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Load data into Java GUI [closed]

.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty height:90px;width:728px;box-sizing:border-box;

2

I am working on a Java GUI which requires data to be read from an XML file. Ultimately i want to deliver a single .exe or .jar file where all the things you need are stored. These databases should get automatically parsed.

My question is: Where and in which format should I store the files so that they are always found automatically?

It should be as if the XML files are written in the sourcecode, but I don't think that this would be a good approach to this problem.

Thanks in advance.

java xml javafx

share|improve this question

asked Mar 22 at 8:21

lospolloslospollos

254

closed as too broad by DanielBarbarian, greg-449, Matteo Baldi, EdChum, Mark Rotteveel Mar 22 at 9:39

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

add a comment | 

2

I am working on a Java GUI which requires data to be read from an XML file. Ultimately i want to deliver a single .exe or .jar file where all the things you need are stored. These databases should get automatically parsed.

My question is: Where and in which format should I store the files so that they are always found automatically?

It should be as if the XML files are written in the sourcecode, but I don't think that this would be a good approach to this problem.

Thanks in advance.

java xml javafx

share|improve this question

asked Mar 22 at 8:21

lospolloslospollos

254

closed as too broad by DanielBarbarian, greg-449, Matteo Baldi, EdChum, Mark Rotteveel Mar 22 at 9:39

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

add a comment | 

I am working on a Java GUI which requires data to be read from an XML file. Ultimately i want to deliver a single .exe or .jar file where all the things you need are stored. These databases should get automatically parsed.

My question is: Where and in which format should I store the files so that they are always found automatically?

It should be as if the XML files are written in the sourcecode, but I don't think that this would be a good approach to this problem.

Thanks in advance.

java xml javafx

share|improve this question

asked Mar 22 at 8:21

lospolloslospollos

254

share|improve this question

asked Mar 22 at 8:21

lospolloslospollos

254

share|improve this question

asked Mar 22 at 8:21

lospolloslospollos

254

asked Mar 22 at 8:21

lospolloslospollos

254

asked Mar 22 at 8:21

lospolloslospollos

254

1 Answer
1

active

oldest

votes

3

If these files shall only be read, you may store them as resources in the class path. If the resource file is in the same directory as your class you may access the resource from within instances of that class like:

final URL myResource = getClass().getResource(nameOfFile);
final InputStream myResourceStream = getClass().getResourceAsStream(nameOfFile);

Static access is possible like:

final URL myResource = MyClass.class.getResource(nameOfFile);

The nameOfFile may also contain a path to navigate withing the package structure:

final Resource myResource = getClass().getResource("subpackage/data.xml");
final Resource myResource = getClass().getResource("/com/myCompany/somePackage/data.xml");

And by the way: i recommend to use the java.nio.* classes to access files. One huge benefit is that this allows you to set custom file system implementations, if needed.

share|improve this answer

edited Mar 22 at 10:58

answered Mar 22 at 8:35

AmadánAmadán

28229

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Thanks, this works fine when executed in my IDE, but does not work when building an .exe artifact and executing the .exe program (it cannot find the directory). Do you know a solution do this problem as well?
  
  – lospollos
  Mar 27 at 8:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @lospollos I am not at all familiar with creating executables from java projects. Maybe your process does not include the resources into your compilation target. Maybe you missed an option for that? Or maybe you need to include a second file with your .exe?
  
  – Amadán
  Mar 27 at 11:54
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Lets forget the .exe thing and assume I want to build a jar application. Do you know how I could set the path in the sourcecode in order to load the databases when delivering my tool? E.g. the databases are in a specific folder and the jar file is in another
  
  – lospollos
  Mar 27 at 15:11
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  If you mean providing the configuration of DB access, then a java properties file as resource may be a good approach. If you mean setting something like the windows PATH variable: for heaven's sake don't! The system configuration is something your application should not take control over.
  
  – Amadán
  Apr 1 at 7:59
  

  
  
  
  

add a comment | 

3

If these files shall only be read, you may store them as resources in the class path. If the resource file is in the same directory as your class you may access the resource from within instances of that class like:

final URL myResource = getClass().getResource(nameOfFile);
final InputStream myResourceStream = getClass().getResourceAsStream(nameOfFile);

Static access is possible like:

final URL myResource = MyClass.class.getResource(nameOfFile);

The nameOfFile may also contain a path to navigate withing the package structure:

final Resource myResource = getClass().getResource("subpackage/data.xml");
final Resource myResource = getClass().getResource("/com/myCompany/somePackage/data.xml");

And by the way: i recommend to use the java.nio.* classes to access files. One huge benefit is that this allows you to set custom file system implementations, if needed.

share|improve this answer

edited Mar 22 at 10:58

answered Mar 22 at 8:35

AmadánAmadán

28229

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Thanks, this works fine when executed in my IDE, but does not work when building an .exe artifact and executing the .exe program (it cannot find the directory). Do you know a solution do this problem as well?
  
