I have a class,in which ther is a func,which opens a properties file. When i write main in the same class & call that function,i am able to open the properties file n read. but, when i am tying to call the same func in my servlet by creating instance to that class, i get file not found exception.
This is the function, which i have written in my class to read properties file. And both my class and servlet are in src folder. I am using eclipse IDE.
Code:
private void readPropertiesFileAndFieldListData() {
String PROP_FILE = "./src/fields.properties";
try {
FileReader reader = new FileReader(PROP_FILE);
BufferedReader br = new BufferedReader(reader);
ArrayList<String> field = new ArrayList<String>();
while ((str = br.readLine()) != null) {
if (!str.startsWith("#") && str.trim().length() > 0) {
// System.out.println(str);
field.add(str);
count++;
}
}
}
You're relying on the current working directory of the disk file system path. The current working directory is dependent on how the application is started and is not controllable from inside your application. Relying on it is a very bad idea.
The normal practice is to put that file in the classpath or to add its path to the classpath. Your file is apparently already in the classpath (you placed it in the src folder), so you don't need to change anything else. You should should just get it from the classpath by the class loader. That's more portable than a disk file system path can ever be.
ClassLoader classLoader = Thread.currentThread().getContextClassLoader();
InputStream input = classLoader.getResourceAsStream("fields.properties");
// ...
See also:
getResourceAsStream() vs FileInputStream
Unrelated to the concrete problem, you're basically reinventing the java.util.Properties class. Don't do that. Use the following construct:
ClassLoader classLoader = Thread.currentThread().getContextClassLoader();
InputStream input = classLoader.getResourceAsStream("fields.properties");
Properties properties = new Properties();
properties.load(input);
// ...
PLease write a small test, for printing the file path of "PROP_FILE" to the log or console.
it seems, that you relativ path is incorrect.
Your relative path starting point can change, depending on where your *.exe file is started.
Your test should print
File tFile = new File(PROP_FILE);
// print tFile.getAbsolutePath()
Its better to get a special class by calling
SomeClass.class.getResource(name)
Eclipse RCP from the Bundle
Bundle.getResource(ResourcePathString)
EDIT:
Please check, whether the resource is part of your *.jar. It could be, that you missed to add it to the build.properties file.
Check whether the file is existing, before you read the properties file.
Related
A lot has been discussed already here about getting a resource.
If there is already a solution - please point me to it because I couldn't find.
I have a program which uses several jars.
To one of the jars I added a properties file under main/resources folder.
I've added the following method to the jar project in order to to read it:
public void loadAppPropertiesFile() {
try {
Properties prop = new Properties();
ClassLoader loader = Thread.currentThread().getContextClassLoader();
String resourcePath = this.getClass().getClassLoader().getResource("").getPath();
InputStream stream = loader.getResourceAsStream(resourcePath + "\\entities.properties");
prop.load(stream);
String default_ssl = prop.getProperty("default_ssl");
}catch (Exception e){
}
}
The problem (?) is that resourcePath gives me a path to the target\test-clasess but under the calling application directory although the loading code exists in the jar!
This the jar content:
The jar is added to the main project by maven dependency.
How can I overcome this state and read the jar resource file?
Thanks!
I would suggest using the classloader used to load the class, not the context classloader.
Then, you have two options to get at a resource at the root of the jar file:
Use Class.getResourceAsStream, passing in an absolute path (leading /)
Use ClassLoader.getResourceAsStream, passing in a relative path (just "entities.properties")
So either of:
InputStream stream = getClass().getResourceAsStream("/entities.properties");
InputStream stream = getClass().getClassLoader().getResourceAsStream("entities.properties");
Personally I'd use the first option as it's briefer and just as clear.
Can you try this:
InputStream stream = getClass().getClassLoader().getResourceAsStream("entities.properties")
Kinda wondering why this happens.
Same code but different result.
File file = new File("src/config/ora2.config.properties");
System.out.println(file.getAbsolutePath());
In a regular java class, it results right path but when in servlet, it returns the installation path of eclipse.
Thanks in advance
EDITED
Properties props = new Properties();
InputStream in = getServletContext().getResourceAsStream("WEB-INF/ora2.config.properties");
if(in != null) {
try {
props.load(in);
props.setProperty("username", "temtem");
FileOutputStream out = new FileOutputStream(in.toString());
props.store(out, null);
out.close();
} finally {
in.close();
}
Code above is not working :(
new File(...) treats the path as relative to the current working directory of the java process, whatever that happens to be. If you're building a servlet for eventual deployment on a standalone server system then you can't necessarily depend on this being a sensible location. Instead you should put this kind of configuration file somewhere under your application's WEB-INF and use the getResource or getResourceAsStream methods of the ServletContext to access it, which resolve relative paths against the location of this specific web application.
