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How to make http call from standalone java application

I'm making a small dictionary kind of app using java swings. I'm using oxford dictionary api for that. Is there any way to make a simple ajax request in java without using servelets and all advanced java concepts. As in android we use http url connection to do this job.I googled a lot for finding this but I could't find a solution as every page is showing results using servelets. But I know core java alone.If it is possible to make ajax call without servelts please help me...Thanks in advance...

Use HttpURLConnection class to make http call.
If you need more help for that then go for offical documentation site of java Here
Example
public class JavaHttpUrlConnectionReader {
public static void main(String[] args) throws IOException{
String results = doHttpUrlConnectionAction("https://your.url.com/", "GET");
System.out.println(results);
}
public static String doHttpUrlConnectionAction(String desiredUrl, String requestType) throws IOException {
BufferedReader reader = null;
StringBuilder stringBuilder;
try {
HttpURLConnection connection = (HttpURLConnection) new URL(desiredUrl).openConnection();
connection.setRequestMethod(requestType);// Can be "GET","POST","DELETE",etc
connection.setReadTimeout(3 * 1000);
connection.connect();// Make call
reader = new BufferedReader(new InputStreamReader(connection.getInputStream()));// Reading Responce
stringBuilder = new StringBuilder();
String line;
while ((line = reader.readLine()) != null) {
stringBuilder.append(line).append("\n");
}
return stringBuilder.toString();
} catch (IOException e) {
throw new IOException("Problam in connection : ", e);
} finally {
if (reader != null) {
try {
reader.close();
} catch (IOException ioe) {
throw new IOException("Problam in closing reader : ", ioe);
}
}
}
}
}
It will make a call and give response as return string. If you want to make POST call the need to do some extra for that :
try{
DataOutputStream wr = new DataOutputStream(connection.getOutputStream());
wr.write(postParam.getBytes());
} catch(IOException e){
}
Note : Here postParam is String type with value somthing like "someId=156422&someAnotherId=32651"
And put this porson befor connection.connect() statement.

How can I load an xml resource file from within an executible Jar file and save it to the folder the jar is located in?

I have a custom java server. It uses an external xml config file.
I have some command line options to help the user, the usual stuff for showing a help file, setting ports, etc...
I've recently added a command to generate a default config file for the server. It's an xml file. After researching my options, packing a default xml file in the jar seemed to be the way to go, but I'm obviously missing something.
So far my code looks like this:
public class ResourceLoader {
private File outFile = null;
private Reader fileReader = null;
private Writer fileWriter = null;
private InputStream is = null;
private char[] buffer = null;
public ResourceLoader() {
outFile = new File("default-server.xml");
}
public void generateDefaultServerXml() {
is = ResourceLoader.class.getResourceAsStream("/default-server.xml");
if (is == null) {
System.out.println("Configuraiton File generation failed. The InputStream is null.");
} else {
fileReader = new InputStreamReader(is);
}
buffer = new char[4096];
FileOutputStream fos;
try {
fos = new FileOutputStream(outFile);
fileWriter = new OutputStreamWriter(fos);
while (fileReader.read(buffer) != -1) {
fileWriter.write(buffer, 0, buffer.length);
fileWriter.flush();
buffer = new char[4096];
}
} catch (FileNotFoundException e) {
// TODO Auto-generated catch block
e.printStackTrace();
} catch (IOException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
try {
fileReader.close();
fileWriter.close();
} catch (IOException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
}
}
The code above works perfectly fine when I run it in eclipse, but intitially, after I export the jar file the server could not locate the default-server.xml file when I run the command from the terminal.
The file itself is located in a package called main.resources along with some other config files and the above class.
I have since moved the ResourceLoader.class to another package. After doing that the server seems to find the xml file in the main.resources package (InputStream is not null) but the resulting generated default-server.xml file is empty.
Again, this all works perfectly well when I run it in eclipse, it's only after I export the project and try issue the command from the terminal that the process fails. What am I doing wrong?
The above class is instantiated, and the generateDefaultServerXml() is called, from the main method of the server.
EDIT: My path for writing default-server.xml was slightly wrong. Now that I've adjusted it the code works exactly as expected when I run it in Eclipse. The resource is read in the correct way, and written to the file in the correct location. But it still doesn't work when I try the same thing from the jar file.

