In my program I have to get directory name&address, in this directory is none, one, or more *.txt(maybe not *.txt files) and I have to read info from all of them.
On C++ is very easy to do that with "findnext, find last, findfirst", so how can I do this in Java?
Requirements - Use only standard Java API and no apache file utils for this.
Most of the answers I found on the internet either dont meet this requirement or load all file names into an array which can consume too much memory when no. of files = 20,000+. how I can do this. Is there also a way to keep track of new files that were added during the execution of the loop in this code ?
ensureFileIsVisible 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.
I've written a program to read files in a directory but I'd like for it to only read the text files of that directory.
import java.io.*;
public class Data {
public static void main(String[] args) throws IOException {
String target_dir = "C:files";
File dir = new File(target_dir);
File[] files = dir.listFiles();
[Code] .....
I'm having another issue. I have 2 java source files(see below). They are place in the same directory. How do I compile them using classpath?
I have already tried 1st attempt :
javac -cp com.companyname.interview.DuplicateReplace.java DuplicateReplaceTest.java [did not work!]
2nd attempt: javac -cp DuplicateReplace.java DuplicateReplaceTest.java [again, did not work!]
package com.companyname.interview;
public class DuplicateReplace { /* code */}
[Code] ....
Within tomcat "webapps" directory, have lots of scanned PDF files (size in KB and MB).Using Java and JSP, I want to search for any user entered text in these scanned PDF files.The result must display the name of those PDF files which has that entered text.
View Replies View RelatedIf I specify the directory in file type or path string type how can I get a list of files that are file type .yml in that folder, so I can loop through and do something with each single file.yml in that specified directory.
View Replies View RelatedI need to design a gui that will find out all the dead symbolic link from the directories. And I used the logic and syntax I find out in java orcle docs. And If I will give the direct path name of the symbolic file then it is running fine with the output and enable to find the smbolic link is active or not. Now the real problem is that , I want that logic of detection should run out through the complete files and subdirectories of the directory. So Then I used the tree walk concept to transver through the complete file system of the directory and embed the logic of the detection inside it for detection of the dead symbolic link. But now it showing erroe which is not approacable for me.
import java.io.File;
import java.io.IOException;
import java.io.PrintStream;
import java.nio.file.Files;
import java.nio.file.Path;
import java.nio.file.Paths;
public class To_Transver {
[Code]...
How can I Cache the data I'm reading from a collection of text files in a directory using a TreeMap? Currently my program reads the data from several text files in a directory and the saves that information in a text file called output.txt. I would like to cache this data in order to use it later. How can I do this using the TreeMap Class? These are the keys,values: TreeMap The data I'd like to Cache is (date from the file, time of the file, current time).
import java.io.*;
public class CacheData {
public static void main(String[] args) throws IOException {
String target_dir = "C:Files";
String output = "C:Filesoutput.txt";
File dir = new File(target_dir);
File[] files = dir.listFiles();
[Code] ....
I am trying to understand how package(directory structure) and JAR files work interchangeably. From my understanding JAR is a package file format typically used to aggregate many Java classes. Since many classes are usually contained inside a package, I should be able to create a JAR file from a package, and use them interchangeably. A single JAR file is easier to handle then multiple classes and folders.
To test my understanding of the concept, I am working with the following project structure,
/development_directory
____/BookPackage
________Book.java
____BookDemo.java
Book.java contains the following code,
package BookPackage;
public class Book{
private String bookname;
public Book(String str){
bookname = str;
}
public String getBookName(){
return bookname;
}
}
and BookDemo.java contains the following code,
import BookPackage.Book;
class BookDemo{
public static void main(String ... args){
Book b = new Book("Lord of the Rings");
System.out.println(b.getBookName());
}
}
From the development_directory, I used the following commands to compile and run the project,
development_directory/ javac BookDemo.java
development_directory/ java BookDemo
...and it compiles and works as expected.Now, I will try to convert the BookPackage folder(package) into single jar file for portability, and I am using the following command to do that,development_directory/ jar -cf bookpkg.jar BookPackage ...this command generates the bookpkg.jar file.
I am going to remove the BookPackage folder to see if bookpkg.jar file really works as a replacement for the BookPackage directory structure.
After removal of BookPackage folder, the development_directory structure looks like this,
/development_directory
____bookpkg.jar
____BookDemo.java
Now, I tried to use the following commands to see if the program works as before,
// to compile
development_directory/ javac -cp ./bookpkg.jar BookDemo.java
...it compiles without any error - this means the replacement of bookpkg.jar for BookPackage(package) folder structure is working - Am I correct?
// to run
development_directory/ java -cp ./bookpkg.jar BookDemo
...And I hit an error,
Error: Could not find or load main class BookDemo
But the BookDemo class is at default namespace - why it JRE won't be able to find the main method?
