i have a folder which contains two types of files (for eg: .csv and .tiff) with same names like, abc.csv and abc.tiff.How to compare the list of files with same names while ignoring their extensions.And my second task is to copy the content from csv file and i have to use that name as file name of .tiff file(Ex: if i have a1 as content in csv file, i have to change .tiff file name from abc.tiff to a1.tiff)
View RepliesI am working with jsoup right now and I am trying to get a range of file extensions such as doc, docx, txt, pdf, and so on. Anyways i have looked through the jsoup api and cant seem to find what I am looking for. The closest thing I have found is
Elements files = doc.select("a[href$=.doc]");
This should work but only grabs one extension. How to grab multiple at one time? My guess would be:
Elements files = doc.select("a[href$=.doc]").select("a[href$=.docx]").select("a[href$=.txt]");
However I'm not sure for certain. Anyways I will provide a link for the api .....
I have two big text file with approx 16 Lakh (1.6 Million) records in each. Both file contains Strings in each line. I would like to compare two files and find the descripencies and print them into separate file.
View Replies View RelatedI would like to create a component to detect the file being modify before process.is it the right way to detect the file modification based on file size value?
Below are the flow:
1. Get the file size of a file
2. Used file size value encrypt it with MD5 algorithm, and say it generated us encrypted value "0123sdf"
3. to avoid user modify the file content, before file process, we take the file and do the encryption with md5 again, if it return value "0123sdf", then we are sure it doesn't have modification.
my question:
a. is it the right approach to detect file modification?
b. what the library advise to use or using java.security.DigestInputStream will do?
I need to make some operations with a file. I struggle to understand why Apache is throwing this error:
(The system cannot find the file specified)
My file is located in the root directory of the project.
And here is how I indicate the path to it:
String filePath = "default.jpg";
InputStream inputStream = new FileInputStream(new File(filePath));
If I put the file in the Eclipse folder, it works... But I need it to work if I put it in my project's root folder.
this is the following task i am trying to do. Write an interface Directory to manipulate entries in a telephone directory for the University. The interface must support the following operations:
1) The ability to insert new entries into the directory. Entries should be stored in alphabetical order of surname.
2) Delete entries from the directory either by name or number
3) Provide a lookup method that will find the extension no of a member of staff given
his/her name. You should try to make this method run as efficiently as possible.
4) Change a person's telephone number
5) Print the telephone directory in a neatly tabulated fashion.
Write a class ArrayDirectory that implements this interface using an array to store the telephone directory information. The class must store the surname and initial for each member of staff and their telephone extension (a four digit number which may start with a zero). You may find it useful to define a class Entry to store information about individual entries. The entries should be read into the array from a text file consisting of multiple lines in the following format:
Surname<tab>Initials<tab>Telephone extension
The part that i am stuck on is trying to do the entry method, delete entries method, loop up method, change persons telephone number and be able to print it.
This is a snippet code that im developing. (no source of it) . I am facing a trouble to extract file in the same directory as the source file (example: extract here option)
public class HelperAction {
public static void unzip(String zipFilePath)throws IOException {
File dir = new File(zipFilePath.substring(0,zipFilePath.length()));
// create output directory if it doesn't exist
if(!dir.exists()) dir.mkdirs();
FileInputStream fis;
[Code] ....
Consider the url-pattern:
<url-pattern> /Beer/* </url-pattern>
The web app structure is: -
webapps
|
-->AdviceApp
|
-->WEB-INF
--> {{Beer}} This can be real or virtual.
My book says that /Beer/* can be a real or a virtual directory. What is the difference between the two and how do I create a virtual directory in tomcat ?
So i have this program that is supposed to execute a command when a button is clicked. I get an error: java.io.IOException: Cannot run program "/Users /brianallison/Documents/Java/RELAP5": error=2, No such file or directory
The actual PATH should be /Users/brianallison/Documents/Java/RELAP5 GUI/issrs/Dist.At least that is where the Java app is and the executable that I am trying to execute when the butotn is clicked. Executable name is relap5.x or relap5.exe depending on the OS.
private void btnRunRelapActionPerformed(java.awt.event.ActionEvent evt) {
String in = " -i " + tfIntdta.getText();
String rst = " - r " + tfRstplt.getText();
String out = " -o " + tfOutdta.getText();
String strip = tfStpdta.getText();
String guistring = "-n gui";
String wd = System.getProperty("user.dir");
String osver = System.getProperty("os.name");
String app = "";
[code].....
