Error No Protocol
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Java.net.malformedurlexception No Protocol In Java
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Java.net.malformedurlexception No Protocol Null
minute: Sign up No protocol exception when reading a URL up vote -3 down vote favorite I am trying to read a URL and then convert it to a string and write the content to a text file, but I get the following exception when I compile the code. Here is my code and my exception: import java.io.*; import java.net.URL; import java.net.URLConnection;
Malformedurlexception Protocol Not Found
public class Main { public static String url = "google.com"; public static String fileName= null; public static String fileConttent ="Something"; public static void main(String[] args) throws Exception { getText(new String(url)); } public static void getText(String url) throws Exception { URL website = new URL("url\n" + " public static void main(String[] args) throws Exception {\n" + " getText(new String(url));\n" + " }\n" + "\n" + " public static void getText(String url) throws Exception {\n" + " URL website = new URL(\"url"); URLConnection connection = website.openConnection(); BufferedReader in = new BufferedReader( new InputStreamReader( connection.getInputStream())); StringBuilder response = new StringBuilder(); String inputLine; while ((inputLine = in.readLine()) != null) response.append(inputLine); in.close(); String toBeWritten = response.toString(); System.out.println(toBeWritten); } public static void createFile(String fileName,String fileContent){ Writer writer = null; try { writer = new BufferedWriter(new OutputStreamWriter( new FileOutputStream("C:\\Users\\Dell\\Documents\\t"+"fileName"), "utf-8")); writer.write(fileContent); } catch (IOException ex) { // report } finally { try {writer.close();} catch (Exception ex) {} } } } This is my exception: Exception in thread "main" java.net.MalformedURLException: no protocol: url at java.net.URL.
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Malformedurlexception Example
the company Business Learn more about hiring developers or posting ads with us Stack Overflow Questions Jobs Documentation malformed url exception Tags Users Badges Ask Question x Dismiss Join the Stack Overflow Community Stack Overflow is a community of 4.7 million programmers, just like you, helping each other. Join them; it http://stackoverflow.com/questions/24155341/no-protocol-exception-when-reading-a-url only takes a minute: Sign up java.net.MalformedURLException: no protocol on URL based on a string modified with URLEncoder up vote 7 down vote favorite 2 So I was attempting to use this String in a URL :- http://site-test.collercapital.com/Meetings/IC/DownloadDocument?meetingId=c21c905c-8359-4bd6-b864-844709e05754&itemId=a4b724d1-282e-4b36-9d16-d619a807ba67&file=\\s604132shvw140\Test-Documents\c21c905c-8359-4bd6-b864-844709e05754_attachments\7e89c3cb-ce53-4a04-a9ee-1a584e157987\myDoc.pdf In this code: - String fileToDownloadLocation = //The above string URL fileToDownload = new URL(fileToDownloadLocation); HttpGet httpget = new HttpGet(fileToDownload.toURI()); But at http://stackoverflow.com/questions/22093864/java-net-malformedurlexception-no-protocol-on-url-based-on-a-string-modified-wi this point I get the error: - java.net.URISyntaxException: Illegal character in query at index 169:Blahblahblah I realised with a bit of googling this was due to the characters in the URL (guessing the &), so I then added in some code so it now looks like so: - String fileToDownloadLocation = //The above string fileToDownloadLocation = URLEncoder.encode(fileToDownloadLocation, "UTF-8"); URL fileToDownload = new URL(fileToDownloadLocation); HttpGet httpget = new HttpGet(fileToDownload.toURI()); However, when I try and run this I get an error when I try and create the URL, the error then reads: - java.net.MalformedURLException: no protocol: http%3A%2F%2Fsite-test.collercapital.com%2FMeetings%2FIC%2FDownloadDocument%3FmeetingId%3Dc21c905c-8359-4bd6-b864-844709e05754%26itemId%3Da4b724d1-282e-4b36-9d16-d619a807ba67%26file%3D%5C%5Cs604132shvw140%5CTest-Documents%5Cc21c905c-8359-4bd6-b864-844709e05754_attachments%5C7e89c3cb-ce53-4a04-a9ee-1a584e157987%myDoc.pdf It looks like I can't do the encoding until after I've created the URL else it replaces slashes and things which it shouldn't, but I can't see how I can create the URL with the string and then format it so its suitable for use. I'm not particularly familiar with all this and was hoping someone might be able to point out to me what I'm missing to get string A into a suitably formatted URL to then use with the correct
here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company Business Learn more about http://unix.stackexchange.com/questions/85782/error-no-protocol-specified-when-running-from-remote-machine-via-ssh hiring developers or posting ads with us Unix & Linux Questions Tags Users Badges Unanswered Ask Question _ Unix & Linux Stack Exchange is a question and answer site for users of Linux, FreeBSD and other Un*x-like operating systems. Join them; it only takes a minute: Sign up Here's how it works: Anybody can ask a question Anybody can answer The best answers are voted up and rise to the top Error `No protocol specified` when running from remote machine no protocol via ssh up vote 3 down vote favorite 4 I have a script, simply to run my Graphical (GUI) Application, as below. #cat gui.sh #!/bin/bash ./gui -display 127.0.0.1:0.0 When I run it from local machine (./gui.sh) it runs perfectly fine. But when I am trying to run it from remote machine via ssh, I got following error. [root@localhost]# ssh -f 192.168.3.77 "cd /root/Desktop/GUI/ && "./gui.sh"" No protocol specified gdm: cannot connect to X server 192.168.3.77:0.0 [root@localhost]# I don't know, which protocol java.net.malformedurlexception no protocol it is asking or am I missing anything? I tried directly by starting the application, without script [ssh -f 192.168.3.77 "cd /root/Desktop/GUI/ && "./gui""], but the result is same. I have tried various combinations like ssh -Y, ssh -fY and more but the result is same! Secondly for my application, there is a must condition that, we have to first go into the directory where the program is located. Any Solutions? ssh x11 share|improve this question edited Aug 7 '13 at 22:44 Gilles 371k696751126 asked Aug 7 '13 at 6:35 Tejas 1,59021432 add a comment| 5 Answers 5 active oldest votes up vote 3 down vote The meaning of the option -display 127.0.0.1:0.0 depends on that gui program, but it's highly likely that it means “display on the X display 127.0.0.1:0.0”. This is the first local X display, accessed over TCP. This is almost certainly wrong for two reasons. First, the local X display should be :0, not 127.0.0.1:0, because including an IP address causes the traffic to go through TCP instead of local access. Going through TCP may not work depending on whether the X server accepts TCP connections. Even if it does, you lose the optimizations that local displays have. The display to use is normally indicated by the DISPLAY environment variable, and that variable tends to be set correctly automatically. (Usually, if DISPLAY has the wrong value, it's because you've been messing with it. The ma