I'm getting 'File not found!' no matter what I do.
FileInputStream fin;
try {
fin = new FileInputStream("foo.txt");
String str = IOUtils.toString(fin);
System.out.println(str);
} catch (FileNotFoundException f) {
System.out.println("File not found!");
}
Do you have a foo.txt in the directory that you are working in?.
If you are using command window and are at a location say, C:\, then your code expects foo.txt to be present there.
If your foo.txt is present in some other path, use the full path in your code.
If you temporarily add this line to your code:
System.out.println(new File("foo.txt").getAbsolutePath());
it should tell you where it expects to find the file. If the file isn't in that location, then you'll either have to specify the path or move the file so that it is.
Make sure you're using the correct path. Try right-clicking and go on to Properties and check the file's path. Copy and paste the path and replace all the \ to either / or \\.
Related
I am facing some problem to fix the right file path.
I have a configuration file and following is the entry in config.properties:
strMasterTestSuiteFilePath="D:\\KeywordDrivenFramework\\MasterTestSuiteFile.xls"
Then i tried to read this property as
Properties prop=new Properties();
prop.load("config.properties")
String strpath=prop.getProperty("strMasterTestSuiteFilePath")
Syso(strpath) //prints above path with single slash as D:\KeywordDrivenFramework\MasterTestSuiteFile.xls
//When i use the same var for File existance check it say not exists
File obj=new File(strpath)
if(obj.exists())
Syso("Exists....")
else
Syso("Does not exist....")
Why it is going to else block, even though the file exists at path?
How to overcome it?
I tried
String str= strpath.replaceAll("\","\\") //but i am getting some syntax error "The method replaceAll(String, String) in the type String is not applicable for the arguments (String)"
Can anyone help me how to overcome this?
Find the code which i am trying, where i am going wrong?
public void LoadMasterTestSuite()
{
String strGlobalConfigSettingFilePath=System.getProperty("user.dir")+"/src/GlobalConfigurationSettings.properties";
FileInputStream objFIS; //Variable to hold FileSystem Objet
File objFile; //Variable to hold File Object
try
{
objFIS = new FileInputStream(new File(strGlobalConfigSettingFilePath));
globalObjProp.load(objFIS);
strMasterTSFilePath=globalObjProp.getProperty("strMasterTestSuiteFilePath");
objAppLog.info("Master Test Suite File Path configured as "+strMasterTSFilePath);
}catch (FileNotFoundException e)
{
objAppLog.info("Unable to find Global Configuration Settings File. So aborting...");
e.printStackTrace();
}catch (IOException e)
{
e.printStackTrace();
}
String str=strMasterTSFilePath.replace("\\", "\\\\");
objFile=new File(str);
System.out.println(str);
if(objFile.exists())
{
LoadTestSuite(strMasterTSFilePath);
}
else
{
objAppLog.info("Master Test Suite File not found at Path: "+strMasterTSFilePath);
System.exit(-1);
}
}
I guess what you wanted to do was:
String str= strpath.replaceAll("\\\\","\\")
\ is a special character in a String. To Treat it like a normal character, put another \ in front of it, so '\' actually is '\\'.
And for your case, I think you want to use .replace() and not .replaceAll().
When you debug, is strpath getting the correct path from the config file? I would also make sure that you're referencing the correct file path (paste the location into Windows Explorer and see if the file exists). If D:\ is a shared drive, use the actual server location.
Also, you can make sure that the extension of the Excel workbook is .xls and not .xlsx.
I've a .txt file ("file.txt") in my netbeans "/build/classes" directory.
In the same directory there is the .class file compiled for the following code:
try {
File f = new File("file.txt");
Scanner sc = new Scanner(f);
}
catch (IOException e) {
System.out.println(e);
}
Debugging the code (breakpoint in "Scanner sc ..") an exception is launched and the following is printed:
java.io.FileNotFoundException: file.txt (the system can't find the
specified file)
I also tried using "/file.txt" and "//file.txt" but same result.
Thank you in advance for any hint
If you just use new File("pathtofile") that path is relative to your current working directory, which is not at all necessarily where your class files are.
If you are sure that the file is somewhere on your classpath, you could use the following pattern instead:
URL path = ClassLoader.getSystemResource("file.txt");
if(path==null) {
//The file was not found, insert error handling here
}
File f = new File(path.toURI());
The JVM will look for the file in the current working directory.
Where this is depends on your IDE settings (how your program is executed).
To figure out where it expects file.txt to be located, you could do
System.out.println(new File("."));
If it for instance outputs
/some/path/project/build
you should place file.txt in the build directory (or specify the proper path relative to the build directory).
Try:
File f = new File("./build/classes/file.txt");
Use "." to denote the current directory
String path = "./build/classes/file.txt";
File f = new File(path);
File Object loads, looking for match in its current directory.... which is Directly in Your project folder where your class files are loaded not in your source ..... put the file directly in the project folder
I am trying to open files with FileInputStream that have whitespaces in their names.
For example:
String fileName = "This is my file.txt";
String path = "/home/myUsername/folder/";
String filePath = path + filename;
f = new BufferedInputStream(new FileInputStream(filePath));
The result is that a FileNotFoundException is being thrown.
I tried to hardcode the filePath to "/home/myUserName/folder/This\\ is\\ my\\ file.txt" just to see if i should escape whitespace characters and it did not seem to work.
Any suggestions on this matter?
EDIT: Just to be on the same page with everyone viewing this question...opening a file without whitespace in its name works, one that has whitespaces fails. Permissions are not the issue here nor the folder separator.
File name with space works just fine
Here is my code
File f = new File("/Windows/F/Programming/Projects/NetBeans/TestApplications/database prop.properties");
System.out.println(f.exists());
try
{
FileInputStream stream = new FileInputStream(f);
}
catch (FileNotFoundException ex)
{
System.out.println(ex.getMessage());
}
f.exists() returns true always without any problem
Looks like you have a problem rather with the file separator than the whitespace in your file names. Have you tried using
System.getProperty("file.separator")
instead of your '/' in the path variable?
No, you do not need to escape whitespaces.
If the code throws FileNotFoundException, then the file doesn't exist (or, perhaps, you lack requisite permissions to access it).
If permissions are fine, and you think that the file exists, make sure that it's called what you think it's called. In particular, make sure that the file name does not contain any non-printable characters, inadvertent leading or trailing whitespaces etc. For this, ls -b might be helpful.
Normally whitespace in path should't matter. Just make sure when you're passing path from external source (like command line), that it doesn't contain whitespace at the end:
File file = new File(path.trim());
In case you want to have path without spaces, you can convert it to URI and then back to path
try {
URI u = new URI(path.trim().replaceAll("\\u0020", "%20"));
File file = new File(u.getPath());
} catch (URISyntaxException ex) {
Exceptions.printStackTrace(ex);
}
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