Using tomcat 6, i created a template inside www/templates/templatefile.html
how do I access it from a servlet ? i want to read the html file and parse it.
I tried getting the real path using request.getRealPath(request.getServletPath())
and from there to go to the templates directory but for some reason it still can't find the file.
Assuming that www is a folder in the public web content root, then you can use ServletContext#getRealPath() to convert a relative web path to an absolute disk file system path as follows:
String relativeWebPath = "/www/templates/templatefile.html";
String absoluteDiskPath = getServletContext().getRealPath(relativeWebPath);
File file = new File(absoluteDiskPath);
InputStream input = new FileInputStream(file);
// ...
Note that this won't work when the WAR is not expanded (Tomcat in default trim does it, but it can be configured not to do that). If all you want to end up is having an InputStream of it, then you'd rather like to use ServletContext#getResourceAsStream() instead.
String relativeWebPath = "/www/templates/templatefile.html";
InputStream input = getServletContext().getResourceAsStream(relativeWebPath);
// ...
Related
I can't figure out why fi.exists() returns false here. I can browse to the file via the browser at contextPath+"/images/default.png
String contextPath = req.getContextPath();
File fi = new File(contextPath+"/images/default.png");
exists = fi.exists();
I think you missunderstood what the context path is.
If you application is deployed on yourdomain.com/app, the context path will be /app.
It is used to tell the client where to look for resources.
When you do contextPath+"/images/default.png", you the path would be dependent of the deployment path (in this case it would be the file /app/images/default.png).
If you want the file next to the installation of your application server, you can use "images/default.png".
If you want to access resource files, you may want to try Thread.currentThread().getContextClassLoader().getResource("images/default.png") instead of files.
If you want to check if a context related resource exist, you can do it as stated here:
boolean exists=req.getServletContext().getResource("images/default.png")!=null;`
or
String path=req.getServletContext().getRealPath("images/default.png");`
Rather than getContextPath(), I needed to use getRealPath():
String path = req.getServletContext().getRealPath("/images/default.png");
File fi = new File(path);
I have a text file, that I am trying to convert to a String array (where each line is an element of the array).
I have found a code here that seems to do this perfectly, the only problem is, even when I use the relative path of the file, I get a FileNotFoundException.
The relative file path:
String path = "android.resource://"+getPackageName()+"/app/src/main/res/raw/"
+ getResources().getResourceEntryName(R.raw.txtAnswers);
When I try to use this path in a reader
BufferedReader reader = null;
try{
reader = new BufferedReader(new FileReader(path));
......
The path is underlined in red and says FileNotFoundException.
Perhaps it is expecting another type of path (not relative?) or did I get the path wrong?
The relative file path
That is not a file path.
Perhaps it is expecting another type of path
res/raw/... is a file on your hard drive. It is not a file on the device.
To access raw resources, use getResources() (called on any Context, like an Activity or Service) to get a Resources object. Then, call openRawResource(), passing in your resource ID, to get an InputStream on that resource's contents.
According to me the best way would be to past that txt file in assets folder of your source code.
To use that File just use the following snippet:
InputStream inputStream = getAssets().open("YOUR_TEXT_FILE");
It returns InputStream which further can be used to read the File & convert the data into a Array
I am writing webApp that use jsp file to call java class directly to get results.
The JSP file is like:
<%
String queryKey = request.getParameter("id");
int jobID = Integer.parseInt(queryKey);
out.println(jobID);
ArrayList<Integer> myTopList = JobRecByBoWJaccard.topJobsByBoW(jobID);
%>
In java classes the files are accessed through:
BufferedReader br = new BufferedReader(new FileReader("WebContent/StopWords/stop-words-english1.txt"));
and
private InputStream modelInputT = new FileInputStream("WebContent/OpenNLP_Models/en-token.bin");
The tomcat cannot find the referenced files and someone said use getResourceAsStream, but that is for servlets. I call the java class directly without any servlets.
private InputStream modelInputT = = this.getClass().getClassLoader().getResourceAsStream("WebContent/OpenNLP_Models/en-token.bin");
This cause java class cannot find the file as well. Need help and how to make changes to these java classes?
The tomcat cannot find the referenced files ...
That is because the pathnames are incorrect unless the JVM's current directory is the parent directory of the "WebContent" directory. When you use FileInputStream to open a file, relative pathname are resolved relative to the current directory of the JVM when it was launched.
... and someone said use getResourceAsStream, but that is for servlets.
No. That's not correct either. That method is not "for servlets". The purpose of that method is to open a resource that is on the class / classloader's classpath. If your "WebContent/StopWords/stop-words-english1.txt" is in the webapp's "/WEB-INFO/classes" or in a JAR file in "/WEB-INFO/lib", then getResourceAsStream will find it.
In your case, it seems like you are talking about the "WebContent" directory that corresponds to the default servlet.
In that case, read this Q&A - How can I get real path for file in my WebContent folder?.
So if you are trying to access those files from within a JSP, it seems as if you should be writing this:
new FileReader("WebContent/StopWords/stop-words-english1.txt")
as this:
new FileReader(getServletContext().getRealPath(
"/StopWords/stop-words-english1.txt"))
This question already has answers here:
How to find the working folder of a servlet based application in order to load resources
(3 answers)
Closed 6 years ago.
I currently have a bunch of images in my .war file like this.
