im having a bit of issues calling a java file, i have created a java file
#ManagedBean(name="pdfSearch")
public class pdfSearch {
public String NewDestination;
public void main(String[] args) throws IOException {
File dir = new File(NewDestination);
String[] extensions = new String[]{"pdf"};// Add more file formats here to disply, could use this later on to display to the user all the files they have uploaded
System.out.println("Getting all .pdf files");
List<File> files = (List<File>) FileUtils.listFiles(dir, extensions, true);
for (File file : files) {
System.out.println(file.getCanonicalPath());
}
}
}
But i can not call this to run from a command button
<p:commandButton action="#{pdfSearch.main}" value="Search" ajax="False"/>
The filter.finder method is not a JSF Action method. That is meant to be a method of the signature
public String methodName()
That won't work in stock-standard JSF. I'd brush up on the JSF documentation for action methods.
Also you have written that you could use .toLowerCase(), if you wanted to search for files that are lowercase only. But this is not the case.
It will just make it see the letters as all lower case if you then wanted to order in them in lexiographical order for example.
Related
I have a method which creates new file after every execution I don't want to hardcode file path in code so I added a new property in application.properties file like
jmeter.jmx.path=D:\\PerformanceTesting\\JMXFiles\\
and instance variable which holds value like
#Value("${jmeter.jmx.path}")
private String jmxPath;
want to get the value of a variable inside method
public void saveAsJmxFile(HashTree projectTree, String fileName) throws IOException {
//TODO
SaveService.saveTree(projectTree, new FileOutputStream(jmxPath+fileName+".jmx"));
}
its not woking for me, but if i hardcode then it i'll work.
public void saveAsJmxFile(HashTree projectTree, String fileName) throws IOException {
//TODO remove hardcoded jmxPath
SaveService.saveTree(projectTree, new
FileOutputStream("D:\\PerformanceTesting\\JMXFiles\\"+fileName+".jmx"));
}
just make sure that the directory is exist
Files.createDirectories(Paths.get(jmxPath));
i'm using java8+ nio here
I downloaded a text file by a click button functionality, using Selenium Java.
then the file is downloaded to a particular location in the system, for example,
C://myAppfiles.
But I can't access that downloaded folder because of some reason. But I have to read that file while downloading.
How to do it? is it possible to read that file from the browser(chrome) using selenium or any other method is available?
so I'd suggest to do the following:
wait until file download is done completely.
After that- try to list all the files in the given directory:
all files inside folder and sub-folder
public static void main(String[]args)
{
File curDir = new File(".");
getAllFiles(curDir);
}
private static void getAllFiles(File curDir) {
File[] filesList = curDir.listFiles();
for(File f : filesList){
if(f.isDirectory())
getAllFiles(f);
if(f.isFile()){
System.out.println(f.getName());
}
}
}
files/folder only
public static void main(String[]args)
{
File curDir = new File(".");
getAllFiles(curDir);
}
private static void getAllFiles(File curDir) {
File[] filesList = curDir.listFiles();
for(File f : filesList){
if(f.isDirectory())
System.out.println(f.getName());
if(f.isFile()){
System.out.println(f.getName());
}
}
}
That will help You to understand if there any files at all (in the given directory).
Dont forget to make paths platform independent (to the folder/ file), like:
//platform independent and safe to use across Unix and Windows
File fileSafe = new File("tmp"+File.separator+"myDownloadedFile.txt");
Also, You might want to check whether file actually exists via Path methods.
import java.nio.file.Files;
import java.nio.file.LinkOption;
import java.nio.file.Path;
import java.nio.file.Paths;
public class Main {
public static void main(String[] args) throws Exception {
Path filePath= Paths.get("C:\\myAppfiles\\downloaded.txt");
System.out.println("if exists: " + Files.exists(firstPath));
}
}
Additionally, path suggests You to check some other options on the file:
The following code snippet verifies that a particular file exists and that the program has the ability to execute the file.
