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Java: Read image from current src directory

I would like to load an image from my current src directory where the java class files are located as well. However, I always get an IOException..
And how can I make sure the file gets loaded properly on Mac/Linux as well on Windows?
My code so far:
String dir = System.getProperty("user.dir") + "/Logo_transparent.png";
File imageFile = new File(dir);
BufferedImage bufferedImage = null;
try {
bufferedImage = ImageIO.read(imageFile);
} catch (IOException e) {
System.out.println(e.getMessage());
System.out.println(dir);
System.out.println();
}
IOException message:
Can't read input file!
(My path is correct - is it because of the space between Google and Drive?)
/Users/myMac/Google Drive/Privat/Programming/Logo_transparent.png
Kind regards and thank you!

I think It's because you didn't create the file, You can create the file if it doesn't exist by using this code
if(!imageFile.exists()) imageFile.createNewFile();
You're code will look like this
String dir = System.getProperty("user.dir") + "/Logo_transparent.png";
File imageFile = new File(dir);
BufferedImage bufferedImage = null;
try {
if(!imageFile.exists()) imageFile.createNewFile();
bufferedImage = ImageIO.read(imageFile);
} catch (IOException e) {
System.out.println(e.getMessage());
System.out.println(dir);
System.out.println();
}
Also you shouldn't concat child files like that instead pass it as a second argument.
File imageFile = new File(System.getProperty("user.dir"), "Logo_transparent.png");

If your image files will be packaged together with your class files (for example in the same .jar) you should not use File but read it as a resource:
bufferedImage = ImageIO.read(this.getClass().getResourceAsStream("/Logo_transparent.png"));
Notice the '/' before the file name. This means to search in the root path of the classpath.
If you specify without / it will search in the package of this (the current class)
this.getClass().getResourceAsStream("Logo_transparent.png")

You can try to build the absolute path to the image like here and read it afterward.

Exported Jar file won't read file inside jar

In the code sample below, when I test the code in Eclipse it works just fine. However, when I export the jar file and test it via the command line, it throws an error: IIOException: Can't read input file!
private BufferedImage img = null;
private String imgSource;
if (img == null)
{
try {
URL url = getClass().getResource("Images/questionMark.png");
System.out.println(url.getPath());
/* This prints: file:/C:/Users/Keno/Documents/javaFile.jar!/javaFile/Images/questionMark.png */
File file = new File(url.getPath());
img = ImageIO.read(file);
imgSource = file.getName();
} catch (IOException e) {
e.printStackTrace();
}
}
The file I want to get is located inside the Images folder which is inside the javaFile package. I've noticed one thing that may indicate the problem.
In the print statement I have, I notice an exclamation sign at the end of the javaFile.jar section. Is that correct? Could that indicate an issue with the file or structure?
Also, just in case someone has a better suggestion as to how I should load the file, I'll tell you my intentions. I would like to load the file from a relative location (Images folder) in the jar. I would like to display it (Already done in my actual code) and also store the location to be passed later on to another function (also done).

try this
public void test() {
try(InputStream is = getClass().getResourceAsStream("Images/questionMark.png")) {
ImageIO.read(is);
} catch (IOException e) {
e.printStackTrace();
}
}

You should try to check if your class is in the same directory than Images inside your jar.
|
|- Your class
|- Images
|- questionMark.png
Also, have you tried using directly your url object ?
File file = new File(url);

Java read file within static method, using ClassLoader gives FileNotFoundException

I want to read a file in my java class. My question is similar to this one, but there are two differences. first, I use a different project layout:
/src/com/company/project
/resources
In the resources folder I have a file called "test.txt":
/resources/test.txt
In the project folder I have a class test.java
/src/com/company/project/test.java
I want mu java class to be able to read the contents of test.txt in a STATIC METHOD. I've tried the following:
private static String parseFile()
{
try
{
ClassLoader classLoader = Thread.currentThread().getContextClassLoader();
String fileURL = classLoader.getResource("test.txt").getFile();
File file = new File(fileURL);
...
}
}
and the following paths:
File file1 = new File("test.txt");
File file2 = new File("/test.txt");
File file3 = new File("/resources/test.txt");
But they all throw a FileNotFoundException when I want to read the file. How can I correctly declare the path to my file in the snippet above with respect to my project setup and the fact that the method needs to be static?