WAR-ROOT
-WEB-INF
-IMAGES
-image1.jpg
-image2.jpg
-index.html
When I generate html via my servlets/jsp/etc I can simple link to
http://host/contextroot/IMAGES/image1.jpg
and
http://host/contextroot/IMAGES/image1.jpg
Not I am writing a servlet that needs to get a filesystem reference to these images (to render out a composite .pdf file in this case). Does anybody have a suggestion for how to get a filesystem reference to files placed in the war similar to how this is?
Is it perhaps a url I grab on servlet initialization? I could obviously have a properties file that explicitly points to the installed directory but I would like to avoid additional configs.
If you can guarantee that the WAR is expanded, then you can use ServletContext#getRealPath() to convert a relative web path to an absolute disk file system which you can further use in the usual Java IO stuff.
String relativeWebPath = "/IMAGES/image1.jpg";
String absoluteDiskPath = getServletContext().getRealPath(relativeWebPath);
File file = new File(absoluteDiskPath);
InputStream input = new FileInputStream(file);
// ...
However, if you can't guarantee that the WAR is expanded (i.e. all resources are still packaged inside WAR) and you're actually not interested on the absolute disk file system path and all you actually need is just an InputStream out of it, then use getServletContext().getResourceAsStream() instead.
String relativeWebPath = "/IMAGES/image1.jpg";
InputStream input = getServletContext().getResourceAsStream(relativeWebPath);
// ...
See also:
getResourceAsStream() vs FileInputStream
Use the getRealPath method of ServletContext.
Ex:
String path = getServletContext().getRealPath("WEB-INF/static/img/myfile.jpeg");
This is relatively straight forward you simply use the class loader to fetch the files from the class plath. :
InputStream is = YourServlet.class.getClassLoader().getResourceAsStream("IMAGES/img1.jpg");
There are a few other getResoruce classes that are worth looking at. Also you don't have to fetch the class loader through the class variable on your servlet. Any class that you happen to know has been loaded by the container should work .
If you know the relative location of the files you could ask the runtime about the exact location using
Thread.currentThread().getContextClassLoader().getResource(<relative-path>/<filename>)
This would give you an URL to the location where the specified image can be found. This URL can be used to read the specified file or you can split it to use the different parts of the URL for further processing.
I need to get real path for file in my WebContent directory, so that framework that I use can access that file. It only takes String file as attribute, so I need to get the real path to this file in WebContent directory.
I use Spring Framework, so solution should be possible to make in Spring.
If you need this in a servlet then use getServletContext().getRealPath("/filepathInContext")!
getServletContext().getRealPath("") - This way will not work if content is being made available from a .war archive. getServletContext() will be null.
In this case we can use another way to get real path. This is example of getting a path to a properties file C:/Program Files/Tomcat 6/webapps/myapp/WEB-INF/classes/somefile.properties:
// URL returned "/C:/Program%20Files/Tomcat%206.0/webapps/myapp/WEB-INF/classes/"
URL r = this.getClass().getResource("/");
// path decoded "/C:/Program Files/Tomcat 6.0/webapps/myapp/WEB-INF/classes/"
String decoded = URLDecoder.decode(r.getFile(), "UTF-8");
if (decoded.startsWith("/")) {
// path "C:/Program Files/Tomcat 6.0/webapps/myapp/WEB-INF/classes/"
decoded = decoded.replaceFirst("/", "");
}
File f = new File(decoded, "somefile.properties");
you must tell java to change the path from your pc into your java project so
if you use spring use :
#Autowired
ServletContext c;
String UPLOAD_FOLDEdR=c.getRealPath("/images");
but if you use servlets just use
String UPLOAD_FOLDEdR = ServletContext.getRealPath("/images");
so the path will be /webapp/images/ :)
In situations like these I tend to extract the content I need as a resource (MyClass.getClass().getResourceAsStream()), write it as a file to a temporary location and use this file for the other call.
This way I don't have to bother with content that is only contained in jars or is located somewhere depending on the web container I'm currently using.
Include the request as a parameter. Spring will then pass the request object when it calls the mapped method
#RequestMapping .....
public String myMethod(HttpServletRequest request) {
String realPath = request.getRealPath("/somefile.txt");
...
You could use the Spring Resource interface (and especially the ServletContextResource): http://static.springsource.org/spring/docs/current/javadoc-api/org/springframework/core/io/Resource.html
This approach uses the resource loader to get the absolute path to a file in your app, and then goes up a few folders to the app's root folder. No servlet context required! This should work if you have a "web.xml" in your WEB-INF folder. Note that you may want to consider using this solely for development, as this type of configuration is usually best stored externally from the app.
public String getAppPath()
{
java.net.URL r = this.getClass().getClassLoader().getResource("web.xml");
String filePath = r.getFile();
String result = new File(new File(new File(filePath).getParent()).getParent()).getParent();
if (! filePath.contains("WEB-INF"))
{
// Assume we need to add the "WebContent" folder if using Jetty.
result = FilenameUtils.concat(result, "WebContent");
}
return result;
}
my solve for: ..../webapps/mydir/ (..../webapps/ROOT/../mydir/)
String dir = request.getSession().getServletContext().getRealPath("/")+"/../mydir";
Files.createDirectories(Paths.get(dir));
try to use this when you want to use arff.txt in your development and production level too
String path=getServletContext().getRealPath("/WEB-INF/files/arff.txt");