Date
1 - 5 of 5
How do I get the path of my application.xgapp?
|
Jacky Miu
I have created an application using Gate Developer. After I have saved the application, I got the usual resources including application.xgapp, application-resources folder, etc. For initialization, I have to read the content of a file stored within the application-resources folder. Now, my question is how I can get the path of this application (or alternatively the path of the application.xgapp file) when I run this application. If I can get the path of the application.xgapp, I can create a relative path to easily access the content of the said file. This is helpful since the application can be saved to any other location. I have tried to use the following code to obtain that path but it was unsuccessful: import java.nio.file.Paths; String currentDir = System.getProperty("user.dir"); Because the currentDir above only provides the path of my Gate program, not the path of the application.xgapp. Thank you in advance of your help! Regards, Jacky Miu |
|
|
|
Ian Roberts
"For initialization" of which
component? If this is in your own custom ProcessingResource then
the standard way to handle this sort of thing is to give your PR
an init parameter of type URL or ResourceReference that points to
the file that you want to load. In your init() method you just
use the normal .openStream() method (implemented by both URL and
ResourceReference) to read the data from the file, or if you need
to access other files in the same folder then you can create new
URLs relative to that one
URL otherFile = new URL(paramValue,
"fileName.txt");
When you save the application as an
xgapp any URL or ResourceReference parameter values will be stored
in the xgapp as a path relative to the xgapp location and resolved
back to an absolute URL when you re-load, so you can move the
xgapp and the target file to any other location and as long as
they remain in the same relative locations it'll all still work.
Ian
On 24/02/2022 06:06, Jacky Miu via
groups.io wrote:
-- Ian Roberts | Department of Computer Science i.roberts@... | University of Sheffield, UK |
|
|
|
Jacky Miu
Hi Ian, Thank you for your advice. I actually want to read a text from a file (e.g. input.txt), then designate the text as a document feature. Following code is used in a jape file: import java.nio.file.Files; import java.nio.file.Paths; data = new String(Files.readAllBytes(Paths.get("C:/Users/input.txt"))); doc.getFeatures().put("input", data); It works fine while I am using Gate Developer. However, if I save the application as an xgapp, the reference to C:/Users/input.txt is no longer correct because the application can be unzipped to any location. Any advice on how I can avoid changing the file path even after the application has been saved as an xgapp? Thanks. Regards, Jacky Miu
On Thursday, February 24, 2022, 05:55:08 PM GMT+8, Ian Roberts <i.roberts@...> wrote:
"For initialization" of which
component? If this is in your own custom ProcessingResource then
the standard way to handle this sort of thing is to give your PR
an init parameter of type URL or ResourceReference that points to
the file that you want to load. In your init() method you just
use the normal .openStream() method (implemented by both URL and
ResourceReference) to read the data from the file, or if you need
to access other files in the same folder then you can create new
URLs relative to that one
URL otherFile = new URL(paramValue,
"fileName.txt");
When you save the application as an
xgapp any URL or ResourceReference parameter values will be stored
in the xgapp as a path relative to the xgapp location and resolved
back to an absolute URL when you re-load, so you can move the
xgapp and the target file to any other location and as long as
they remain in the same relative locations it'll all still work.
Ian
On 24/02/2022 06:06, Jacky Miu via
groups.io wrote:
I have created an application
using Gate Developer. After I have saved the application, I
got the usual resources including application.xgapp,
application-resources folder, etc.
For initialization, I have to
read the content of a file stored within the
application-resources folder. Now, my question is how I can
get the path of this application (or alternatively the path of
the application.xgapp file) when I run this application. If I
can get the path of the application.xgapp, I can create a
relative path to easily access the content of the said file.
This is helpful since the application can be saved to any
other location.
I have tried to use the
following code to obtain that path but it was unsuccessful:
import
java.nio.file.Paths;
String currentDir =
System.getProperty("user.dir");
Because the currentDir above
only provides the path of my Gate program, not the path of the
application.xgapp. Thank you in advance of your help!