Properties props = new Properties();
InputStream in = getServletContext().getResourceAsStream("/WEB-INF/ora2.config.properties");
if(in != null) {
try {
props.load(in);
} finally {
in.close();
}
}
If you absolutely must access the properties as a java.io.File rather than a URL or input stream (e.g. if you have to pass the path to third party code that can't accept a stream or a Properties object directly) then ServletContext also has a. getRealPath method you can use, but that requires that the web application be expanded as a real directory on disk, it won't work if you're running directly from a WAR file. The getResource methods don't have this restriction.
The directory structure of my application is as follows:-
My App
++++++ src
++++++++com
++++++++++readProp.java
++++++++resource
++++++++++message.properties
I am trying to read the file as follows:-
public Static final string FilePath="resource.message.properties"
Here the code to read the file. I tried using the following two techniques but to no use...
File accountPropertiesFile = new File(FacesContext.getCurrentInstance()
.getExternalContext().getRequestContextPath()
+ FilePath);
properties.load(externalContext.getResourceAsStream(FilePath));
But none yeild any sucess while reading through the Bean class. please help...
Your properties file is in the classpath. The java.io.File only understands the local disk file system structure. This is not going to work. You need to get it straight from the classpath by the classloader.
Here's a kickoff example:
ClassLoader classLoader = Thread.currentThread().getContextClassLoader();
InputStream input = classLoader.getResourceAsStream("/resources/messages.properties");
if (input != null) {
Properties properties = new Properties();
try {
properties.load(input);
} finally {
input.close();
}
}
I don't know if this is your problem, but you should try using slashes instead of periods, since they're stored as actual folders in the filesystem.
I have a project with 2 packages:
tkorg.idrs.core.searchengines
tkorg.idrs.core.searchengines
In package (2) I have a text file ListStopWords.txt, in package (1) I have a class FileLoadder. Here is code in FileLoader:
File file = new File("properties\\files\\ListStopWords.txt");
But I have this error:
The system cannot find the path specified
Can you give a solution to fix it?
If it's already in the classpath, then just obtain it from the classpath instead of from the disk file system. Don't fiddle with relative paths in java.io.File. They are dependent on the current working directory over which you have totally no control from inside the Java code.
Assuming that ListStopWords.txt is in the same package as your FileLoader class, then do:
URL url = getClass().getResource("ListStopWords.txt");
File file = new File(url.getPath());
Or if all you're ultimately after is actually an InputStream of it:
InputStream input = getClass().getResourceAsStream("ListStopWords.txt");
This is certainly preferred over creating a new File() because the url may not necessarily represent a disk file system path, but it could also represent virtual file system path (which may happen when the JAR is expanded into memory instead of into a temp folder on disk file system) or even a network path which are both not per definition digestable by File constructor.
If the file is -as the package name hints- is actually a fullworthy properties file (containing key=value lines) with just the "wrong" extension, then you could feed the InputStream immediately to the load() method.
Properties properties = new Properties();
properties.load(getClass().getResourceAsStream("ListStopWords.txt"));
Note: when you're trying to access it from inside static context, then use FileLoader.class (or whatever YourClass.class) instead of getClass() in above examples.
The relative path works in Java using the . specifier.
. means same folder as the currently running context.
.. means the parent folder of the currently running context.
So the question is how do you know the path where the Java is currently looking?
Do a small experiment
File directory = new File("./");
System.out.println(directory.getAbsolutePath());
Observe the output, you will come to know the current directory where Java is looking. From there, simply use the ./ specifier to locate your file.
For example if the output is
G:\JAVA8Ws\MyProject\content.
and your file is present in the folder "MyProject" simply use
File resourceFile = new File("../myFile.txt");
Hope this helps.
The following line can be used if we want to specify the relative path of the file.
File file = new File("./properties/files/ListStopWords.txt");
InputStream in = FileLoader.class.getResourceAsStream("<relative path from this class to the file to be read>");
try {
BufferedReader reader = new BufferedReader(new InputStreamReader(in));
String line = null;
while ((line = reader.readLine()) != null) {
System.out.println(line);
}
} catch (Exception e) {
e.printStackTrace();
}
try .\properties\files\ListStopWords.txt
I could have commented but I have less rep for that.
Samrat's answer did the job for me. It's better to see the current directory path through the following code.
File directory = new File("./");
System.out.println(directory.getAbsolutePath());
I simply used it to rectify an issue I was facing in my project. Be sure to use ./ to back to the parent directory of the current directory.
./test/conf/appProperties/keystore
While the answer provided by BalusC works for this case, it will break when the file path contains spaces because in a URL, these are being converted to %20 which is not a valid file name. If you construct the File object using a URI rather than a String, whitespaces will be handled correctly:
URL url = getClass().getResource("ListStopWords.txt");
File file = new File(url.toURI());
Assuming you want to read from resources directory in FileSystem class.
