I want to dynamically reference an XSD from a bean, how is this possible? I already added the XSD to the project, so it's located somewhere in the GlassFish domain.
Use the ExternalContext.
If you want to load the resource in the bean, do it via getResource or getResourceAsStream:
InputStream stream = FacesContext.getCurrentInstance().getExternalContext()
.getResourceAsStream("/foo.xsd");
If you want to return a URL to the resource, use getRequestContextPath to get the path relative to the host root:
ExternalContext ext = FacesContext.getCurrentInstance()
.getExternalContext();
String path = ext.getRequestContextPath();
path += path.endsWith("/") ? "foo.xsd" : "/foo.xsd";
String url = ext.encodeResourceURL(path);
Related
I need to save file from javabean or servlet, and I'm having trouble finding relative path, I tried:
(from servlet)
ServletContext servCont = this.getServletContext();
String contextPath = servCont.getRealPath(File.separator);
System.out.println("REAL PATH: "+ contextPath);
this gives me:
REAL PATH: E:\Web\.metadata\.plugins\org.eclipse.wst.server.core\tmp0\wtpwebapps\Saloni\
and project folder is:
E:\Web\Saloni
and from bean (bean is called Salon)
String path = Salon.class.getResource("Salon.class").getPath();
and got basically the same thing
/E:/Web/.metadata/.plugins/org.eclipse.wst.server.core/tmp0/wtpwebapps/Saloni/WEB-INF/classes/beans/Salon.class
If I just put file name into FileOutputStream file gets saved in eclipse workspace.
I read somewhere that I'm supposed to get to WEB-INF somehow but can't do that ..
I am parsing the HttpRequest inside a web application. My servlet extends HttpServlet
I need to find the path till WEB-INF inside a WAR file so that file present inside WEB-INF can be accessed.
I cannot use Realpath so there has to be some alternative.
tried using
ServletContext servletContext = getServletContext();
InputStream input = servletContext.getResourceAsStream("/WEB-INF");
but everytime
Sysem.out.println(input) returns null
Please suggest.
You can get the root web directory of your application like this:
servletContext.getRealPath(File.separator));
So assuming your WEB-INF is in there, you could do this:
servletContext.getRealPath(File.separator + "/WEB-INF/whatever.file"));
It returns a String path, so to actually get the file:
new File(servletContext.getRealPath(File.separator + "/WEB-INF/whatever.file")));
I am working on a JSF website and I need some help. I have an XML file that I am trying to read from by backing bean but I don't know how to find the path to it. It is in my resources folder (resources/movies.xml). How do I do this?
If it is indeed the /resources folder of the public web content where you usually store the static web resources like CSS/JS/images, then you can use ExternalContext#getResourceAsStream() to get an InputStream of it.
ExternalContext externalContext = FacesContext.getCurrentInstance().getExternalContext();
InputStream input = externalContext.getResourceAsStream("/resources/movies.xml");
// ...
This is how I get path of folder in webapp resources folder:
ServletContext servletContext = (ServletContext) FacesContext.getCurrentInstance().getExternalContext().getContext();
String resHomeImgPath = servletContext.getRealPath("resources/img/home");
This question already has answers here:
Recommended way to save uploaded files in a servlet application
(2 answers)
Closed 6 years ago.
Need to get absolute path in java class file, inside a dynamic web application...
Actually i need to get path of apache webapps folder... where the webapps are deployed
e.g. /apache-root/webapps/my-deployed-app/WebContent/images/imagetosave.jpg
Need to get this in a java class file, not on jsp page or any view page...
any ideas?
Actually i need to get path of apache webapps folder... where the webapps are deployed
e.g. /apache-root/webapps/my-deployed-app/WebContent/images/imagetosave.jpg
As mentioned by many other answers, you can just use ServletContext#getRealPath() to convert a relative web content path to an absolute disk file system path, so that you could use it further in File or FileInputStream. The ServletContext is in servlets available by the inherited getServletContext() method:
String relativeWebPath = "/images";
String absoluteDiskPath = getServletContext().getRealPath(relativeWebPath);
File file = new File(absoluteDiskPath, "imagetosave.jpg");
// ...
However, the filename "imagetosave.jpg" indicates that you're attempting to store an uploaded image by FileOutputStream. The public webcontent folder is the wrong place to store uploaded images! They will all get lost whenever the webapp get redeployed or even when the server get restarted with a cleanup. The simple reason is that the uploaded images are not contained in the to-be-deployed WAR file at all.
You should definitely look for another location outside the webapp deploy folder as a more permanent storage of uploaded images, so that it will remain intact across multiple deployments/restarts. Best way is to prepare a fixed local disk file system folder such as /var/webapp/uploads and provide this as some configuration setting. Finally just store the image in there.
String uploadsFolder = getItFromConfigurationFileSomehow(); // "/var/webapp/uploads"
File file = new File(uploadsFolder, "imagetosave.jpg");
// ...
See also:
What does servletcontext.getRealPath("/") mean and when should I use it
Where to place and how to read configuration resource files in servlet based application?