You should use the class loader of the class which is in the same JAR as the resource instead of the TCCL. And then you need to specify the name of the resource with a full path. And it is typically not good to access those as files. Just open it directly for read (or copy it to a temp file if you need to):
InputStream is =
Project.class.getClassLoader().getResourceAsStream("/resource/test.txt");
BTW: if you simply want to open a file, you need to use a relative file name. This is searched relative to the start dir, which is normally the project main dir (in eclipse):
File resource = new File("resource/test.txt");
(but this wont work if you package it up as a JAR).

After endless trials, I gave up on ClassLoader and getResource methods of any kind.
Absolutely nothing worked, especially if the opening attempt was made from another project. I always ended up getting the bin folder instead of the src folder.
So I devised the following work around:
public class IOAccessory {
public static String getProjectDir() {
try {
Class<?> callingClass = Class.forName(Thread.currentThread().getStackTrace()[2].getClassName());
URL url = callingClass.getProtectionDomain().getCodeSource().getLocation();
URI parentDir = url.toURI().resolve("..");
return parentDir.getPath();
} catch (ClassNotFoundException e) {
e.printStackTrace();
} catch (URISyntaxException e) {
e.printStackTrace();
}
return "";
}
}
The getProjectDir method returns the physical path of the project from which it was called, e.g. C:/workspace/MyProject/.
After that, all you need to do is concatenate the relative path in MyProject of your resource file to open the stream:
public void openResource() throws IOException {
InputStream stream = null;
String projectDir = IOAccessory.getProjectDir();
String filePath = "resources/test.txt";
try {
stream = new FileInputStream(projectDir + filePath);
open(stream);
} catch(Exception e) {
e.printStackTrace();
} finally {
if (stream != null)
stream.close();
}
}
This technique works whether the openResource method is static or non-static, and whether it is called from within the project or from another project on the build path.

It really depends on how your IDE generates output from your project. Typically, classloaders load resources relative to the invoking classes, but if treated right, 'resources' will just end up in the 'root' of your output folder hierarchy, and you can access them accordingly.
For example, if I recreate your code in IntelliJ IDEA, in a class called com/acme/TestClass.class, the following output structure is generated within the IDE when building. This assumes I have "test.txt" sitting in a folder I called "resources", and that folder is specified as being a "resources root":
/com
/acme
TestClass.class
test.txt
The text file ends up in the output folder's root, so accessing it is simple. The following code works for me when I attempt to load the file in a static method within TestClass:
ClassLoader cl = TestClass.class.getClassLoader();
InputStream is = cl.getResourceAsStream("test.txt");

The only thing not covered in the other answers is that your URL conversion to file might not work correctly. If the directories above your project contain a characters that must be decoded then your call to 'getResource("test.txt").getFile()' is not giving you a valid java.io.File path.

I load shader for openGL ES from static function.
Remember you must use lower case for your file and directory name, or else the operation will be failed
public class MyGLRenderer implements GLSurfaceView.Renderer {
...
public static int loadShader() {
// Read file as input stream
InputStream inputStream = MyGLRenderer.class.getResourceAsStream("/res/raw/vertex_shader.txt");
// Convert input stream to string
Scanner s = new Scanner(inputStream).useDelimiter("\\A");
String shaderCode = s.hasNext() ? s.next() : "";
}
...
}
Another method to convert input stream to string.
byte[] bytes;
String shaderCode = "";
try {
bytes = new byte[inputStream.available()];
inputStream.read(bytes);
shaderCode = new String(bytes);
}
catch (IOException e) {
e.printStackTrace();
}

android - reading user created file

Couldn't find too much about how to read a file that isn't somewhere on the SD card or storage, but rather right there in the android project directory.
I keep getting a FileNotFoundException. This is how I declare the file,
File SPPolicy = new File("SHPR_policy");
I've gotten the same error when putting it in the src/ directory, the src/[[package]]/ directory and the main project directory, and I get this error:
java.io.FileNotFoundException: /SHPR_policy: open failed: ENOENT (No such file or directory)
Is there a certain place I have to put this file? Is it because my file doesn't have an extension (I noticed the "/" before SHPR_policy but I didn't think it would be a problem because Eclipse let me create a file without an extension)?