Regards,
Jacky Miu
-- Ian Roberts | Department of Computer Science i.roberts@... | University of Sheffield, UK |
|
|
|
Ian Roberts
To be honest, I think your best bet in
this situation is still to turn this logic into a proper PR rather
than a JAPE. This would let you take advantage of the URL
relativisation trick in xgapps and would also mean that you could
make it read the file from disk just once (at PR init time) and
store the string in a variable that could then be assigned to
every document, rather than having to read the file off disk
repeatedly for every individual document. I've made an example of
how you could do this at https://github.com/ianroberts/filereader
Ian
On 25/02/2022 07:35, Jacky Miu via
groups.io wrote:
-- Ian Roberts | Department of Computer Science i.roberts@... | University of Sheffield, UK |
|
|
|
Jacky Miu
Dear Ian, The filereader works fine, and fully accomplishes my task! Thank you so much for your help! Best regards, Jacky
On Friday, February 25, 2022, 06:33:24 PM GMT+8, Ian Roberts <i.roberts@...> wrote:
To be honest, I think your best bet in
this situation is still to turn this logic into a proper PR rather
than a JAPE. This would let you take advantage of the URL
relativisation trick in xgapps and would also mean that you could
make it read the file from disk just once (at PR init time) and
store the string in a variable that could then be assigned to
every document, rather than having to read the file off disk
repeatedly for every individual document. I've made an example of
how you could do this at https://github.com/ianroberts/filereader
Ian
On 25/02/2022 07:35, Jacky Miu via
groups.io wrote:
Hi Ian,
Thank you for your advice.
I actually want to read a
text from a file (e.g. input.txt), then designate the text as
a document feature. Following code is used in a jape file:
import java.nio.file.Files;
import java.nio.file.Paths;
data = new
String(Files.readAllBytes(Paths.get("C:/Users/input.txt")));
doc.getFeatures().put("input", data);
It works fine while I
am using Gate Developer. However, if I save the application
as an xgapp, the reference to C:/Users/input.txt is no
longer correct because the application can be unzipped to
any location. Any advice on how I can avoid changing the
file path even after the application has been saved as an
xgapp? Thanks.
Regards,
Jacky Miu
On Thursday, February 24, 2022, 05:55:08 PM GMT+8, Ian
Roberts <i.roberts@...> wrote:
"For
initialization" of which component? If this is in
your own custom ProcessingResource then the standard
way to handle this sort of thing is to give your PR an
init parameter of type URL or ResourceReference that
points to the file that you want to load. In your
init() method you just use the normal .openStream()
method (implemented by both URL and ResourceReference)
to read the data from the file, or if you need to
access other files in the same folder then you can
create new URLs relative to that one
URL otherFile
= new URL(paramValue, "fileName.txt");
When you save
the application as an xgapp any URL or
ResourceReference parameter values will be stored in
the xgapp as a path relative to the xgapp location and
resolved back to an absolute URL when you re-load, so
you can move the xgapp and the target file to any
other location and as long as they remain in the same
relative locations it'll all still work.
Ian
On
24/02/2022 06:06, Jacky Miu via groups.io wrote:
I have created an application using
Gate Developer. After I have saved the
application, I got the usual resources including
application.xgapp, application-resources folder,
etc.
For initialization, I have to read
the content of a file stored within the
application-resources folder. Now, my question is
how I can get the path of this application (or
alternatively the path of the application.xgapp
file) when I run this application. If I can get
the path of the application.xgapp, I can create a
relative path to easily access the content of the
said file. This is helpful since the application
can be saved to any other location.
I have tried to use the following
code to obtain that path but it was unsuccessful:
import java.nio.file.Paths;
String currentDir =
System.getProperty("user.dir");
Because the currentDir above only
provides the path of my Gate program, not the path
of the application.xgapp. Thank you in advance of
your help!
Regards,
Jacky Miu
-- Ian Roberts | Department of Computer Science i.roberts@... | University of Sheffield, UK
-- Ian Roberts | Department of Computer Science i.roberts@... | University of Sheffield, UK |
|
|