Open a URL from Servlet

Hi,
          I'm running a servlet and wants to open a URL for a file, how do I do that ?
          (To get the same effect as if I open the file from a browser)
          thanks in advance..
          ..Per

If you mean to read the contents of a URL, then see URL.getConnection. If
          you mean to download a file to a browser, then search this newsgroup for
          "download".
          Cameron Purdy
          Tangosol, Inc.
          http://www.tangosol.com
          +1.617.623.5782
          WebLogic Consulting Available
          "Per Lovdinger" <[email protected]> wrote in message
          news:3a9a7650$[email protected]..
          > Hi,
          >
          > I'm running a servlet and wants to open a URL for a file, how do I do that
          >
          > (To get the same effect as if I open the file from a browser)
          >
          > thanks in advance..
          > ..Per
          >
          >

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Want to open a URL from java application?

I want to open a URL (say http://yahoo.com) from my java application without using servlets. PLzzz help asap

Include these packages:
import java.net.*;
import java.io.*;
Proxy Setting If any:
//getProxy your proxy address default "No Proxy".
if (!getProxy.equals("No Proxy"))
System.getProperties().put("proxySet","true");
System.getProperties().put("proxyHost",getProxy);
System.getProperties().put("proxyPort","80");
Main code:
URL url = new URL("http://mail.yahoo.com");
URLConnection connection1 = url.openConnection();
String filetext=postURL(connection1);
Function postURL:
public String postURL(URLConnection connection)
try
BufferedReader httpResponse = new BufferedReader(new InputStreamReader(connection.getInputStream()));
String line = new String();
StringBuffer buffer = new StringBuffer();
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buffer.append(line);
httpResponse.close();
return buffer.toString();
catch(Exception e)
System.out.println("Error:"+e);
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If I open another URL from my homepage, the "go back one page" button is grayed out so I can't get back to my home page.

This problem only started recently.
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Start Firefox in [[Safe Mode]] to check if one of the add-ons is causing the problem (switch to the DEFAULT theme: Tools > Add-ons > Themes).
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Opening an URL from a JAVA application

hi,
iam new to Java.
Currently iam designing a simle GUI which takes input(URL) from user in a test field and has a button.Upon clicking button i want to open the webPage specified by that URL in a default webBrowser.
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advanced thanks for your help.
vd.rach

getting an error on that line as follows: "unreported
exception; must be caught or declared to be thrown".exec() throws IOException so catch it.
try {
Runtime.getRuntime.exec(...);
catch(IOException e) {
System.err("Cannot launch the browser: " + e.getMessage());
}On the side note: I found javaworld's [url http://www.javaworld.com/javatips/jw-javatip66_p.html]solution most useful in serving this purpose in a project, even though its a little bit platform specific.

Issue in opening an URL from UI

Hi Experts,
When clicking button in an assignment block of campaign screen, it should open an URL (SRM screen) in new window.
I have written the following code in htm page inside script tag.
window.open("lv_url").focus();
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Hi Ramya,
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One solution is to generate a guid and pass it with the url so that the url will differ (since the guid differes the url will not be the same). Check the below sample code,
<%
CALL FUNCTION 'GUID_CREATE'
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CONCATENATE 'guid=' lv_guid INTO lv_query.
CALL METHOD cl_crm_web_utility=>create_url
    EXPORTING
      iv_path                     = lc_path
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      iv_no_cookie            = 'X'
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Regards,
Arun

Opening a file from Servlet

Hi,
I`m trying to open a file from Tomcat Servlet init method but it fails with java.io.FileNotFoundException: test.properties (The system cannot find the file specified).
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What is the error saying???
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java.lang.NullPointerException
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at com.evermind[Oracle Application Server Containers for J2EE 10g (10.1.2.0.2)].server.http.ServletRequestDispatcher.invoke(ServletRequestDispatcher.java:826)
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at com.evermind[Oracle Application Server Containers for J2EE 10g (10.1.2.0.2)].server.http.HttpRequestHandler.processRequest(HttpRequestHandler.java:830)
at com.evermind[Oracle Application Server Containers for J2EE 10g (10.1.2.0.2)].server.http.AJPRequestHandler.run(AJPRequestHandler.java:224)
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Suresh

getServletContext().getRequestDispatcher("http://<my_host>/forms/frmservlet?config=test&session_id="+session.getId()).forward(request,response)
From the Javadocs for ServletContext.getRequestDispather():
http://java.sun.com/j2ee/1.4/docs/api/javax/servlet/ServletContext.html#getRequestDispatcher(java.lang.String)
The pathname must begin with a "/" and is interpreted as relative to the current context root. Use getContext to obtain a RequestDispatcher for resources in foreign contexts. This method returns null if the ServletContext cannot return a RequestDispatcher.
-steve-

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Opening a URL from an application

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my problem is that i need to open that certain URL ( predefined) from a running application and not an applet, it hasn't worked for me, This is the only problem. is it possible ? please i need your help with this , i'd really appreciate it if someone would help ... thanks a lot for your time.

hii guys am really sorry for the confusion ,the problem was actually with the java security poilcy not with the code as i thought.. i thought my code was wrong but it turned out to be correct.. here it is for anyone who's interested... but thanks anyway for everything... tc
Runtime rt = Runtime.getRuntime();
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Edited by: Paul Thomson on Feb 18, 2010 4:00 PM

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Open a URL from Servlet

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