I'm studying Tomcat architecture. I wanted to fix org.apache.jasper.servlet.JspServlet and confirm its result.
For example i add the following code in JspServlet.java init() method.
FileWriter test = new FileWriter("c:\\worktest.txt");
test.write("test!!");
test.close();
After edit code, compile it and repackage jasper.jar file. And then I replace old jasper with new thing. of course restart tomcat.
I think that adding code will work and write text file in C drive. But result doesn't exist.
I've already check .jsp file, but it doesn't have any problem. because of creating JAVA and CLASS file.
There's something wrong with my way?
Related
so I did run into one very weird issue. The idea is simple: create temp dir, place some files in it and then try to access them. Now the problem is that calling File.createTempDir() or Files.createTempDirectory(prefix) creates new file inside AppData/Local/temp with shortened path, so the full path to folder looks something like C:/Users/FirstNam~1/AppData/Local/Temp/myFolder/myFile.txt instead of C:/Users/FirstName LastName/AppData/Local/Temp/myFolder.myFile.txt.
The difference is that generated path inside java contains FirstNam~1 instead of FistName SecondName. Java then throws exception File Not Found.
When I try to copy and paste shortened path into file explorer I get an error saying that file does not exist, but if I do change shortened path to full one then file opens and it works as intended.
Is there any way to fix it? Ether by forcing java to use full path names or enabling something in windows? I did Enable NTFS long paths policy, but it did not help.
This is happening when using java 8/11 and windows 10 running on VM, project is using AGP and gradle. Temp file is created inside groovy file that extends Plugin<Project>
Just when I lose hope and create a ticket, couple hours after that I find the answer. So, java has method Path.toRealPath() which solves this ~1 issue. After using this method paths no longer contain shortening and are correctly resolved.
EDIT: looks like java is doing everything correct and paths are actually valid, problem did come from library that I'm using and it's a bug with it.
Yesterday, I had a problem because I couldn't manage to open a xml file (it owuld give me a FileNotFoundException) located in the ressources folder of my .jar file, which I managed to open on eclipse using the following lines of code. You can see my old problem here. This was my code with the problem :
File xmlFile = new File("ressources/emitter.xml");
ConfigurableEmitter emitter = ParticleIO.loadEmitter(xmlFile);
Someone told me it that one way was to use getClassLoader().getRessourceAsStream method to open a xml file in a .jar file that was exported
InputStream i= this.getClass().getClassLoader().getResourceAsStream("ressources/emitter.xml");
ConfigurableEmitter emitter = ParticleIO.loadEmitter(i);
Unfortunately, that solution only works when I export my project into a .jar file, so if I want to go back debugging my program, I have to take the old code that would only works on eclipse.
My question is: is there any better way to do this without having to change my code if I want to export it or if I want to debug it?
Thank you
edit :
Thank you all, it works perfectly fine now
my problem was that I put my ressources folder like that :
+project
+src
+ressources
+emitter.xml
InputStream i= this.getClass().getClassLoader().getResourceAsStream("/ressources/emitter.xml");
The above should work in both cases (Note is is /resources/.... This is assuming say your directory structure is below:
MyProject
+src
+ressources
emitter.xml
Place the file alongside your source files, then you can use the getResourceAsStream() method in both cases. Don't forget to update the path (which should be the package name of your class, but with slashes instead of dots).
My question is: is there any better way to do this without having to
change my code if I want to export it or if I want to debug it?
Yes, use Maven. Maven will handle that and it hooks into Eclipse beautifully (NetBeans too!) What you do is place the resource in src/main/resources and then you can have Eclipse run the test goal of the Maven project or you can just run mvn test from the command line. Another advantage of using Maven here is that you can also have src/test/resources/emitter.xml which overrides the one in src/main with environment-specific test instructions and it won't affect your deployment.
InputStream i= getClass().getClassLoader().getResourceAsStream("ressources/emitter.xml");
or
InputStream i= getClass().getResourceAsStream("/ressources/emitter.xml");
(note the absolute positioning)
both work when the class is in the same jar, on the same class path.
In the jar the names must be case sensitive, but as the jar already works. Ensure that the ressources directory is on the class path too, or copied to the target directory.
As "ressources" is probably configured yourself (not named "resources" as in English), you probably need to add it to the build somehow.
I know this question is not new but I am posting this after going through enough googling around.
I have a jar file which I don't control. I mean I have not written the Java class in there and all I know is that there is a class named "hist" in it which takes two arrays and gives out the output. Now I have to use PHP to call this class "hist" in the jar file named "histvol.jar". I have installed the PHP/Java Bridge and it is installed correctly.
But I don't understand how to call this jar file in PHP and where to place this jar file.
This is what I did:
Installed tomcat and php-javabridge
Placed the jarfile "histvol.jar" under Tomcat/webapps/JavaBridgeTemplate621/webinf/lib/
Went to xampp/htdocs and created a file named testjava.php
<?php
require_once("http://localhost:8080/JavaBridgeTemplate621/java/Java.inc");
$System = java("java.lang.System");
$myclass=java("histvol");
echo $System->getProperties();
?>
Class not found exception is expected because I am not calling it anywhere in PHP, but how do I call it?
I am lost, please help (I don't know a word of Java).
Ok I got it finally, I just had to do this:-
$myclass=new java("histvolone.histvol");
instead of
$myclass=new java("histvol");
It worked !
I had problems while finding the path of file(s) in Netbeans..
Problem is already solved (checked answer).
Today I noticed another problem: When project is finished,
I have to execute the generated .jar to launch the program, but it doesn't work because an error occurs: NullPointer (where to load a file) when accessing/openning jar outside Netbeans.
Is it possible to open a file with the class file in Java/Netbeans which works in Netbeans and even in any directory?
I've found already some threads about my problem in site but none was helpful.
Code:
File file = new File(URLDecoder.decode(this.getClass().getResource("file.xml").getFile(), "UTF-8"));
The problem you have is that File only refer to files on the filesystem, not files in jars.
If you want a more generic locator, use a URL which is what getResource provides. However, usually you don't need to know the location of the file, you just need its contents, in which case you can use getResourceAsInputStream()
This all assumes your class path is configured correctly.
Yes, you should be able to load a file anywhere on your file system that the java process has access to. You just need to have the path explicitly set in your getResource call.
For example:
File file = new File(URLDecoder.decode(this.getClass().getResource("C:\\foo\\bar\\file.xml").getFile(), "UTF-8"));
I'm new to help file creation in java. I have created a help file "sample.chm" with a 3rd party tool, added it to a java program with package name as "help" calling with runtime class and build the jar. When I run the jar file it is giving me an error that the "file cannot be found, null pointer Exception". I have given a relative path to identify the file like "../help/sample.chm" still it is not working and I tried with various classes to ientify the path. But still the same error.
Request you to please help me in fixing it.
The jar can be placed in different systems and should open this help file with out any issues.
I hope my explanation is sufficient you to identify the problem.
Regards,
Chandu
If you have a file inside a jar, you can't access it as you normally would. You can access it like this:
URL helpFile=Thread.currentThread().getContextClassLoader().getResource("help/sample.chm");
The method used above (getResource) will return a URL; if you want, you can get it as an InputStream as well by using getResourceAsStream instead.
At least a workaround unless a better solution pops up. Use the this.getClass().getClassLoader().getResource way to get an inputstream to the help file inside the jar.
Copy the bytes to a new help file in the target systems temp folder and use this extracted file with the external help file viewer.