I am trying to read a file inside my jar from another class inside the jar. However I am continually getting the same error: Caught: class java.io.FileNotFoundException while attempting to read metrics: metrics.yml
At first I had my code do something like this, assuming it was from the path of the class:
String yamlPath = ".." + File.separator + ".." + File.separator + ".." + File.separator + ".." + File.separator + "myYaml.yml";
InputStream in = new FileInputStream(new File(yamlPath));
InputStreamReader isr = new InputStreamReader(in);
BufferedReader input = new BufferedReader(isr);
yamlObj = (HashMap) javaYAML.load(input);
I also did this assuming it might take the path from the base of the jar:
String yamlPath = "myYaml.yml";
InputStream in = new FileInputStream(new File(yamlPath));
InputStreamReader isr = new InputStreamReader(in);
BufferedReader input = new BufferedReader(isr);
yamlObj = (HashMap) javaYAML.load(input);
I then noticed this thread How to read a file from jar in Java? and found out that I needed a "/" before my path. I tried both methods above using the slash as well.
String yamlPath = File.seperator + ".." + File.separator + ".." + File.separator + ".." + File.separator + ".." + File.separator + "myYaml.yml";
OR
String yamlPath = File.seperator + "myYaml.yml";
I am completely lost on what to do about this error now. Is there something I am not getting about the jar structure? Why can my file not be found. Thanks in advance for any help / information.
Sorry, I forgot to mention where it is in the JAR:
The class is in the following path: com/a/b/c/myclass.class
The yaml is in the following path: myYaml.yml
File inside Jar is no longer a File, Change inputStream creation with this
InputStream in = YourClass.class.getResourceAsStream("myYaml.yml");
Assuming your .yml file is at root of jar
InputStream is = this.getClass().getClassLoader().getResourceAsStream("file");
Somehow for me, i have to add '/' in front of the file name.
InputStream resourceAsStream = Xx.class.getResourceAsStream("/test.yml");
From java doc:
Before delegation, an absolute resource name is constructed from the given resource name using this algorithm:
If the name begins with a '/' ('\u002f'), then the absolute name of the resource is the portion of the name following the '/'.
Otherwise, the absolute name is of the following form:
modified_package_name/name
Where the modified_package_name is the package name of this object with '/' substituted for '.' ('\u002e').
And after i checked more detailed information, my solution is same as
InputStream in = Xx.class.getClassLoader().getResourceAsStream("test.yml");
Related
Recently we added checking with Fortify to our code during build.
Following there are 2 lines of code, both lines cause 2 critical errors in Fortify: "Path manipulation".
Notice that I added the call to normalize() just to avoid that, and for the same reason I used a Path instead of String.
But now, instead of having 1 error as previously, I have 2 errors (in both lines).
Path normalizedPath = Paths.get(aPath + "/" + aStringFromDB + "/" + aFileName + ".txt").normalize();
BufferedOutputStream b = new BufferedOutputStream(new FileOutputStream(new File(String.valueOf(normalizedPath))));
What am I doing wrong?
It came out that the "normalize" of Path was not considered by Fortify.
The solution was to import and use the Apache normalize():
// import org.apache.commons.io.FilenameUtils;
String normalizedPath = FilenameUtils.normalize((aPath + "/" + aStringFromDB + "/" + aFileName + ".txt");
BufferedOutputStream b = new BufferedOutputStream(new FileOutputStream(new File(normalizedPath)));
I have a JFilechooser to select a filename and path to store some data. But I also want to store an additional file in the same path, same name but different extension. So:
File file = filechooser.getSelectedFile();
String path = file.getParent();
String filename1 = file.getName();
// Check the extension .ext1 has been added, else add it
if(!filename1.endswith(".ext1")){
filename2 = filename1 + ".ext2";
filename1 += ".ext1";
}
else{
filename2 = filename1;
filename2 = filename2.substring(0, filename2.length-4) + "ext2";
}
// And now, if I want the full path for these files:
System.out.println(path); // E.g. prints "/home/test" withtout the ending slash
System.out.println(path + filename1); // E.g. prints "/home/testfilename1.ext1"
Of course I could add the "/" in the middle of the two strings, but I want it to be platform independent, and in Windows it should be "\" (even if a Windows path file C:\users\test/filename1.ext1 would probably work).
I can think of many dirty ways of doing this which would make the python developer I'm carrying inside cry, but which one would be the most clean and fancy one?
You can use the constants in the File class:
File.separator // e.g. / or \
File.pathSeparator // e.g. : or ;
or for your path + filename1 you can do
File file = new File(path, filename1);
System.out.println(file);
Just use the File class:
System.out.println(new File(file.getParent(), "filename1"));
You can use:
System.getProperty("file.separator");
I am trying to read a file inside my jar from another class inside the jar. However I am continually getting the same error: Caught: class java.io.FileNotFoundException while attempting to read metrics: metrics.yml
At first I had my code do something like this, assuming it was from the path of the class:
String yamlPath = ".." + File.separator + ".." + File.separator + ".." + File.separator + ".." + File.separator + "myYaml.yml";
InputStream in = new FileInputStream(new File(yamlPath));
InputStreamReader isr = new InputStreamReader(in);
BufferedReader input = new BufferedReader(isr);
yamlObj = (HashMap) javaYAML.load(input);
I also did this assuming it might take the path from the base of the jar:
String yamlPath = "myYaml.yml";
InputStream in = new FileInputStream(new File(yamlPath));
InputStreamReader isr = new InputStreamReader(in);
BufferedReader input = new BufferedReader(isr);
yamlObj = (HashMap) javaYAML.load(input);
I then noticed this thread How to read a file from jar in Java? and found out that I needed a "/" before my path. I tried both methods above using the slash as well.