Path file = ...;
boolean isRegularExecutableFile = Files.isRegularFile(file) &
Files.isReadable(file) & Files.isExecutable(file);
Once You face any exception- feel free to post it here.
Hope this helps You
I want to delete the file which is opened and done writing but not closed. Please refer to code below:
Class A (can't be changed):
import java.io.FileOutputStream;
public class A {
public void run(String file) throws Exception {
FileOutputStream s = new FileOutputStream(file);
}
}
Class B:
import java.io.File;
import java.nio.file.Files;
import java.nio.file.Paths;
public class B {
public static void main(String[] args) throws Exception {
String path = "D:\\CONFLUX_HOME\\TestClient\\Maps\\test\\newTest.txt";
A a = new A();
a.run(path);
File f = new File(path);
Files.delete(Paths.get(f.getAbsolutePath()));
}
}
In Class A , just open the stream without closing the file.
In class B , calling A's run method and then try to delete the file.
Since the file is still opened. I'm unable to delete the file.
Error is :
The process cannot access the file because it is being used by another process.
Actual Scenario is :
We are loading the jars dynamically. Classes inside jar are creating the file. When there is an exception, a file gets created whose size will be 0 bytes. We need to delete this file. Since the file is not closed during the exception, we can't delete the file.
We could fix the issue if we could close the streams in the jar classes, but we can't modify the jars that create the files as they are client specific jars.
Please suggest how to delete the opened file, without modifying the code in class A.
Make sure you close the file, even if there was an Exception when writing to it.
E.g.
public void run(String file) throws Exception {
FileOutputStream s = null;
try {
s = new FileOutputStream(file);
} finally {
try {
s.close();
} catch(Exception e) {
// log this exception
}
}
}
You have to close the file before any delete operation as firstly its a bad practice and second is it will lead to memory leaks.
If you are using Tomcat, it is possible to set AntiLockingOption and antiJARLocking in $CATALINA_HOME/conf/context.xml for Windows:
<Context antiJARLocking="true" antiResourceLocking="true" >
Important note:
The antiResourceLocking option can stop JSPs from redeploying when they are edited requiring a redeploy.
Read more about this option:
http://tomcat.apache.org/tomcat-7.0-doc/config/context.html
antiResourceLocking:
If true, Tomcat will prevent any file locking. This will significantly impact startup time of applications, but allows full webapp hot deploy and undeploy on platforms or configurations where file locking can occur. If not specified, the default value is false.
Pass the resource as a parameter and it becomes the caller's responsibility to clear up the resources
public void run(FileOutputStream stream) throws Exception {
...
}
caller:
try(FileStream stream = new FileStream(path)){
A a = new A();
a.run(stream);
}catch(Exception e){
.. exception handling
}
Updated according to OPs comment.
Another approach could be to subclass A and override run().
public static void main(String[] args) throws Exception {
String path = "D:\\CONFLUX_HOME\\TestClient\\Maps\\test\\newTest.txt";
A a = new A() {
#Override
public void run(String file) throws Exception {
FileOutputStream s = new FileOutputStream(file);
s.close();
}
};
a.run(path);
File f = new File(path);
Files.delete(Paths.get(f.getAbsolutePath()));
System.out.println("foo");
}
I don't think you'll find a pure java solution to this problem. One option is to install Unlocker (being careful to hit "Skip" on all the junkware) and invoke it from your code.
If you have UAC enabled, you'll also need to be running your java in an elevated process (e.g. start command prompt as Administrator). Then, assuming unlocker is in C:\Program Files\Unlocker:
Process p = new ProcessBuilder("c:\\Program Files\\Unlocker\\Unlocker.exe",path,"-s").start();
p.waitFor();
And after that you can delete the file as before. Or you could use "-d" instead of "-s" and Unlocker will delete the file for you.
