I am invoking a local servlet from a jsp, the servlet simply returns a json string:
URL url = new URL("http://myapp.appspot.com/myservlet");
URLConnection conn = url.openConnection();
conn.setConnectTimeout(5000);
InputStream is = conn.getInputStream();
StringWriter writer = new StringWriter();
IOUtils.copy(is, writer, "UTF-8");
String jsonStr = writer.toString();
Can I do this with a relative path so that it works both locally and on the deployed instance?
You could use JSTL with the tag
<c:import>
Alternatively, for your posted code, you could use
String requestURL = request.getRequestURL().toString();
String servletPath = request.getServletPath();
String serverPath = requestURL.substring(0,requestURL.indexOf(servletPath));
URL url = new URL(serverPath + "/myservlet");
You mean like this?
String urlString = "http://localhost/myservlet/";
localhost is an alias for 127.0.0.1, which is always "the local computer".
ServletRequest.getServerPort() will let you know the port where the user connected.
Depending on where this is happening, what you really might want to use is ServletRequest.getRequestDispatcher(), which bypasses the network layer, and stays inside your servlet container.
You can wrap the HttpResponse, and send that through to the RequestDispatcher, then extract the String that was produced with something like this:
http://goo.gl/kRW1b
Related
I'm trying to send a video url in a http POST request but it's not working for me, I think I've (almost?) the necessary code to make it work, or else I'm missing something very simple?
public void postVideoURL() throws IOException {
String encodedUrl = URLEncoder.encode("http://video.ted.com/talks/podcast/DavidBrooks_2011.mp4", "UTF-8");
URL obj = new URL("http://10.50.0.105:8060/launch/dev?url="+encodedUrl);
HttpURLConnection con = (HttpURLConnection) obj.openConnection();
//add request header
con.setRequestMethod("POST");
// Send post request
con.setDoOutput(true);
OutputStreamWriter wr = new OutputStreamWriter(con.getOutputStream());
System.out.println(con.getResponseCode());
System.out.println(con.getResponseMessage());
wr.flush();
wr.close();
wr.write("");
}
Any tips to lead me to the right direction?
Here is what I'm trying to do but in C#
using System.Net;
using System.Web;
class Program
{
static void Main()
{
string rokuIp = "192.168.0.6";
string channelId = "dev";
string videoUrl = "http://video.ted.com/talks/podcast/DavidBrooks_2011.mp4";
// Note that the query string parameters should be url-encoded
string requestUrl = $"http://{rokuIp}:8060/launch/{channelId}?url={HttpUtility.UrlEncode(videoUrl)}";
using (var wc = new WebClient())
{
// POST the query string with no data
wc.UploadString(requestUrl, "");
}
}
}
The following Post command to use in terminal works, this is essentially what I want to do, but in Java:
curl -d "" "http://10.50.0.46:8060/launch/12?url=http%3A%2F%2Fvideo.ted.com%2Ftalks%2Fpodcast%2FDavidBrooks_2011.mp4"
You are never writing anything to the output stream. You have to call wr.write() to write your data to the stream.
Also, you can't encode the URL like that inside the String. You need to concatenate the two Strings together after you've encoded the url separately. Like this:
String encodedUrl = URLEncoder.encode("http://video.ted.com/talks/podcast/DavidBrooks_2011.mp4");
URL obj = new URL("http://10.50.0.105:8060/launch/dev?url="+encodedUrl);
I have devices that publish an html page when you connect via their ip address. For example, if I were to go to "192.168.1.104" on my computer, i would see the html page the device publishes. I am trying to scrape this html, but I am getting some errors, specifically a MalformedURLException at the first line of my method. I have posted my method below. I found some code for getting html and tweaked it for my needs. Thanks
public String getSbuHtml(String ipToPoll) throws IOException, SocketTimeoutException {
URL url = new URL("http", ipToPoll, -1, "/");
URLConnection con = url.openConnection();
con.setConnectTimeout(1000);
con.setReadTimeout(1000);
Pattern p = Pattern.compile("text/html;\\s+charset=([^\\s]+)\\s*");
Matcher m = p.matcher(con.getContentType());
String charset = m.matches() ? m.group(1) : "ISO-8859-1";
BufferedReader r = new BufferedReader(
new InputStreamReader(con.getInputStream(), charset));
String line = null;
StringBuilder buf = new StringBuilder();
while ((line = r.readLine()) != null) {
buf.append(line).append(System.getProperty("line.separator"));
}
return buf.toString();
}
EDIT: The above code has been changed to reflect constructing a new URL to work properly with an ip. However, when I try and get the contentType from the connection, it is null.
