url = new java.net.URL()

Question

url = new java.net.URL(s) doesn't work for me.

I have a string C:\apache-tomcat-6.0.29\webapps\XEPServlet\files\m1.fo and need to make a link and give it to my formatter for output, but malformed url recieved. It seems that it doesn't make my string to url. I want also mention, that file m1.fo file is in files folder, in my webapp\product\, and I gave the full path to string like: getServletContext().getRealPath("files/m1.fo"). What I am doing wrong? How can I recieve the url link?

@davogotland -- it should NOT be expected to be at 100%. One of his questions has no answer, and some of them don't seem to have any clear winner for an answer. 0% is a disincentive, but don't punish someone for not having 100% or even 75%. — Jason S
– Jason S, Commented Jan 14, 2011 at 22:01

Jcs · Accepted Answer · 2011-01-14 21:59:28Z

5

It is possible to get an URL from a file path with the java.io.File API :

String path = "C:\\apache-tomcat-6.0.29\\webapps\\XEPServlet\\files\\m1.fo";
File f = new File(path);
URL url = f.toURI().toURL();

answered Jan 14, 2011 at 21:59

Jcs

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2 Comments

dkarp Over a year ago

You're going to need to escape those backslashes.

user207421 Over a year ago

You don't need backslashes in filenames at all in Java. Use /.

Martin Ždila · Accepted Answer · 2011-01-14 21:59:36Z

1

Try: file:///C:/apache-tomcat-6.0.29/webapps/XEPServlet/files/m1.fo

answered Jan 14, 2011 at 21:59

Martin Ždila

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Community · Accepted Answer · 2013-02-06 13:48:53Z

1

It isn't preferable to write file:/// . Indeed it works on windows system,but in unix - there were problems. Instead of using

myReq.put("xml", new String []{"file:" + System.getProperty("file.separator") + 
                        getServletContext().getRealPath(DESTINATION_DIR_PATH) + 
                        System.getProperty("file.separator") + xmlfile});

you can write

myReq.put("xml", new String [] {getUploadedFileURL (xmlfile)} );

, where

public String getUploadedFileURL(String filename) {
    java.io.File filePath = new java.io.File(new 
            java.io.File(getServletContext().getRealPath(DESTINATION_DIR_PATH)), 
            filename);

    return filePath.toURI().toURL().toString();

edited Feb 6, 2013 at 13:48

CommunityBot

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answered Jan 20, 2011 at 7:00

Vardges

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1 Comment

Payedimaunt Over a year ago

Ok +1 for me, but maybe the solution could be exposed more clearly, I guess.

Konstantin Komissarchik · Accepted Answer · 2011-01-14 21:58:59Z

0

A file system path is not a URL. A URL is going to need a protocol prefix for one. To reference file system use "file:" in front of your path.

answered Jan 14, 2011 at 21:58

Konstantin Komissarchik

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3 Comments

Vardges Over a year ago

I set the path file:///C:/apache-tomcat-6.0.29/webapps/XEPServlet/files/m1.fo and also used this: java.net.URL url = null; try{ url = new java.net.URL(s); } catch(java.net.MalformedURLException e){ errors = true; errorReport += "malformed url: '"+s+"'\n"; } return url;

Vardges Over a year ago

Hi. I have already started my webservice, but it formats the file with static path. Now I want to keep the same path, but take any file. My uploader gives the path like file:///C:\Documents and Settings\Administrator\Desktop\m1.fo Question is : How can I take just the name from this path- file:///C:\Documents and Settings\Administrator\Desktop\m1.fo ??? Thank you in advance

Konstantin Komissarchik Over a year ago

Take a look at the URL class javadoc. There are a number of methods that allow you to retrieve various parts of the URL.

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