I am trying to unzip a zip file to a directory but I've tried several different codes and none of them work.The zip structure is like this:
test.zip
New Text Doc.txtNew Folder
New Text Doc.txt
and this is the code I am using
try {
// Open the zip file
ZipFile zipFile = new ZipFile(outputFileName);
Enumeration<?> enu = zipFile.entries();
while (enu.hasMoreElements()) {
ZipEntry zipEntry = (ZipEntry) enu.nextElement();
[code]....
I am working on a java logic game and I want it to be able to work on other people's Mac's, so I tried to figure out how to make it create a folder in which it can create files. The file creating is going fine however the folder never seems to create.Here is the code I attempted to use:
File logicFile = new File("/Library/Application Support/LogicGameSupport");
if(!logicFile.exists()){
if(logicFile.mkdir()){
System.out.println("Directory success");
}else{
System.out.println("Directory failed");
}
}
Here my code . Iam trying to open text file in window_pane it will open when i click on save it will ask destination to save file if i save example like Read.txt it will save...when i go and check in that folder Read.txt file won't found...what is the error ....
Java Code:
import javax.swing.*;
import javax.swing.filechooser.FileFilter;
import javax.swing.filechooser.FileNameExtensionFilter;
import java.io.*;
import java.net.MalformedURLException;
import java.awt.*;
import java.awt.event.*;
public class Read extends JFrame
[Code] .....
Here my code am trying to open text file in window_pane it will open when i click on save it will ask destination to save file if i save example like Read.txt it will save...when i go and check in that folder Read.txt file won't found..
import javax.swing.*;
import javax.swing.filechooser.FileFilter;
import javax.swing.filechooser.FileNameExtensionFilter;
import java.io.*;
import java.net.MalformedURLException;
import java.awt.*;
import java.awt.event.*;
[code]....
Are the WebDAV extensions related to REST principles? What I mean by that is, REST is an abstraction of the WWW architecture and the WebDAV is just an extension to the HTTP. So can we say that when I speak in terms of REST API's, I can restrict myself to HTTP and not HTTP extensions like WebDAV?
View Replies View RelatedensureFileIsVisible not working.
JFileChooser fileChooser = new JFileChooser();
fileChooser.setCurrentDirectory(new File(currentDirectory));
File f = new File(currentDirectory);
ArrayList<String> names = new ArrayList<>(Arrays.asList(f.list()));
for (String s : names)
if (s.startsWith(yearofIssueTXT.getText())) {
fileChooser.setSelectedFile(new File(s));
fileChooser.ensureFileIsVisible(fileChooser.getSelectedFile());
break;
}
int result = fileChooser.showOpenDialog(addNewCoinBody);
The file is not visible because it is at almost the end of the files in the directory. I want the filechooser to automatically scroll so that the file is seen in the list.
Is there a way I can package a file INTO a jar file from a directory, not the other way around?
View Replies View RelatedI don't really understand the question is the thing. Its about calling a file in the java program. I read the book and i can see how to call a file from a directory and how to save it and edit it but I don't understand what the instructor is asking for. Mostly what files need called and println.
Assuming the object that manages output will be called writer, write a statement that opens a text file called home.html for output by println statements.
To be file-writable and file-readable, an object must be an instance of a class that implements___________.
Write a statemnt that opens a file for input of objects and assigns a reference to the connection to ObjectIn. Assume the file is in the current directory, with the name automobiles. data.
I'm new to Methods and do not really know how to write a method header. What would an appropriate header look like for the following instance? getFilename: This method takes no parameters. It asks the user for a filename. If that file exists, it returns the filename. If it does not exist, it asks the user for another filename. It continues to ask for filenames until the user gives the name of a file that exists.
Also, how would I make a return tag for this and call it into the main method.
In Java® identifiers, you are allowed letters and numbers (also _ $£¢€ etc, but you should avoid them in normal identifiers). So you cannot have spaces. You cannot write public class Hello World because the javac tool will see World as a separate identifier and not understand what it means and will fail to compile the code. You must write public class HelloWorld instead. And because the class is labelled public you must call the source file Hello World. java. Since you can't have two classes with the same [fully‑qualified] name, you cannot write two public classes in the same source file.
“What about names of source files?” somebody will ask. Well, some file systems will permit spaces in file names; ext4 will and I suspect so will NTFS. Can you write file names with spaces in? You would have to have a different name of the class inside the file, because you can't have spaces, and you therefore cannot make the class public, but maybe you can write a package‑private class with a different name.