A URL (Uniform Resource Locator) must have a resource to locate (index.html) along with the means of network communication (http://). So an example of valid URL can be
http://192.168.1.104:8080/app/index.html
Merely 192.168.1.104 doesn't represent a URL
You need to add http:// to the front of your String that you pass into the method.
Create your URL as follows:
URL url = new URL("http", ipToPoll, -1, "/");
And since you're reading a potentially long HTML page I suppose buffering would help here:
BufferedReader r = new BufferedReader(
new InputStreamReader(con.getInputStream(), charset));
String line = null;
StringBuilder buf = new StringBuilder();
while ((line = r.readLine()) !- null) {
buf.append(line).append(System.getProperty("line.separator"));
}
return buf.toString();
EDIT: In response to your contentType coming null problem.
Before you inspect any headers like with getContentType() or retrieve content with getInputStream() you need to actually establish a connection with the URL resource by calling
URL url = new URL("http", ipToPoll, "/"); // -1 removed; assuming port = 80 always
// check your device html page address; change "/" to "/index.html" if required
URLConnection con = url.openConnection();
// set connection properties
con.setConnectTimeout(1000);
con.setReadTimeout(1000);
// establish connection
con.connect();
// get "content-type" header
Pattern p = Pattern.compile("text/html;\\s+charset=([^\\s]+)\\s*");
Matcher m = p.matcher(con.getContentType());
When you call openConnection() first (it wrongly suggests but) it doesn't establish any connection. It just gives you an instance of URLConnection to let you specify connection properties like connection timeout with setConnecTimeout().
If you're finding this hard to understand it may help to know that it's analogous to doing a new File() which simply represents a File but doesn't create one (assuming it doesn't exist already) unless you go ahead and call File.createNewFile() (or pass it to a FileReader).
I'm trying to write a test program to test my web service. I'm sending a JSON object to my web service via the GET method but it's not working. My test URL looks like this:
http://testserver:8080/mydir/{"filename":"test.jpg", "Path":"test/2/2"}
I'm thinking the "/" in the path are causing me problems since the program works fine once I remove them.
Per REST how to pass values containing "/" as path parameter in the URI?, I've tried to use java.net.URLEncoder.encode but that isn't helping. Here's a snippet of my test program:
// some code from main method
<snip snip>
String url = "http://testserver:8080/mydir/";
String JSON = "{\"filename\":\"test.jpg\",\"Path\":\"test/2/2\"}";
String enc_JSON = URLEncoder.encode(JSON,"UTF-8");
String testGet = url + enc_JSON;
String out2 = TestCode.httpGet(testGet);
<snip snip>
// code from httpGet method
public static String httpGet(String serverURL) {
URL url;
HttpURLConnection conn;
try {
url = new URL (serverURL);
conn = (HttpURLConnection) url.openConnection();
conn.setRequestMethod("GET");
conn.setUseCaches(false);
conn.setDoInput(true);
conn.setDoOutput(true);
conn.connect();
// failing at line below
InputStream input = conn.getInputStream();
<snip snip>
The result of my program is I get an HTTP response code: 400. Did I forget to add something in my code in the httpGet() method that's causing it to fail or am I doing something illegal in my URL due to the JSON object being tacked on at the end with the "/" in the path location?
Thanks in advance for you help.
For REST APIs, JSON objects are typically sent (POST) or returned in the body of the request. They are not typically encoded as part of the URL.
For a GET request, you can either pass the information as segments in the url or as querystring parameters.
As segments in the url:
/resourcetype/{path}/{filename}
http://testserver:8080/resourcetype/test/2/2/test.jpg
As querystring params:
/resourcetype?path={path}&file={filename}
http://testserver:8080/resourcetype?path=test/2/2&filename=test.jpg
I need to make a struts 2 application. In the one view of this app, I have to get the view of another application through the URL provided for example (http://localhost:8080/hudson/)...
Now.
1. How to connect with the other application? (Can it be done with Apache HttpURLClient? OR any other way please guide. )
2 .If it can be done with Apache HttpURLClient, then how to render the Response object in stuts2 framework.
Please help. Many thanks in advance.
