I am working on reports. Now I have a jrxml file which I have to load so that I can parse it. However, the problem is, I want to load the file with its name directly
String path = "myreport.jrxml";
But I am unable to find that where should I place that file in my project so that the above path is valid. In simple java project, we just place our desired file in our project folder where src folder is placed, but it is not working in vaadin project. So where should I place my file.
Thanks.
Are you deploying your project to a web server? Then you would need to find out what is the default working directory of that web server.
You could use:
JRXmlLoader.load("/myreport.jrxml");
or if you want to test if the resource is found and take another tries you could load it as a resource and pass it load method from JRXmlLoader as follows:
JRXmlLoader.load(getClass().getResourceAsStream("/myreport.jrxml"));
Related
I'm developing a web service using axis2 & tomcat . There I use a data.xml file to store some information. When I give the Absolute Path for the data.xml file, everything works fine. But I am looking for a way to give the file path in relative to the source file. To achive that I have tried several methods.
(the deployed aar file is located under C:/tomcat/webapps/axis2/WEB-INF/service.aar )
This is my folder structure.
+Project
|-src
|-data
|-data.xml
I have added the data folder to the build path.
I have tried to include the file as ./data/data.xml
But it failed. Can anyone suggest me a best/recommended way to do this?
-Regards
After a long research, I found that it is impossible to do it. It should be a absolute path for the xml file.
I am creating a project using jsp/servlet in which I am trying to create java file and class file inside the project itself. But I am able to do this for only my system because the path I give their is like : C:\Users\MySystem\Desktop\Test\.. which works only for my system. What should I do so that if I have to run this project in another system I don't have to change path again and again.
Well if it is maven project just put your resources files under src/main/resources
and you can read them using this lines.
String path = Thread.currentThread().getContextClassLoader()
.getResource("yourFileName").getPath();
System.out.println(path);
Or even this way you can do it.
String pathOfTheFile = getServletContext().getResource("yourFile").getPath();
and don't forget to put the file under web-content or webapp folder
I am working on a web app i have java files in it which uses certain files.I want to specify these files using relative path in java so that it doesn't produce mobility issue.But Where should i place a file in a web app so that i can use relative path.? I have tried placing the files under source package, web folder, directly under the web-application.Please help.Thanks in advance
The simplest way to get the current directory of a java application is :
System.out.println(new File(".").getAbsolutePath());
Like that you can consider the given path as the root of your application.
Cheers,
Maxime.
Read the file as a resource. Put it somewhere in the src. For instance
src/resources/myresource.txt
Then you can just do
InputStream is = getClass().getResourceAsStream("/resources/myresource.txt");
Note: if you are using maven, then you are more accustomed to something like this
src/main/resources/myresource.txt
With maven, everything in the main/resources folder gets built to the root, so you would leave out the resources in your path
InputStream is = getClass().getResourceAsStream("/myresource.txt");
I am using eclipse Indigo IDE ..I am developing a jsp application (myProject) using eclipse in which i have a property file (myProject\webcontent\db.properties) for configuring database credentials.
I am trying to access this file from a class (myProject\src\samplePackage\sampleDBConnect.java). I have exported samplePackage.jar into *myProject\webContent\WEB-INF\lib*.
I have a Test.jsp page which calls a method in sampleDBConnect.java.
When i am trying to run this Test.jsp page, the current directory path shows C:\Documents and Settings\username. I have my project in some other drive(E:).
Can someone tell me how to access the properties file....
getClass().getResourceAsStream("db.properties");
This assumes your db.properties is in the same directory "samplePackage".
If you want to keep it in src directory or say resource directory then use
getClass().getClassLoader().getResourceAsStream("db.properties");
Since your java code belongs to the same web project you don't have to create a jar of the same project and place that in the WEB-INF/lib folder
move db.properties to myProject\src dir,
java code:
InputStream input = sampleDBConnect.class.getClassLoader().getResourceAsStream("db.properties");
I'm working with a project that is setup using the standard Maven directory structure so I have a folder called "resources" and within this I have made a folder called "fonts" and then put a file in it. I need to pass in the full String file path (of a file that is located, within my project structure, at resources/fonts/somefont.ttf) to an object I am using, from a 3rd party library, as below, I have searched on this for a while but have become a bit confused as to the proper way to do this. I have tried as below but it isn't able to find it. I looked at using ResourceBundle but that seemed to involve making an actual File object when I just need the path to pass into a method like the one below (don't have the actual method call in front of me so just giving an example from my memory):
FontFactory.somemethod("resources/fonts/somefont.ttf");
I had thought there was a way, with a project with standard Maven directory structure to get a file from the resource folder without having to use the full relative path from the class / package. Any advice on this is greatly appreciated.
I don't want to use a hard-coded path since different developers who work on the project have different setups and I want to include this as part of the project so that they get it directly when they checkout the project source.
This is for a web application (Struts 1.3 app) and when I look into the exploded WAR file (which I am running the project off of through Tomcat), the file is at:
<Exploded war dir>/resources/fonts/somefont.ttf
Code:
import java.io.File;
import org.springframework.core.io.*;
public String getFontFilePath(String classpathRelativePath) {
Resource rsrc = new ClassPathResource(classpathRelativePath);
return rsrc.getFile().getAbsolutePath();
}
In your case, classpathRelativePath would be something like "/resources/fonts/somefont.ttf".
You can use the below mentioned to get the path of the file:
String fileName = "/filename.extension"; //use forward slash to recognize your file
String path = this.getClass().getResource(fileName).toString();
use/pass the path to your methods.
If your resources directory is in the root of your war, that means resources/fonts/somefont.ttf would be a "virtual path" where that file is available. You can get the "real path"--the absolute file system path--from the ServletContext. Note (in the docs) that this only works if the WAR is exploded. If your container runs the app from the war file without expanding it, this method won't work.
You can look up the answer to the question on similar lines which I had
Loading XML Files during Maven Test run
The answer given by BobG should work. Though you need to keep in mind that path for the resource file is relative to path of the current class. Both resources and java source files are in classpath