Simplest way to serve static data from outside the application server in a Java web application
If you have a javax.servlet.ServletContext you can call:
servletContext.getRealPath("/images/imagetosave.jpg")
to get the actual path of where the image is stored.
ServletContext can be accessed from a javax.servlet.http.HttpSession.
However, you might want to look into using:
servletContext.getResource("/images/imagetosave.jpg")
or
servletContext.getResourceAsStream("/images/imagetosave.jpg")
String path = MyClass.getClass().getProtectionDomain().getCodeSource().getLocation().getPath();
This should return your absolute path based on the class's file location.
Method -1 :
//step1 : import java.net.InetAddress;
InetAddress ip = InetAddress.getLocalHost();
//step2 : provide your file path
String filepath="GIVE YOUR FILE PATH AFTER WEB FOLDER something like /images/grid.png"
//step3 : grab all peices together
String a ="http://"+ip.getHostAddress()+":"+request.getLocalPort()+""+request.getServletContext().getContextPath()+filepath;
Method - 2 :
//Step : 1-get the absolute url
String path = request.getRequestURL().toString();
//Step : 2-then sub string it with the context path
path = path.substring(0, path.indexOf(request.getContextPath()));
//step : 3-provide your file path after web folder
String finalPath = "GIVE YOUR FILE PATH AFTER WEB FOLDER something like /images/grid.png"
path +=finalPath;
MY SUGGESTION
keep the file which you want to open in the default package of your source folder and open the file directly to make things simple and clear.
NOTE : this happens because it is present in the class path of your IDE if you are coding without IDE then keep it in the place of your java compiled class file or in a common folder which you can access.
HAVE FUN
I was able to get a reference to the ServletContext in a Filter. What I like best about this approach is that it occurs in the init() method which is called upon the first load the web application meaning the execution only happens once.
public class MyFilter implements Filter {
protected FilterConfig filterConfig;
public void init(FilterConfig filterConfig) {
String templatePath = filterConfig.getServletContext().getRealPath(filterConfig.getInitParameter("templatepath"));
Utilities.setTemplatePath(templatePath);
this.filterConfig = filterConfig;
}
I added the "templatepath" to the filterConfig via the web.xml:
<filter>
<filter-name>MyFilter</filter-name>
<filter-class>com.myapp.servlet.MyFilter</filter-class>
<init-param>
<param-name>templatepath</param-name>
<param-value>/templates</param-value>
</init-param>
</filter>
<filter-mapping>
<filter-name>MyFilter</filter-name>
<url-pattern>/*</url-pattern>
</filter-mapping>
You can get path in controller:
public class MyController extends MultiActionController {
private String realPath;
public ModelAndView handleRequest(HttpServletRequest request,
HttpServletResponse response) throws Exception {
realPath = getServletContext().getRealPath("/");
String classPath = realPath + "WEB-INF/classes/" + MyClass.ServletContextUtil.class.getCanonicalName().replaceAll("\\.", "/") + ".java";
// add any code
}
You could write a ServletContextListener:
public class MyServletContextListener implements ServletContextListener
{
public void contextInitializedImpl(ServletContextEvent event)
{
ServletContext servletContext = event.getServletContext();
String contextpath = servletContext.getRealPath("/");
// Provide the path to your backend software that needs it
}
//...
}
and configure it in web.xml
<listener>
<listener-class>my.package.MyServletContextListener</listener-class>
</listener>
I am creating a Swing application with a JEditorPane that should display an HTML file named url1.html stored locally in the page folder in the root folder of the project.
I have instantiated the following String object
final String pagePath = "./page/";
and in order to be displayed by the JEditorPane pane I have created the following URL object:
URL url1 = new URL("file:///"+pagePath+"url1.html");
However when the setPage() method is called with the created URL object as a parameter:
pagePane.setPage(url1);
it throws me a java.io.FileNotFoundException error.
It seems that there is something wrong with the way url1 has been constructed. Anyone knows a solution to this problem?
The solution is to find an absolute path to url1.html make an object of java.io.File on it, and then use toURI().toURL() combination:
URL url1 = (new java.io.File(absolutePathToHTMLFile)).toURI().toURL();
Assuming if the current directory is the root of page, you can pass a relative path to File:
URL url1 = (new java.io.File("page/url1.html")).toURI().toURL();
or
URL url1 = (new java.io.File(new java.io.File("page"), "url1.html")).toURI().toURL();
But this will depend on where you run the application from. I would make it taking the root directory as a command-line argument if it is the only configurable option for the app, or from a configuration file, if it has one.
The another solution is to put the html file as a resource into the jar file of your application.
To load a resource from the classpath (as khachik mentioned) you can do the following:
URL url = getClass().getResource("page/url1.html");
or from a static context:
URL url = Thread.currentThread().getContextClassLoader().getResource("page/url1.html");
So in the case above, using a Maven structure, the HTML page would be at a location such as this:
C:/myProject/src/main/resources/page/url1.html
I would try the following
URL url = new URL("file", "", pagePath+"url1.html");
I believe by concatenating the whole string, you are running into problems. Let me know, if that helped