You can use java.net package to resolve the issue. Example Code :
URL urlApi = new URL(requestUrl);
HttpURLConnection httpURLConnection = (HttpURLConnection) urlApi.openConnection();
httpURLConnection.setRequestMethod(<requestMethod>); // GET or POST
httpURLConnection.setDoOutput(true);
//in case HTTP POST method un-comment following to write request body
//DataOutputStream ds = new DataOutputStream(httpURLConnection.getOutputStream());
//ds.writeBytes(body);
InputStream content = (InputStream) httpURLConnection.getInputStream();
BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(content));
StringBuilder stringBuilder = new StringBuilder(100);
String line = null;
while ((line = bufferedReader.readLine()) != null) {
stringBuilder.append(line);
}
String serverResult = stringBuilder.toString();
//now you have a string representation of the server result page
//do what you need
Hope this helps
I have to do the same, i`m using struts2 but i have doing a servlet. In struts.xml you have to put you can get the content of the url with httpauarlconnection or with httpclient (apache) and and put it at servlet response.
I have this but i have problems with the relative links of the html because it try to resolve with my domain name (the name of the servlet that make the work).
I have a list of URLs which I need to get the content of.
The URL is with special characters and thus needs to be encoded.
I use Commons HtpClient to get the content.
when I use:
GetMethod get = new GetMethod(url);
I get a " Invalid "illegal escape character" exception.
when I use
GetMethod get = new GetMethod();
get.setURI(new URI(url.toString(), false, "UTF-8"));
I get 404 when trying to get the page, because a space is turned to %2520 instead of just %20.
I've seen many posts about this problem, and most of them advice to build the URI part by part. The problem is that it's a given list of URLs, not a one that I can handle manually.
Any other solution for this problem?
thanks.
What if you create a new URL object from it's string like URL urlObject = new URL(url), then do urlObject.getQuery() and urlObject.getPath() to split it right, parse the Query Params into a List or a Map or something and do something like:
EDIT: I just found out that HttpClient Library has a URLEncodedUtils.parse() method which you can use easily with the code provided below. I'll edit it to fit, however is untested.
With Apache HttpClient it would be something like:
URI urlObject = new URI(url,"UTF-8");
HttpClient httpclient = new DefaultHttpClient();
List<NameValuePair> formparams = URLEncodedUtils.parse(urlObject,"UTF-8");
UrlEncodedFormEntity entity;
entity = new UrlEncodedFormEntity(formparams);
HttpPost httppost = new HttpPost(urlObject.getPath());
httppost.setEntity(entity);
httppost.addHeader("Content-Type","application/x-www-form-urlencoded");
HttpResponse response = httpclient.execute(httppost);
HttpEntity entity2 = response.getEntity();
With Java URLConnection it would be something like:
// Iterate over query params from urlObject.getQuery() like
while(en.hasMoreElements()){
String paramName = (String)en.nextElement(); // Iterator over yourListOfKeys
String paramValue = yourMapOfValues.get(paramName); // replace yourMapOfNameValues
str = str + "&" + paramName + "=" + URLEncoder.encode(paramValue);
}
try{
URL u = new URL(urlObject.getPath()); //here's the url path from your urlObject
URLConnection uc = u.openConnection();
uc.setDoOutput(true);
uc.setRequestProperty("Content-Type","application/x-www-form-urlencoded");
PrintWriter pw = new PrintWriter(uc.getOutputStream());
pw.println(str);
pw.close();
BufferedReader in = new BufferedReader(new
InputStreamReader(uc.getInputStream()));
String res = in.readLine();
in.close();
// ...
}
If you need to manipulate with request URIs it is strongly advisable to use URIBuilder shipped with Apache HttpClient.
try it out
GetMethod get = new GetMethod(url.replace(" ","%20")).toASCIIString());
Please use the URLEncoder class.
I used it in an exact scenario and it worked just fine for me.
What I did is to use the URL class, to get the part that comes after the host
(for example - at www.bla.com/mystuff/bla.jpg this would be "mystuff/bla.jpg" - you should URLEncode only this part, and then consturct the URL again.
For example, if the orignal string is "http://www.bla.com/mystuff/bla foo.jpg" then:
Encode - "mystuff/bla foo.jpg" and get "mystuff/bla%20foo.jpg" and then attach this to the host and protocol parts:
"http://www.bla.com/mystuff/bla%20foo.jpg"
I hope this helps