You current line ResourceLoader.class.getResourceAsStream("/default-server.xml") means that you are trying to load a resource named default-server.xml from the root of your classpath, or put simpler, from the root of your jar file. This means that xml file should NOT be in any package inside the jar file.
When you assemble your jar file and then run jar tf my.jar on it, do you see your default-server.xml file? Does it reside in some package or in the root of the jar file?

The problem here is since you are packaging the application as a jar. The procedure to call an external resource is quite different.
You need to have a folder structure as
root
--your jar
--your xml file
Your code shallwork if the application is using an default-server.xml file inside the jar.
Otherwise, Replace below line in your code if you want to use an external default xml file.
is = new FileInputStream("./default-server.xml");
If the output file you want at root location the use below code
public ResourceLoader() {
outFile = new File("./default-server.xml");
}
Alternate code as per discussion
public class ResourceLoader {
public void generateDefaultServerXml() {
try {
String defaultxmltext =readFileToString("/default-server.xml");
writeFileFromInputString(defaultxmltext);
} catch (IOException e) {
//exception
}
}
public static void writeFileFromInputString(String everything) throws IOException {
try (BufferedWriter writer = new BufferedWriter(new FileWriter("./default-server.xml"))) {
everything = everything.replaceAll("\n", System.getProperty("line.separator"));
writer.write(everything);
}
}
public static String readFileToString(String path) throws IOException {
String everything = null;
try (BufferedReader br = new BufferedReader(new FileReader(path))) {
StringBuilder sb = new StringBuilder();
String line = br.readLine();
while (line != null) {
sb.append(line);
sb.append(System.lineSeparator());
line = br.readLine();
}
everything = sb.toString();
}
return everything;
}
}
Hope this helps

consider your file located on src/main/resources try this
getClass().getClassLoader().getResource(fileName)

well as far as i can see your main problem is that you are passing the wrong path, since you mentioned the xml is under main.resources you will need to add this to the path when trying to load the file, here is a sample piece of code that should work for you
Scanner sc = null;
PrintWriter writer = null;
try {
sc = new Scanner(getClass().getResourceAsStream("main/resources/server.xml"));
writer = new PrintWriter("./default_server.xml", "UTF-8");
while(sc.hasNextLine()) {
writer.println(sc.nextLine());
}
} catch (Exception e) {
} finally {
if(sc != null) {
sc.close();
}
if(writer != null){
writer.close();
}
}

Cannot read data from file

I am trying to read values from CSV file which is present in package com.example.
But when i run code with the following syntax:
DataModel model = new FileDataModel(new File("Dataset.csv"));
It says:
java.io.FileNotFoundException:Dataset.csv
I have also tried using:
DataModel model = new FileDataModel(new File("/com/example/Dataset.csv"));
Still not working.
Any help would be helpful.
Thanks.

If this is the FileDataModel from org.apache.mahout.cf.taste.impl.model.file then it can't take an input stream and needs just a file. The problem is you can't assume the file is available to you that easily (see answer to this question).
It might be better to read the contents of the file and save it to a temp file, then pass that temp file to FileDataModel.
InputStream initStream = getClass().getClasLoader().getResourceAsStream("Dataset.csv");
//simplistic approach is to put all the contents of the file stream into memory at once
// but it would be smarter to buffer and do it in chunks
byte[] buffer = new byte[initStream.available()];
initStream.read(buffer);
//now save the file contents in memory to a temporary file on the disk
//choose your own temporary location - this one is typical for linux
String tempFilePath = "/tmp/Dataset.csv";
File tempFile = new File(tempFilePath);
OutputStream outStream = new FileOutputStream(tempFile);
outStream.write(buffer);
DataModel model = new FileDataModel(new File(tempFilePath));
...

