I am using JSP on NetBeans.
In the java code, I am trying to read data from a file. First I open the file by specifying its path.
Because my code runs on a server (GlassFish), I would like to have my file path independent of the machine where it runs. Therefore, I want to start the path with the folder name that contains the file which is saved on the root of the project directory. I tried so hard to achieve that but I couldn't.
I read online and I found this way, but it still doesn't work:
<%
//building the tree here.
GraphBuilder tree = new GraphBuilder("${pageContext.request.contextPath}\\src\\java\\Database\\OptimizedFullTermFile.pad");
%>
Can anyone help? Thank you.
Just read it from the classpath. Given that you're using the typical src folder representing the Java source code (the Java package/class structure), I assume that the file OptimizedFullTermFile.pad is placed in the Java package java.Database (eek, a capital in package name? lowercase it all). In that case, it's already in the classpath and thus you can just get it straight from there as follows:
InputStream input = Thread.currentThread().getContextClassLoader().getResourceAsStream("java/Database/OptimizedFullTermFile.pad");
// ...
As to your failed attempt: the EL ${pageContext.request.contextPath} isn't ever going to work in a scriptlet. Even if it did, it's not the right thing, it returns the context path in the webapp URL which is absolutely not part of the local disk file system path, let alone the classpath. Using scriptlets is strongly discouraged since a decade, by the way.
See also:
getResourceAsStream() vs FileInputStream
Where to place and how to read configuration resource files in servlet based application?
You can try the following code to get your directory file in project from the following code.
this.getServletContext().getRealPath("")+[your file path in project web directory]
Related
what I want to ask seems so simple and crazy but since I am so beginner I dare to ask you guys.
I want to give relative address to read a file in eclipse java. my java file is in common package and json file is in resources package in the same project. but I do not know how to provide relative address to that.
BufferedReader in = new BufferedReader(new FileReader("/?/file.json"));
so I have a project:
> src/main/java
>com.project.cc.restful.common
>com.project.cc.restful.resources
any help?
thanks!
What you need is a path that is relative to your working directory. The working directory is a configurable parameter. In eclipse the default is usually the root folder of the project (not the source code folder!). It can be configured in the "Run Configurations.." menu.
To be sure, run your application once with System.out.println(System.getProperty("user.dir")) to see the absolute path to your working directory. Once you have that, use ../../Resources (or something similar) to get to the resources directory using a relative path.
I have some stuff under src/main/resources path.
Specifically I have a folder with report templates called reports.
I understand that when the application is deployed/run all files and folders under src/main/resources go to the classpath, namely my project's WEB-INF/classes.
This means that a folder WEB-INF/classes/reports will be created in my server.
Now I want to access my reports as paths, not as inputstream, because my reporting code in java supports a filepath and not an inputstream. So I have to be able to get the WEB-INF/classes/reports absolute path (or relative, I don't care as long as it is right).
Reading some answers regarding similar questions, I have already tried the following things:
getClass().getResource(".").getPath(); --> this returns the exact path of the class I am currently at in my classpath, namely: C:\Tools\JBoss Application Server 7.1.1\standalone\deployments\myProject.war\WEB-INF\classes\aaa\bbb\ccc\ddd
getClass().getClassLoader().getResource(".").getPath(); --> this returns: C:\Tools\JBoss Application Server 7.1.1\modules\sun\jdk\main\service-loader-resources, which is completely irrelevant.
I want something to return C:\Tools\JBoss Application Server 7.1.1\standalone\deployments\myProject.war\WEB-INF\classes
If it is not possible, I will get the first path and go as many folders back as needed to reach classes folder.
Thank you.
You need ServletContext.getRealPath(String) method.
getServletContext().getRealPath("/WEB-INF")
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
I am trying to create a PDF file using struts 2.Action class location is as follows.
/home/Jagan/MATCH/Jagan/src/ActionClasses/PDFFile.java
Here workspace starts from
/MATCH
In PDFFile.java.I am writing as given below and it is working fine.
pdfwriter=PdfWriter.getInstance(document,new
FileOutputStream("/home/Jagan/xyz.pdf"));
But i have to create this under the folder
/home/Jagan/MATCH/Jagan/PDFs
I should not use /home/Jagan/ as it will become hardcode if i have to run this application in other system.
I tried
pdfwriter=PdfWriter.getInstance(document,new
FileOutputStream("../../../PDFs/xyz.pdf"));
But it is not creating file.Even if it works it is not feasible solution (because "../../../" does not work in windows ).
Please suggest me a good way to specify path for creating file.
Adding to the question.
I have to provide download option for downloading this created file in JSP page.Which struts tag should i use. Please provide me syntax for that
downloading files has 2 aproaches:
1. you can return it with outstream result like is explained here
2. as you are trying to save the file first at filesystem then access it from another url.
Answer to your question is you should get servletContext.getRealPath("/WEB-INF"), and after that everything is relative to WEB-INF.
I should not use /home/Jagan/ as it will become hardcode if i have to run this application in other system.
Correct, Use following instead
System.getProperty("user.home");///home/Jagan/, it will return you path to your home dir
I know that you can use java.util.Properties to read Java properties files.
See: Java equivalent to app.config?
Is there a standard place to put this file? In .NET we put application.exe.config in the same directory as application.exe. The application looks for it here by default.
Java can be made to look for a properties file in the class path but I am struggling to understand the filename/path structure to use and how to use either a standard .properties format or XML format file.
Assuming I have an API packaged in org_example_api.jar (the root package is org.example.api). I don't want to put the properties file inside the jar as it should be editable by the user. I want the user to be able to put the required configuration properties in either a .properties or .xml file somewhere relative to the classpath so I can find it without needing to know anything about the ir file system structure.
Will this work on all systems:
/classpath/org_example_api.jar
/classpath/org/example/api/config.properties OR
/classpath/org/example/api/config.xml
Code:
java.util.Properties = ? //NEED SOME HELP HERE
This purely depends on the type of application you are developing.
1) If it is a web application the best place is inside the WEB-INF/classes/ folder.
2) If you are developing a standalone application there are many approaches. From your example I think the following structure will work.
/<dist>/org_example_api.jar
/<dist>/config.xml
/<dist>/run.sh
In the run.sh you can start the java application providing the current directory also in the classpath. Something like this.
java -cp .:org_example_api.jar ClassToExecute
3) If it is an API distribution it is up to the end user. You can tell the user that they can provide the config.xml in the classpath which should follow some predefined structure. You can look at Log4J as an example in this case.
The world is wide open to you here. The only best practice is what works best for you:
Whatever program the user is running can require the path to the properties file as an argument
Your application can be configured to look in the current directory for config.properties.
If the file can't be found, you could maybe fall back to the user.home directory, or fall back to wherever your application is installed.
Personally I usually have my applications attempt to read properties files from the classpath - but I'm not in a world where I have end-users update/change the file.
Whatever option you choose, just make sure you clearly document it for your users so they know which file to edit and where it needs to be!
You can put the properties file in a directory or JAR in your CLASSPATH, and then use
InputStream is = getClass().getResourceAsStream("/path/goes/here");
Properties props = new Properties();
props.load(is);
(I noticed you mentioned this in your OP, but others may find the code useful.)