program put the path file to it. Like this:
in = new BufferedReader(new FileReader(curDir+"\\bin\\items.txt"));
Where curDir is the user's current directory (found with System.getProperty("user.dir")).
Now, problem is, the program runs fine when I run it from Eclipse, but when I try to make it a runnable JAR and then run it, the program runs, but the info from the text file does not load. It look like Eclipse is not putting the text file with the JAR.
EDIT: Solved-ish the problem? So the JAR file needs to the in a folder with all the original files? I am so confused, what is a JAR file then?
A more robust way to get a file whether you are running from Eclipse or a JAR is to do
MyClass.getResource("items.txt")
where MyClass is a class in the same package (folder) as the resource you need.
If Eclipse is not putting the file in your JAR you can go to
Run -> Run Configurations -> -> Classpath tab -> Advanced -> Add Folders
Then add the folder containing your file to the classpath. Alternatively, export the Ant script and create a custom build script.
To the point, the FileReader can only read disk file system resources. But a JAR contains classpath resources only. You need to read it as a classpath resource. You need the ClassLoader for this.
Assuming that Foo is your class in the JAR which needs to read the resource and items.txt is put in the classpath root of the JAR, then you should read it as follows (note: leading slash needed!):
InputStream input = Foo.class.getResourceAsStream("/items.txt");
reader = new BufferedReader(new InputStreamReader(input, "UTF-8"));
// ...
Or if you want to be independent from the class or runtime context, then use the context class loader which operates relative to the classpath root (note: no leading slash needed!):
ClassLoader classLoader = Thread.currentThread().getContextClassLoader();
InputStream input = classLoader.getResourceAsStream("items.txt");
reader = new BufferedReader(new InputStreamReader(input, "UTF-8"));
// ...
(UTF-8 is of course the charset the file is encoded with, else you may see Mojibake)
Get the location of your jar file
Firstly create a folder(say myfolder) and put your files inside it
Consider the following function
public String path(String filename)
{
URL url1 = getClass().getResource("");
String ur=url1.toString();
ur=ur.substring(9);
String truepath[]=ur.split("myjar.jar!");
truepath[0]=truepath[0]+"myfolder/";
truepath[0]=truepath[0].replaceAll("%20"," ");
return truepath[0]+filename;
}//This method will work on Windows and Linux as well.
//You can alternatively use the following line to get the path of your jar file
//classname.class.getProtectionDomain().getCodeSource().getLocation().getPath();
Suppose your jar file is in D:\Test\dist
Then path() will return /D:/Test/dist/myfolder/filename
Now you can place 'myfolder' inside the folder where your jar file is residing
OR
If you want to access some read-only file inside your jar you should copy it to one
of your packages and can access it as
yourClassname.getResource("/packagename/filename.txt");