Save the file in raw folder and try the below code.
try
{
Resources res = getResources();
InputStream in_s = res.openRawResource(R.raw.test);
byte[] b = new byte[in_s.available()];
in_s.read(b);
txtHelp.setText(new String(b));
} catch (Exception e) {
// e.printStackTrace();
txtHelp.setText("errr.");
}

Use these piece of code to check if the file was already created:
File f;
f=new File("myfile");
if(!f.exists()){
f.createNewFile();
}
Also, if you want to allocate your file in the external directory you can use these code:
File newxmlfile = new File( Environment.getExternalStorageDirectory() + "/new.xml");
XmlSerializer serializer = Xml.newSerializer();
try {
newxmlfile.createNewFile();
} catch (Exception e) {
Log.e("IOException", "exception in createNewFile() method", e);
}
And, did you added the permission in the manifest to be able to manipulate files?
<uses-permission android:name="android.permission.WRITE_EXTERNAL_STORAGE" />

You should place the file in the assets folder and use this:
InputStream ims = getAssets().open("SHPR_policy");

Getting file path for local Android project files

I want to programmatically access a specific file which will be included in my project folder. Is there a way to do this? If so, where in my project folder do I put the file, and what is some simple code to get its file path?
private void saveFileToDrive() {
Thread t = new Thread(new Runnable() {
#Override
public void run() {
try {
java.io.File spreadsheet = new java.io.File("Untitled spreadsheet.xlsx");
String filePath = spreadsheet.getAbsolutePath();
System.out.println("file path is"+filePath);
URL fileURL = getClass().getClassLoader().getResource("Untitled spreadsheet.xlsx");
String filePath2 = fileURL.getPath();
System.out.println("file path2 is"+filePath2);
java.io.File fileContent = new java.io.File(filePath);
FileContent mediaContent = new FileContent("application/vnd.ms-excel", fileContent);
File body = new File();
body.setTitle(fileContent.getName());
body.setMimeType("application/vnd.ms-excel");
File file = service.files().insert(body, mediaContent).setConvert(true).execute();
if (file != null) {
showToast("File uploaded: " + file.getTitle());
}
else
;
} catch (UserRecoverableAuthIOException e) {
startActivityForResult(e.getIntent(), REQUEST_AUTHORIZATION);
} catch (IOException e) {
e.printStackTrace();
}
}
});
t.start();
}

Put the file in root folder of your project. Then get the File URL, Path and other details as:
File file = new File("test.txt");
String filePath = file.getAbsolutePath();
EDIT: Alternate way (if the file is in your classpath e.g. put the file in "src" folder, and make sure its moved in "bin" or "classes" folder after compilation):
URL fileURL = getClass().getClassLoader().getResource(fileName);
String fileName = fileURL.getFile();
String filePath = fileURL.getPath();

This depends a lot on what type of file you want to access. You can put the file in either assets or an appropriate subdirectory of res (see Difference between /res and /assets directories).

So you want to access a file internal to your app; and you want to do so directly, rather, that is, from an Android Context (and then with a [android.|<package_name>.]R.<resource_type>.<resource_name>).
You have two choices as to location: the res/raw folder or assets/ folder (outside of the res parent).
To choose between the two note from https://developer.android.com/guide/topics/resources/providing-resources.html
Arbitrary files to save in their raw form. To open these resources with a raw InputStream, call Resources.openRawResource() with the resource ID, which is R.raw.filename.
However, if you need access to original file names and file hierarchy, you might consider saving some resources in the assets/ directory (instead of res/raw/). Files in assets/ aren't given a resource ID, so you can read them only using AssetManager.
To access a file in res/raw/ directly rather, that is, from an Android Context (and then with a [android.|<package_name>.]R.<resource_type>.<resource_name>) you can do something like this:
File file = new File("app/src/main/res/raw/country_data_from_world_bank.xml");
BufferedReader bufferedReader = new BufferedReader(new FileReader(file));