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.
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;
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 ?
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();
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:
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
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..
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));
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 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.
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.
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.
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)
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
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...
JFrame parentFrame = new JFrame(); File f12=new File("E: ewfile.txt"); JFileChooser fileChooser = new JFileChooser(); fileChooser.setSelectedFile(f12); fileChooser.setDialogTitle("Specify a file to save"); int userSelection = fileChooser.showSaveDialog(parentFrame);
[code]....
I want to save selected text file as html file in target directory....
If 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.
I'm trying to make a search engine that will search my computer given a path with any options i decide. So if i decides to look for the file name "resume" with extension "pdf", it will give me the files in the directory, including subfiles.
import java.io.*; import java.util.*; import java.text.*; public class mySearch { private final static int path = 0; private final static int name = 1;
I'm using a PrimeFaces UploadedFile xhtml page to select a csv file to read and write using a managed bean (SuperCSVParser.java). The file is read and written to an entity class which then persists the data to a database. The application works fine if I specify a file path on the physical server and select a csv file on that file path. But for the production version I want the user to select ANY file name from ANY directory on their local system.
I know about the FacesContext methods and I've looked at some methods from the java.io File class. Most of these methods are about getting the path from the server, where I want to 'pass' the path String from the client machine to allow the uploaded file to go through. When I try with the below code I get:
java.io.FileNotFoundException: data.csv (The system cannot find the file specified)
I'd like to know what I'm doing as I prefer not to explicitly declare a path in the final app. I'm almost sure that's possible.
/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?
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 ?