On ext4, you have to write out the name of the file and the shell will interpret the space as meaning there are two different file, so you have to escape the space.
campbell@campbellsComputer:~/java$ gedit My First Class.java
// My First Class.java
class Foo
{
public static void main(String... args)
{
System.out.println("Hello, World!");
}
}
campbell@campbellsComputer:~/java$ javac My First Class.java
campbell@campbellsComputer:~/java$ java Foo
I'm new to Methods and do not really know how to write a method header. What would an appropriate header look like for the following instance? getFilename: This method takes no parameters. It asks the user for a filename. If that file exists, it returns the filename. If it does not exist, it asks the user for another filename. It continues to ask for filenames until the user gives the name of a file that exists.
Also, how would I make a return tag for this and call it into the main method.
My problem is that I want to load an Image from the source directory in Canavas' paint() method.
My code is currently (in a try/catch)
g.drawImage(ImageIO.read(new File(imageLocation)), imageX, imageY, null);
Where imageLocation leads to an image on my HDD ("C:UsersVladdeDocumentsFolders-Filesfolder.png")
I'm trying to add a CLASSPATH to a manifest file linking (not sure If that's the right word) to an sqljdbc4.jar file.
If I copy the file Into my root folder and use CLASSPATH: sqljdbc4.jar It works fine.
Class-Path: sqljdbc4.jar
Main-Class: com.ncntech.go
If I try CLASSPATH: C:Program FilesMicrosoft JDBC Driver 4.0 for SQL Serversqljdbc_4.0enusqljdbc4.jar I receive a "Could not find or load main class" error.
Class-Path: C:Program FilesMicrosoft JDBC Driver 4.0 for SQL Serversqljdbc_4.0enusqljdbc4.jar
Main-Class: com.ncntech.go
Is the syntax correct for specifying a directory path ?
I need to write the exact directory path like C:LisaestUpdate to a properties file.
I am trying it by
FileInputStream in = new FileInputStream(test.properties);
Properties props = new Properties();
props.load(in);
in.close();
FileOutputStream out = new FileOutputStream(newprop.properties);
props.setProperty("myDirectory","C:Lisa estUpdate" );
props.store(out, null);
out.close();
but the properties file is updated as C:Lisa estUpdate
Extra comes before :.
How can I remove that.
Even I tried it with an command but got same output.
How the cacerts file if populated within the java install directory?
Are there standard certificates that come with the jre install? Are they pulled from the OS somehow?
I have a text file containing numbers and names as shown in the example below:
61133128805241Albert Rosenberg
64763645543509Joel
79256337754219Michael Burton
I'm making a program that will read the file and put the numbers in a list of int arrays and names in another list of strings. In my program i created two classes. One will receive the numbers and the other will receive the names. But i only can read the numbers! How can I read everything and separate into two different lists?
try( BufferedReader br = new BufferedReader( new FileReader( jFileChooser1.getSelectedFile().getAbsolutePath() ) ) ) {
hist = new Historic();//Will recieve the array of numbers.
while( br.ready() ) {
int line[] = new int[ 7 ];
[Code] ...
I'm trying to do a program that send a single file from a client to a default directory of a Server. The program use the datagramSocket and datagramPacket for the transfer via UDP. The client sent the packet of the file correctly and the server start the receiving but every time the Server class crashes after it's receiving 4/5 packet (exactly 8192 byte)
Then i put the code of the 2 class and the 2 output.
OUTPUT CLIENT:
PROGRAM TRANSFER PACKAGES
name of file is : Doc1.pdf
his dimension: 11408 byte
*** start sending
- 1024 byte
- 2048 byte
- 3072 byte
- 4096 byte
- 5120 byte
- 6144 byte
- 7168 byte
- 8192 byte
- 9216 byte
- 10240 byte
- 11264 byte
- 11408 byte
*** FILE SUCCESFULL SENDED***
BUILD SUCCESSFUL (total time: 3 seconds)
OUTPUT SERVER
PROGRAM TRANSFER PACKAGES
START NEW CONNECTION
Directory: C:prova
Server waiting in port: 9876
Waiting Client...
Name of file is : Doc1.pdf
Dimension : 11408
[Code] ....
In this point the program go in loop and didn't anything. I don't understand how can i resolve it...
SERVER
package udp_pacchetti;
import java.io.ByteArrayInputStream;
import java.io.File;
import java.io.FileInputStream;
import java.io.FileNotFoundException;
import java.io.FileOutputStream;
[Code] ...