public class ReadCVS {
public static void main(String[] args) {
ReadCVS obj = new ReadCVS();
obj.run();
}
public void run() {
String csvFile = "file path of csv";
BufferedReader br = null;
String line = "";
String cvsSplitBy = ",";
try {
br = new BufferedReader(new FileReader(csvFile));
while ((line = br.readLine()) != null) {
// Do stuff here
}
} catch (FileNotFoundException e) {
e.printStackTrace();
} catch (IOException e) {
e.printStackTrace();
} finally {
if (br != null) {
try {
br.close();
} catch (IOException e) {
e.printStackTrace();
}
}
}
System.out.println("Done");
}
}

CSV file which is present in package com.example
You can use getResource() or getResourceAsStream() to access the resource from within the package. For example
InputStream is = getClass().getResourceAsStream("/com/example/Dataset.csv");//uses absolute (package root) path
BufferedReader br = new BufferedReader(new InputStreamReader(is));
//read from BufferedReader
(note exception handling and file closing are omitted above for brevity)

RSS reader openStream()

I am new to Java , but really want to become better at it. I'm trying to write a simple RSS reader. Here's the code:
import java.io.*;
import java.net.*;
public class RSSReader {
public static void main(String[] args) {
System.out.println(readRSS("http://www.usnews.com/rss/health-news"));
}
public static String readRSS(String urlAddress){
try {
URL rssUrl = new URL(urlAddress);
BufferedReader in = new BufferedReader(new InputStreamReader(rssUrl.openStream()));
String sourceCode = "";
String line;
while((line = in.readLine())!=null){
if(line.contains("<title>")){
int firstPos = line.indexOf("<title>");
String temp = line.substring(firstPos);
temp = temp.replace("<title>","");
int lastPos = temp.indexOf("</title>");
temp = temp.substring(0,lastPos);
sourceCode +=temp+"\n";
}
}
System.out.println("YAAAH"+sourceCode);
in.close();
return sourceCode;
} catch (MalformedURLException ue) {
System.out.println("Malformed URL");
} catch (IOException ioe) {
System.out.println("WTF?");
}
return null;
}
}
But it is catching IOException all the time, and I see "WTF".
I realised that the whole program fails when OpenStream() starts its' work.
I don't know how to fix it.

As indicated, you would need to set your proxy parameters/credentials right before you establish a connection.
Set proxy username and password only in case your proxy is authenticated.
public static String readRSS(String urlAddress) {
System.setProperty("http.proxyHost", YOUR_PROXY_HOST);
System.setProperty("http.proxyPort", YOUR_PROXY_PORT);
//Below 2 for authenticated proxies only
System.setProperty("http.proxyUser", YOUR_USERNAME);
System.setProperty("http.proxyPassword", YOUR_PASSWORD);
try {
...
I tested your method behind a proxy and it works perfectly, after setting the parameters i.e.

Need to write JUnit Test case

I am a fresher java developer.
But I dont have much knowledge about writing Junit test cases.
I am going to have a job test soon.
For which they want me to write a program
To read HTML from any website say "http://www.google.com" ( You
can use any API of inbuilt APIs in Java like URLConnection )
Print on console the HTML from the url above and save it to a file (
web-content.txt) in local machine.
JUnit test cases for above
programme.
I have completed the first two steps as below:
import java.io.*;
import java.net.*;
public class JavaSourceViewer{
public static void main (String[] args) throws IOException{
System.out.print("Enter url of local for viewing html source code: ");
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String url = br.readLine();
try {
URL u = new URL(url);
HttpURLConnection uc = (HttpURLConnection) u.openConnection();
int code = uc.getResponseCode();
String response = uc.getResponseMessage();
System.out.println("HTTP/1.x " + code + " " + response);
InputStream in = new BufferedInputStream(uc.getInputStream());
Reader r = new InputStreamReader(in);
int c;
FileOutputStream fout=new FileOutputStream("D://web-content.txt");
while((c = r.read()) != -1){
System.out.print((char)c);
fout.write(c);
}
fout.close();
} catch(MalformedURLException ex) {
System.err.println(url + " is not a valid URL.");
} catch (IOException ie) {
System.out.println("Input/Output Error: " + ie.getMessage());
}
}
}
Now I need help with 3rd step.

