This question already has answers here:
Programmatically list open projects in an eclipse workspace from outside of eclipse
(2 answers)
Closed 6 years ago.
I need a way to get the directory path of an eclipse project by deserializining the .location file found in the eclipe workspace at:
.metadata\.plugins\org.eclipse.core.resources.projects\myproject\.location
Its contents look something like:
#±‹#¼ %–磓¾ URI//file:/D:/proj/myproject ÀXûó#¼ QóŒ{»wÆ
so I would like to programmatically get the "D:/proj/myproject" string out of it.
Bonus points if the process doesn't use the Eclipse API.
Eclipse read this file in the method LocalMetaArea.readPrivateDescription(..).
It uses a SafeChunkyInputStream and DataInputStream for reading it.
It contains UTF8 strings and some integers.
For the complete code see
org.eclipse.core.internal.resources.LocalMetaArea at GrepCode.com
Related
This question already has answers here:
Why do my java files look strange in Eclipse?
(5 answers)
Closed 1 year ago.
I have two class .java but in pdcController.java I cannot do anything:
Look the image:
Look the icon are different in BandoService works all, but in .pdfController I cannot do anything (for example I cannot suggest for method or error import). pdfController and BandoServiceare in the same project.
This is not the source file, check the file path, you're probably not in the correct package.
I guess you have several modules in your maven project and you opened the file from the parent module.
If you see the file only in the parent module it's time to git fetch.
This question already has answers here:
Problem with IntelliJ -> Cannot create class with name "Main"
(2 answers)
Closed 2 years ago.
I cannot create java class file with name TokenProperties.java. I also tried restart and clean cache but doesn't help. Class with different name I can crate.
Here is error pop-up.
File will added but marked as plain text file, and no java file.
Do you have idea whats is going on? Thank you.
It kind of looks like you create a TXT file with the name TokenProperties.java when you want to create a JAVA file with the name TokenProperties similar to how DemoAAplication was created.
Right click com.example.demoa package.
Select New
Select Java Class
I wasn't able to fix it so I reinstalled Intellij and now it works.
This question already has answers here:
How to read text file from classpath in Java?
(17 answers)
Closed 3 years ago.
I want to read the configuration file outside the jar, how should I do?
I tried classloader.getResourceAsStream("config.xml") and succeeded while I am running the code inside Intellij. Now I want to build the jars to a folder and place the config.xml under the same folder, not inside the jar, but the program fails to detect the config.xml.
Is there a graceful way of reading the config.xml instead of using File with relative path in the code, which doesn't work while debugging/running inside the IDE?
Yes, turn it into a system property, and provide it to anything running any Java process/application.
Let's say you have a config.xml file located inside /some/path/down/the/line/, then you can do: java --classpath ... -Dapp.config=/some/path/down/the/line/config.xml tld.domain.Application.
Then all you have to do in your Java code is to reference that name/path: final String configFile = System.getProperty("app.config");, and use any well-known routine to read it from there.
Basically, you have to make sure the file/path/location is provided somehow to the Java classpath.
This question already has answers here:
How to create a folder in Java?
(8 answers)
Closed 6 years ago.
I would like to write a Java Programm which should be able to create a Folder in Windows, but the Client should be a able to decide on the Name of the Folder. How can I do that?
Thanks for any suggestions!
A small google search would have given you the answer. If you want to name it a specific way, just grab the name from a JTextField or something like that.
Create a File object and use mkdir() to create a new folder.
File folder;
folder = new File("path/to/new/folder");
folder.mkdir();
Obviously you will need to check if the folder already exists and you will need to use an inputted string for the name of the folder.
This question already has answers here:
getResourceAsStream() vs FileInputStream
(6 answers)
Closed 4 years ago.
I had created a java application which creates a jasper report whenever user clicks the print button it provides a report.
code:
String
srcfile1="C:\\Users\\VINO\\Documents\\NetBeansProjects\\rework\\src\\rework\\report1.jasper";
JasperPrint firstsecondlinked = JasperFillManager.fillReport(srcfile1,map,cons);
I copied the same to the colleague but it doesn't works fine giving an error states that file not found.
As to create a application how to name the source file in the string?
You can do the following.
Specify the paths in some properties file and read it by loading that file. While deploying your application to some other machine, change the path before deployment.
application-resources.properties
report1.path=''
If it's a web-application, you can access any resource inside your application by using request.getContextPath(). You will not need to hard code it like C:\Users\VINO.. Only the relative path will suffice.