And, the way I am trying to replace package folder structure with a single JAR file - am I getting the concept correctly?
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.
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 ?
How to find the directories of the files. Unfortunately, I have not been able to grasp the concept behind this idea. If I find to see if a particular file is there, can I not go to the file directly rather then having to use CMD (command). What is the purpose of CMD?
Secondly, I took a c++ class and now I'm in my second course which is integration the materials from c++ in to java. We are learning about classes. I have not been able to understand what classes do for you...and I noticed that in c++ (visual studio) the main started like Java Code: int_tmain(...) mh_sh_highlight_all('java'); but the topic we covered yesterday started like Java Code:
public class helloworld
{
public class helloworld(String)
{}
} mh_sh_highlight_all('java');
Program is to find the most frequently used words across all the input files, where each word must appear at least once in each file.
View Replies View RelatedJava Code:
if(Directory.getText().matches("")){
JOptionPane.showMessageDialog(null, "Please search for the directory");
}
else if(fileName.getText().matches("")){
[Code].....
This is actually my question:
Java Code:
1. billing-20140527[09].txt has
a)XGMS,2014-05-27 10:08:04,122,PLAYER_VERIFY,VERIFY to LBA,0x580000,0xC0000,253040.
b)XGMS,2034-05-27 30:08:04,122,PLAYER_VERIFY,VERIFY to LBA,0x580000,0xC0000,253040.
2. billing-20140527[10].txt has
a)XCGS,2014-05-27 10:08:04,122,PLAYER_VERIFY,VERIFY to LBA,0x580000,0xC0000,253040.
b)HELO
[B]I try to find the number 1 in both text files, if lets say I input the text file name is billing, I can find the number 1 in both text file and output them:[/B]
a) XGMS,2014-05-27 10:08:04,122,PLAYER_VERIFY,VERIFY to LBA,0x580000,0xC0000,253040.
b) XCGS,2014-05-27 10:08:04,122,PLAYER_VERIFY,VERIFY to LBA,0x580000,0xC0000,253040.
[B]However, if I specify the text file name: billing-20140527[09].txt and find the number 1 inside the text file, it will only output:[/B]
a) XGMS,2014-05-27 10:08:04,122,PLAYER_VERIFY,VERIFY to LBA,0x580000,0xC0000,253040. mh_sh_highlight_all('java');
So far, I can't get this program to run,
What step to know to develop software..
View Replies View RelatedIn a program I created, I'm using a text file that contains some texts needed for the program. The method relevant to this is something like the following.
private String wordgen(){
try {
BufferedReader reader = new BufferedReader(new FileReader("src/Resources/adjectives.txt"));
Random rand = new Random();
int low = rand.nextInt(400);
String fil="";
int i=0;
while(i!=low){
[Code]...
The program runs fine in netbeans project but once the jar is created it does not corporate with the text file. ("null" is returned) How can I attach text files to jar and exe?
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 need to transformation the txt files into xml files, but each row txt files don't have same elements, for example the first book is composite one author
<books>
<book>Title</book>
<price>price</price>
<author>Author</author>
</books>
but the second book is composite two author
<books>
<book>Title</book>
<price>price</price>
<author>Author</author>
<author>Author2</author>
</books>
I would have the xml files with the number of authors variables, I can use JAXB for my problem?or not?
I see you can import in 2 ways:
import java.util.*;
or
import java.util.Scanner/Random/etc;
if I used the first one it imports all the librarys within that (directory?). If so would that add to a bigger file size?
I have a directories in UNIX:
/home/t_bmf/Java/HelloWorld/src/helloworld :will contain a .java file
/home/t_bmf/Java/HelloWorld/bin :will contain all .class file
Let say a have a code:
package helloworld;
public class HelloWorld {
public static void main(String[] arg) {
System.out.println("Hello World");
}
}
a command to compile this even outside the directory /home/t_bmf/Java/HelloWorld/src/helloworld
javac -d /home/t_bmf/Java/HelloWorld/bin /home/t_bmf/Java/HelloWorld/src/helloworld/HelloWorld.java
This will generate a directory /home/t_bmf/Java/HelloWorld/bin/helloworld and file inside this is HelloWorld.class
To run this program I must be in directory /home/t_bmf/Java/HelloWorld/bin and using this command:
java helloworld.HelloWorld
Question:
I already how to run the HelloWorld.class, but I must be in helloworld /home/t_bmf/Java/HelloWorld/bin to run it. Is there's a way to run the class even when I am not in directory /home/t_bmf/Java/HelloWorld/bin? Let's say I'm in /home/t_bmf, can I still run the HelloWorld.class?
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]....