You need to extract a method therefore like this
package abc.def;
import java.io.*;
import java.net.*;
public class JavaSourceViewer {
public static void main(String[] args) throws IOException {
System.out.print("Enter url of local for viewing html source code: ");
BufferedReader br =
new BufferedReader(new InputStreamReader(System.in));
String url = br.readLine();
try {
FileOutputStream fout = new FileOutputStream("D://web-content.txt");
writeURL2Stream(url, fout);
fout.close();
} catch (MalformedURLException ex) {
System.err.println(url + " is not a valid URL.");
} catch (IOException ie) {
System.out.println("Input/Output Error: " + ie.getMessage());
}
}
private static void writeURL2Stream(String url, OutputStream fout)
throws MalformedURLException, IOException {
URL u = new URL(url);
HttpURLConnection uc = (HttpURLConnection) u.openConnection();
int code = uc.getResponseCode();
String response = uc.getResponseMessage();
System.out.println("HTTP/1.x " + code + " " + response);
InputStream in = new BufferedInputStream(uc.getInputStream());
Reader r = new InputStreamReader(in);
int c;
while ((c = r.read()) != -1) {
System.out.print((char) c);
fout.write(c);
}
}
}
Ok, now you are able to write the JUnit-Test.
package abc.def;
import java.io.ByteArrayOutputStream;
import java.io.IOException;
import java.net.MalformedURLException;
import org.junit.Test;
import junit.framework.TestCase;
public class MainTestCase extends TestCase {
#Test
public static void test() throws MalformedURLException, IOException{
ByteArrayOutputStream baos = new ByteArrayOutputStream();
JavaSourceViewer.writeURL2Stream("http://www.google.de", baos);
assertTrue(baos.toString().contains("google"));
}
}

First move your code from the main method to other methods in the
JavaSourceViewer.
Second make these methods return something that
you can test. e.g. the file name of the output file or the Reader.
Then create a class for the unit test
import org.junit.* ;
import static org.junit.Assert.* ;
public class JavaSourceViewerTest {
#Test
public void testJavaSourceViewer() {
String url = "...";
JavaSourceViewer jsv = new JavaSourceViewer();
// call you methods here to parse the site
jsv.xxxMethod(...)
....
// call you checks here like:
// <file name to save to output data from your code, actual filename> - e.g.
assertEquals(jsv.getOutputFile(), "D://web-content.txt");
....
}
}
To execute the Junit use an IDE (like eclipse) or put the junit.jar file in the classpath and from the console:
java org.junit.runner.JUnitCore JavaSourceViewerTest

Your steps are not right. Doing it in this way would definitely help, 3, then 1, and then 2. Doing it this way will force you to think in terms of functionalities and units. And the result code will be testable, without doing anything special. Test first also guide the design of your system, besides providing you a safety net.
P.S. Never try to write the code before the test. It's simply not natural neither it gives much value, as you can see.
Now, to test the 1st unit, you can compare the string, the html from google.com, with some existing string. But that test case will break if Google changes it's page. Other way, is to just check the HTTP code from the header, if that's 200, you are fine. Just an idea.
For the second one, you can compare the string, you read from the web page, to string you wrote to the file, by reading the file.

String str = "";
URL oracle = new URL("http://www.google.com/");
BufferedReader in = new BufferedReader(
new InputStreamReader(oracle.openStream()));
File file = new File("C:/Users/rohit/Desktop/rr1.txt");
String inputLine;
FileWriter fw = new FileWriter(file.getAbsoluteFile());
BufferedWriter bw = new BufferedWriter(fw);
while ((inputLine = in.readLine()) != null) {
System.out.println(inputLine);
bw.write(inputLine);
}
bw.close();
in.close();