  – lospollos
  Mar 27 at 8:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @lospollos I am not at all familiar with creating executables from java projects. Maybe your process does not include the resources into your compilation target. Maybe you missed an option for that? Or maybe you need to include a second file with your .exe?
  
  – Amadán
  Mar 27 at 11:54
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Lets forget the .exe thing and assume I want to build a jar application. Do you know how I could set the path in the sourcecode in order to load the databases when delivering my tool? E.g. the databases are in a specific folder and the jar file is in another
  
  – lospollos
  Mar 27 at 15:11
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  If you mean providing the configuration of DB access, then a java properties file as resource may be a good approach. If you mean setting something like the windows PATH variable: for heaven's sake don't! The system configuration is something your application should not take control over.
  
  – Amadán
  Apr 1 at 7:59
  

  
  
  
  

add a comment | 

3

If these files shall only be read, you may store them as resources in the class path. If the resource file is in the same directory as your class you may access the resource from within instances of that class like:

final URL myResource = getClass().getResource(nameOfFile);
final InputStream myResourceStream = getClass().getResourceAsStream(nameOfFile);

Static access is possible like:

final URL myResource = MyClass.class.getResource(nameOfFile);

The nameOfFile may also contain a path to navigate withing the package structure:

final Resource myResource = getClass().getResource("subpackage/data.xml");
final Resource myResource = getClass().getResource("/com/myCompany/somePackage/data.xml");

And by the way: i recommend to use the java.nio.* classes to access files. One huge benefit is that this allows you to set custom file system implementations, if needed.

share|improve this answer

edited Mar 22 at 10:58

answered Mar 22 at 8:35

AmadánAmadán

28229

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Thanks, this works fine when executed in my IDE, but does not work when building an .exe artifact and executing the .exe program (it cannot find the directory). Do you know a solution do this problem as well?
  
  – lospollos
  Mar 27 at 8:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @lospollos I am not at all familiar with creating executables from java projects. Maybe your process does not include the resources into your compilation target. Maybe you missed an option for that? Or maybe you need to include a second file with your .exe?
  
  – Amadán
  Mar 27 at 11:54
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Lets forget the .exe thing and assume I want to build a jar application. Do you know how I could set the path in the sourcecode in order to load the databases when delivering my tool? E.g. the databases are in a specific folder and the jar file is in another
  
  – lospollos
  Mar 27 at 15:11
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  If you mean providing the configuration of DB access, then a java properties file as resource may be a good approach. If you mean setting something like the windows PATH variable: for heaven's sake don't! The system configuration is something your application should not take control over.
  
  – Amadán
  Apr 1 at 7:59
  

  
  
  
  

add a comment | 

If these files shall only be read, you may store them as resources in the class path. If the resource file is in the same directory as your class you may access the resource from within instances of that class like:

final URL myResource = getClass().getResource(nameOfFile);
final InputStream myResourceStream = getClass().getResourceAsStream(nameOfFile);

Static access is possible like:

final URL myResource = MyClass.class.getResource(nameOfFile);

The nameOfFile may also contain a path to navigate withing the package structure:

final Resource myResource = getClass().getResource("subpackage/data.xml");
final Resource myResource = getClass().getResource("/com/myCompany/somePackage/data.xml");

And by the way: i recommend to use the java.nio.* classes to access files. One huge benefit is that this allows you to set custom file system implementations, if needed.

share|improve this answer

edited Mar 22 at 10:58

answered Mar 22 at 8:35

AmadánAmadán

28229

final URL myResource = getClass().getResource(nameOfFile);
final InputStream myResourceStream = getClass().getResourceAsStream(nameOfFile);

final URL myResource = MyClass.class.getResource(nameOfFile);

final Resource myResource = getClass().getResource("subpackage/data.xml");
final Resource myResource = getClass().getResource("/com/myCompany/somePackage/data.xml");

share|improve this answer

edited Mar 22 at 10:58

answered Mar 22 at 8:35

AmadánAmadán

28229

answered Mar 22 at 8:35

AmadánAmadán

28229

answered Mar 22 at 8:35

AmadánAmadán

28229