I have File object which looks like this
public class FileTO implements Serializable{
private String fileName;
private String filePath;
public String getFileName() {
return fileName;
}
public void setFileName(String fileName) {
this.fileName = fileName;
}
public String getFilePath() {
return filePath;
}
public void setFilePath(String filePath) {
this.filePath = filePath;
}
Of course there are lot other objects in my struts action response, which I am not listing here.
After the action completes, the filePath will be populated with the actual path of the file where it resides so that the file can be downloaded. I want to display the fileName and filePath in a <s:a> tag.
Goal is to have href point to filePath. I tried play with OGNL i.e #, %{}, $() and none seems to display the link properly.
Eg:
<s:a href="?????????"> Click to the get the File </s:a>
The question is a bit unclear but from what I understand, you're looking for s:url tag.
Link
<s:a href="?????????"> Click to the get the File </s:a>
If the file is in an accessible folder:
<s:a href="http://server.with.files.com/path/to/file/fileName.txt">
Click to the get the File
</s:a>
If the file is in a folder that is protected, not accessible, or is inside the webapplication, or comes from a database, etc... you should call an Action (or a Servlet, if not in Struts2), that should read the file, and return a Stream result. Read:
How to use Stream result from Action
How to download a file from Action / Servlet
To understand the OGNL syntax, instead, read this answer.
i get the error "AWT-EventQueue-0 java.lang.IllegalArgumentException: URI is not hierarchical".
-I'm trying to use the java.awt.Desktop api to open a text file with the OS's default application.
-The application i'm running is launched from the autorunning jar.
I understand that getting a "file from a file" is not the correct way and that it's called resource. I still can't open it and can't figure out how to do this.
open(new File((this.getClass().getResource("prova.txt")).toURI()));
Is there a way to open the resource with the standard os application from my application?
Thx :)
You'd have to extract the file from the Jar to the temp folder and open that temporary file, much like you would do with files in a Zip-file (which a Jar basically is).
You do not have to extract file to /tmp folder. You can read it directly using `getClass().getResourceAsStream()'. But note that path depend on where your txt file is and what's your class' package. If your txt file is packaged in root of jar use '"/prova.txt"'. (pay attention on leading slash).
I don't think you can open it with external applications. As far as i know, all installers extract their compressed content to a temp location and delete them afterwards.
But you can do it inside your Java code with Class.getResource(String name)
http://download.oracle.com/javase/6/docs/api/java/lang/Class.html#getResource(java.lang.String)
Wrong
open(new File((this.getClass().getResource("prova.txt")).toURI()));
Right
/**
Do you accept the License Agreement of XYZ app.?
*/
import java.awt.Dimension;
import javax.swing.*;
import java.net.URL;
import java.io.File;
import java.io.IOException;
class ShowThyself {
public static void main(String[] args) throws Exception {
// get an URL to a document..
File file = new File("ShowThyself.java");
final URL url = file.toURI().toURL();
// ..then do this
SwingUtilities.invokeLater( new Runnable() {
public void run() {
JEditorPane license = new JEditorPane();
try {
license.setPage(url);
JScrollPane licenseScroll = new JScrollPane(license);
licenseScroll.setPreferredSize(new Dimension(305,90));
int result = JOptionPane.showConfirmDialog(
null,
licenseScroll,
"EULA",
JOptionPane.OK_CANCEL_OPTION);
if (result==JOptionPane.OK_OPTION) {
System.out.println("Install!");
} else {
System.out.println("Maybe later..");
}
} catch(IOException ioe) {
JOptionPane.showMessageDialog(
null,
"Could not read license!");
}
}
});
}
}
There is JarFile and JarEntry classes from JDK. This allows to load a file from JarFile.
JarFile jarFile = new JarFile("jar_file_Name");
JarEntry entry = jarFile.getJarEntry("resource_file_Name_inside_jar");
InputStream stream = jarFile.getInputStream(entry); // this input stream can be used for specific need
If what you're passing to can accept a java.net.URLthis will work:
this.getClass().getResource("prova.txt")).toURI().toURL()