Create A File in the WebContent Folder From Java

Question

I wanted to know how to create a file in the WebContent folder of a dynamic web project using Java?

The basic question remaining is how to get the path of the WebContent folder.

Note: No servlet is to be used!

Edit:

Okay, i am trying to create a new xml file from java method. I want the file to be created in the WebContent folder so that the file is created even when the application is deployed.

I am using Jboss, maven, JSF to create the dynamic web project. I need the xml file to pass data to highcharts. Please note that i will be using this method only.

Overview:

Create xml file on request

XML file to created in the WebContent folder itself

Use this xml file to pass data

can you please more elobarate your problem how you are trying ? — Hemant Metalia, Feb 09 '12 at 12:07
Not a good idea, especially since alot of newer app servers are using virtual filesystems to deploy your web content. — Perception, Feb 09 '12 at 12:11
You cannot do this without violating the servlet spec, making your web application depend on vendor specific side effects. — Thorbjørn Ravn Andersen, Feb 09 '12 at 12:12
There is no WebContent folder. WebContent is the name of the source of the webapp in Eclipse. When deployed on the server, there could be a folder, but there could also be a war file, which is not writable. And if you redeploy the app, you will lose all the created files. Store your files elsewhere. — JB Nizet, Feb 09 '12 at 12:18
This is a matter of persisting data between invocations. Use a database or the Settings API for that. — Thorbjørn Ravn Andersen, Feb 09 '12 at 12:21
@JBNizet so i cannot create any file dynamically in the webapp? — yash, Feb 09 '12 at 12:21
You can, but not in the webapp's folder. You have your database, and the whole file system at your disposition. — JB Nizet, Feb 09 '12 at 12:28
@JBNizet How will i access the file from client side using the $.get() method in highcharts? — yash, Feb 09 '12 at 12:36
By having a servlet that reads the file from the file system, and writes it to the HTTP response output stream. — JB Nizet, Feb 09 '12 at 12:45
@JBNizet servlet with JSF? please note that am not using JSP or anything else.. JSF facelets, JBoss AS 7, Maven, Oracle — yash, Feb 09 '12 at 12:49
So what? A JEE webapp can define and use servlets even if it uses JSF. And I guess you could implement it using a JSF action (or whatever JSF calls them). Anything that can access an HttpServletResponse object and writes to its output stream is fine. — JB Nizet, Feb 09 '12 at 12:52
@JBNizet thanks for your answer but i think it is going out of the real question. JEE and dynamic web, different.. i think either i am failing to make you understand or you are not understanding me completely. — yash, Feb 09 '12 at 13:03
what is "dynamic web". This is not a standard Java term. Eclipse use the term "Dynamic web project" to refer to an Eclipse project to generate JEE web application. And you can define a servlet in a "Dynamic web project". JSF is a specification that is part of the global JEE specification. And JSF apps can only be deployed in a web container that supports JEE webapps. — JB Nizet, Feb 09 '12 at 13:07
No. Use google. It's 15 lines of basic IO to read a file, and write what you read to a stream. I won't do the work for you. Google for Java IO tutorial. — JB Nizet, Feb 09 '12 at 13:47

kornero · Answer 1 · 2012-02-09T12:22:28.800

Glassfish solution. AbstractSearchPageBean - any of your class

private static final String WEB_INF = "WEB-INF";

public static String getWebPath() {
    final String webInfPath = getWebInfPath();
    return webInfPath.substring(0, webInfPath.indexOf(WEB_INF) - 1);
}

public static String getWebInfPath() {
    String filePath = "";

    java.net.URL url = AbstractSearchPageBean.class.getResource("AbstractSearchPageBean.class");
    if (url != null) {
        String className = url.getFile();
        filePath = (className.contains(WEB_INF)) ? className.substring(0, className.indexOf(WEB_INF) + WEB_INF.length()) : className;
    }
    return filePath.replace("%20", " ");
}

// Create file in webapp/xml directory
private void createXmlFile(String xml) {
    try {
        String fileName = System.currentTimeMillis() + ".xml";
        File file = new File(Settings.getWebPath() + File.separatorChar + "xml" + File.separatorChar + fileName);
        logger.debug("parseXML(): Creating file: " + file.getAbsolutePath());
        if (file.createNewFile()) {
            FileWriter fw = new FileWriter(file);
            fw.write(this.parseXML(xml));
            fw.flush();
            fw.close();
            logger.debug("parseXML(): file saved to the: " + Settings.getAPPLICATION_DOMAIN() + '/' + fileName);
        } else {
            logger.warn("parseXML(): Can't create file: " + file.getAbsolutePath());
        }
    } catch (IOException ioe) {
        logger.error("parseXML(): Bad save file: ", ioe);
    }
}

Create A File in the WebContent Folder From Java

1 Answers1