String yamlPath = File.seperator + ".." + File.separator + ".." + File.separator + ".." + File.separator + ".." + File.separator + "myYaml.yml";
OR
String yamlPath = File.seperator + "myYaml.yml";
I am completely lost on what to do about this error now. Is there something I am not getting about the jar structure? Why can my file not be found. Thanks in advance for any help / information.
Sorry, I forgot to mention where it is in the JAR:
The class is in the following path: com/a/b/c/myclass.class
The yaml is in the following path: myYaml.yml
File inside Jar is no longer a File, Change inputStream creation with this
InputStream in = YourClass.class.getResourceAsStream("myYaml.yml");
Assuming your .yml file is at root of jar
InputStream is = this.getClass().getClassLoader().getResourceAsStream("file");
Somehow for me, i have to add '/' in front of the file name.
InputStream resourceAsStream = Xx.class.getResourceAsStream("/test.yml");
From java doc:
Before delegation, an absolute resource name is constructed from the given resource name using this algorithm:
If the name begins with a '/' ('\u002f'), then the absolute name of the resource is the portion of the name following the '/'.
Otherwise, the absolute name is of the following form:
modified_package_name/name
Where the modified_package_name is the package name of this object with '/' substituted for '.' ('\u002e').
And after i checked more detailed information, my solution is same as
InputStream in = Xx.class.getClassLoader().getResourceAsStream("test.yml");
How can I get file size in Java if I have a relative path to a file such as:
String s = "/documents/19/21704/file2.pdf/0929c695-d023-49d7-a8ff-65ccea46bebc"
I tried with two diferent strings:
String[] separatedPath = s.split("/");
List<String> wordList = Arrays.asList(separatedPath);
String ret = "/" + wordList.get(1) + "/" + wordList.get(2) + "/" + wordList.get(3)+ "/" + wordList.get(4);
s = ret;
In this case s="/documents/19/21704/file2.pdf";
In second case s="/documents/19/21704/file2.pdf/0929c695-d023-49d7-a8ff-65ccea46bebc"
I tried with:
File file1 = new File(s);
long filesize = file1.length();
and with:
String filePath = new File(s).toURI().getPath();
File file2 = new File(filePath);
long filesize2 = file1.length();
and also with (if the problem is in not providing full path):
String absolutePath = FileUtil.getAbsolutePath(file1);
File file3 = new File(absolutePath);
long filesize3 = file3.length();
byte[] bytes1=FileUtil.getBytes(file1);
byte[] bytes2=FileUtil.getBytes(file2);
byte[] bytes3=FileUtil.getBytes(file3);
I am always getting in debug that filesizes in all cases are 0.
Maybe is worth noticing that the three attributes of file1 and file2 and file3 are always:
filePath: which is always null;
path: "/documents/19/21704/liferay-portlet-development.pdf"
prefixLength: 1
Since I am also using Liferay I also tried their utility.
long compId = article.getCompanyId();
long contentLength = DLStoreUtil.getFileSize(compId, CompanyConstants.SYSTEM, s);
I also should notice that in my .xhtml view I can access the file with:
<a target="_blank"
href="/documents/19/21704/file2.pdf/0929c695-d023-49d7-a8ff-65ccea46bebc">
file2.pdf
</a>
Pdf opens in a new window. So it is stored on my server.
What am I doing wrong here? That I cant get the file size from bean?
Any answer would be greatly appreciated.
What am I doing wrong here?
In Java, you can use the File.length() method to get the file size in bytes.
File file =new File("c:\\java_xml_logo.jpg");
if(file.exists()){
double bytes = file.length();
}
System.out.println("bytes : " + bytes);
The problem is that your "relative" path is expressed as an absolute path (begining with "/", which is read as FS root).
A relative file path should look like:
documents/19/21704/file2.pdf/0929c695-d023-49d7-a8ff-65ccea46bebc
./documents/19/21704/file2.pdf/0929c695-d023-49d7-a8ff-65ccea46bebc
Or, you could get your application root folder File and compose the absolute path:
File rootFolder =new File("path to your app root folder");
File myfile=new File(rootFolder, "/documents/19/21704/file2.pdf/0929c695-d023-49d7-a8ff-65ccea46bebc");
In Java, I have a File object representing a folder:
String folderName = "/home/vektor/folder";
File folder = new File(folderName);
Now I want to create another File representing a file in this folder. I want to avoid doing a string concatenation like this:
String fileName = "test.txt";
File file = new File(folderName + "/" + fileName);
Because if I go deeper in creating this structure, I will come up with something like this:
File deepFile = new File(folderName + "/" + anotherFolderName + ... + "/" + fileName);
I would instead like to do something like
File betterFile = folder.createUnder(fileName);
Or even:
File otherFile = SomeFileUtils.createFileInFolder(folder, fileName);
Do you know of such solution?
Note: It's quite OK to use "/" because Java will translate it to "\" for Windows, but it is not clean - I should use something like "file.separator" from System.getProperties().
Look at the Javadoc for File and you will see that the constructor takes a File object as parent.
Use the following form:
File deepFile = new File(folder, fileName);
I would use
String folderName =
String fileName =
File under = new File(folderName, fileName);
or
File folderFile =
String fileName =
File under = new File(folderFile, fileName);
simple as that ;)