String file = "dummy.txt";
var path = Paths.get("src/com/company/fs/resources/", file);
System.out.println(path);
System.out.println(Files.readString(path));
Note: Leading . is not needed.
I wanted to parse 'command.json' inside src/main//js/Simulator.java. For that I copied json file in src folder and gave the absolute path like this :
Object obj = parser.parse(new FileReader("./src/command.json"));
For me actually the problem is the File object's class path is from <project folder path> or ./src, so use File file = new File("./src/xxx.txt"); solved my problem
For me it worked with -
String token = "";
File fileName = new File("filename.txt").getAbsoluteFile();
Scanner inFile = null;
try {
inFile = new Scanner(fileName);
} catch (FileNotFoundException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
while( inFile.hasNext() )
{
String temp = inFile.next( );
token = token + temp;
}
inFile.close();
System.out.println("file contents" +token);
If text file is not being read, try using a more closer absolute path (if you wish
you could use complete absolute path,) like this:
FileInputStream fin=new FileInputStream("\\Dash\\src\\RS\\Test.txt");
assume that the absolute path is:
C:\\Folder1\\Folder2\\Dash\\src\\RS\\Test.txt
String basePath = new File("myFile.txt").getAbsolutePath();
this basepath you can use as the correct path of your file
if you want to load property file from resources folder which is available inside src folder, use this
String resourceFile = "resources/db.properties";
InputStream resourceStream = ClassLoader.getSystemClassLoader().getResourceAsStream(resourceFile);
Properties p=new Properties();
p.load(resourceStream);
System.out.println(p.getProperty("db"));
db.properties files contains key and value db=sybase
If you are trying to call getClass() from Static method or static block, the you can do the following way.
You can call getClass() on the Properties object you are loading into.
public static Properties pathProperties = null;
static {
pathProperties = new Properties();
String pathPropertiesFile = "/file.xml";
// Now go for getClass() method
InputStream paths = pathProperties.getClass().getResourceAsStream(pathPropertiesFile);
}
Im trying to write a program to read a text file through args but when i run it, it always says the file can't be found even though i placed it inside the same folder as the main.java that im running.
Does anyone know the solution to my problem or a better way of reading a text file?
Do not use relative paths in java.io.File.
It will become relative to the current working directory which is dependent on the way how you run the application which in turn is not controllable from inside your application. It will only lead to portability trouble. If you run it from inside Eclipse, the path will be relative to /path/to/eclipse/workspace/projectname. If you run it from inside command console, it will be relative to currently opened folder (even though when you run the code by absolute path!). If you run it by doubleclicking the JAR, it will be relative to the root folder of the JAR. If you run it in a webserver, it will be relative to the /path/to/webserver/binaries. Etcetera.
Always use absolute paths in java.io.File, no excuses.
For best portability and less headache with absolute paths, just place the file in a path covered by the runtime classpath (or add its path to the runtime classpath). This way you can get the file by Class#getResource() or its content by Class#getResourceAsStream(). If it's in the same folder (package) as your current class, then it's already in the classpath. To access it, just do:
public MyClass() {
URL url = getClass().getResource("filename.txt");
File file = new File(url.getPath());
InputStream input = new FileInputStream(file);
// ...
}
or
public MyClass() {
InputStream input = getClass().getResourceAsStream("filename.txt");
// ...
}
Try giving an absolute path to the filename.
Also, post the code so that we can see what exactly you're trying.
When you are opening a file with a relative file name in Java (and in general) it opens it relative to the working directory.
you can find the current working directory of your process using
String workindDir = new File(".").getAbsoultePath()
Make sure you are running your program from the correct directory (or change the file name so that it will be relative to where you are running it from).
If you're using Eclipse (or a similar IDE), the problem arises from the fact that your program is run from a few directories above where the actual source is located. Try moving your file up a level or two in the project tree.
Check out this question for more detail.
The simplest solution is to create a new file, then see where the output file is. That is the correct place to put your input file into.
If you put the file and the class working with it under same package can you use this:
Class A {
void readFile (String fileName) {
Url tmp = A.class.getResource (fileName);
// Or Url tmp = this.getClass().getResource (fileName);
File tmpFile = File (tmp);
if (tmpFile.exists())
System.out.print("I found the file.")
}
}
It will help if you read about classloaders.
say I have a text file input.txt which is located on the desktop
and input.txt has the following content
i came
i saw
i left
and below is the java code for reading that text file
public class ReadInputFromTextFile {
public static void main(String[] args) throws Exception
{
File file = new File(
"/Users/viveksingh/desktop/input.txt");
BufferedReader br
= new BufferedReader(new FileReader(file));
String st;
while ((st = br.readLine()) != null)
System.out.println(st);
}
}
output on the console:
i came
i saw
i left