I have a problem in my hand.
I have a URL, And when i initiate the connect to this url and execute url.getContent().
The response is of type sun.net.www.protocol.http.HttpURLConnection$HttpInputStream
I tried to assign the output to HttpURLConnectionHttpInputStream h = url.getContent(). But i was unsuccessful.
I imported corresponding libraries to code, but still no luck.
If i inspect the url.getContent() in eclipse, it also shows the variable thei$0 in it.
All i need is a URL in this$0. But till now i am unable to retreive it.
In this$0 there is a variable names url and i am trying to fetch it.
I also have hard time understand this$0 and hoe to retrieve it.
After using the streams i get some non readable output
Regards
Dheeraj Joshi
You should use the openStream method of the url class.
Code snippet:
InputStream in = url.openStream();
BufferedReader reader = new BufferedReader(new InputStreamReader(in));
String line = reader.readLine();
If the output is not in a readable string format, then use:
InputStream in = url.openStream();
byte[] buffer = new byte[512];
int bytesRead = in.read(buffer);
I found the answer.
The problem statement: When i execute a URL the response had another URL in it and i needed to fetch it.
Solution:
java.net.URLConnection urlconn = url.openConnection();
java.net.HttpURLConnection conn = (java.net.HttpURLConnection)urlconn;
conn.connect();
conn.getContent();
URL newurl = conn.getURL();
System.out.println(newurl.toString());
The response can be get using getContent() and. The connection object will have a delegate with the new URL. The new URL can be fetched using getURL method.
Regards
Dheeraj Joshi
Related
I want to get a query result from Stack Exchange API using my Java program. For example, I want to pass this URL and get the data of the question with id 805107. I have tried but only got the resulted web page content. I did not get the query result, i.e. the question data, although the resulted page shows the question data.
url = new URL ("https://api.stackexchange.com/docs/questions-by-ids#order=desc&sort=activity&ids=805107&filter=default&site=stackoverflow&run=true");
byte[] encodedBytes = Base64.encodeBase64("root:pass".getBytes());
String encoding = new String (encodedBytes);
HttpURLConnection connection = (HttpURLConnection) url.openConnection();
connection.setRequestMethod("GET");
connection.setDoInput (true);
connection.setRequestProperty ("Authorization", "Basic " + encoding);
connection.connect();
InputStream content = (InputStream)connection.getInputStream();
BufferedReader in = new BufferedReader (new InputStreamReader (content));
String line;
while ((line = in.readLine()) != null) {
System.out.println(line);
}
As Stephen C said, you need to use the query URL, not the URL of the documentation. You can find the query URL in the "Try it" part of the documentation page. Try using
url = new URL ("https://api.stackexchange.com/2.2/questions/805107?order=desc&sort=activity&site=stackoverflow")
It will return the result you are looking for as JSON like it is displayed on the documentation page.
I'm trying to get data off of a URL, but the information I need takes a few seconds to load, and only shows as LOADING in the HTML until it does load, so when I use this code I can't pull the data I need.
URL url = new URL("https://www.cardservices.uga.edu/fs_mobile/");
URLConnection con = url.openConnection();
InputStream is = con.getInputStream();
BufferedReader br = new BufferedReader(new InputStreamReader(is));
String line = null;
while ((line = br.readLine()) != null){
System.out.println(lineNumber +": "+ line);
}
How could I go about allowing the URL to load for a set amount of time before pulling the HTML off of it?
The webpage you are calling probably call an ajax call to fetch the data, thats why you won't get it using your approach.
You have 2 options to get that data:
Use browser's inspect elements(F12 in chrome) and in "network" tab, get that ajax call, and use it instead of the URL you are using in your code.
Call your URL using a headless library(e.g ghoustjs) and after page is load crawl the data.
IMO I would choose option 1
Here is a working alternate,
URL url = new URL("https://www.cardservices.uga.edu/fs_mobile/index.php/dashboard/occupancies/"); //This is the AJAX call that goes to load the data into webpage. You can get this from inspecting the network calls.
URLConnection con = url.openConnection();
InputStream is = con.getInputStream();
BufferedReader br = new BufferedReader(new InputStreamReader(is));
String line = null;
while ((line = br.readLine()) != null){
System.out.println(line);
}
Which basically gives you the JSON response containing the percentage.
Hope it helps.
Also, you can use Selenium for performing wait if you are so curious to get the exact HTML output.
I'm trying to do a "POST" method in Java. I create my output with the OrientDB method like this:
"http://xxxxxxxxxxx:2480/command/mydb/sql/CREATE VERTEX V SET name = ' datoAletarorio'"
I need to use the write and flush methods to send the command.
My DB is empty with this method.
Where is my error? Here is my code:
//...
PrintWriter out = null;
//...
conexion = (HttpURLConnection) url.openConnection();
conexion.setDoOutput(true);
conexion.setRequestMethod("POST");
out = new PrintWriter(conexion.getOutputStream());
conexion.connect();
//...
String cumuloDatos1 = "http://xxxxxxxxxxx:2480/command/mydb/sql/CREATE VERTEX V SET name = ' datoAletarorio'"
out.write(cumuloDatos1);
out.flush();
//..
conexion.disconnect();
Thank in advance.
The docs says:
The command-text can appear in either the URL or the content of the
POST transmission. Where the command-text is included in the URL, it
must be encoded as per normal URL encoding.
So you probably have to encode the URL before sending the request:
String cumuloDatos1 =
"http://xxxxxxxxxxx:2480/command/mydb/sql/" +
"CREATE%20VERTEX%20V%20SET%20name%20%3D%20%27%20datoAletarorio%27"
Anyway, you should see messages in the logs for a 400 or similiar in the server, if the request isn't valid.
String album = "http://picasaweb.google.com/data/feed/api/user/"+email;
HttpURLConnection con = (HttpURLConnection) new URL(albumUrl).openConnection();
// request method, timeout and headers
con.setRequestMethod("GET") ;
con.setReadTimeout(15000);
con.setRequestProperty("Authorization", "GoogleLogin auth="+auth);
con.setRequestProperty("GData-Version", "2");
// set timeout and that we will process output
con.setReadTimeout(15000);
con.setDoOutput(true);
// connnect to url
con.connect();
// read output returned for url
BufferedReader reader = new BufferedReader(new InputStreamReader(con.getInputStream()));
Problem : Everytime i call con.getInputStream() it gives me file not found exception.
But when i load the same url in the desktop browser then it is displaying correct data.
I am confused why on android it is throwing exception.
Thanks in advance.
Did you get this? Maybe you just missed the https
below example uses default for authenticated user and the experimental fields list.
url = "https://picasaweb.google.com/data/feed/api/user/default?kind=album&access=public&fields="
+ URLEncoder
.encode("entry(title,id,gphoto:numphotosremaining,gphoto:numphotos,media:group/media:thumbnail)",
"UTF-8");
https://developers.google.com/picasa-web/docs/2.0/developers_guide_protocol#ListAlbums
I want to post among other data a String variable to a PHP file by using the HttpConnection stuff. The first data is the byte[] data returned from a recordstore. So it should be posted alone. So how to post the String variable also ?
You can pass the data to a PHP file using GET or POST methods.
Get method is the easy way to pass simple data. Using GET you can add the variable to the URL
Example:
192.168.1.123/myproject/uploads/treatphoto.php?myVariable1=MyContent&myVariable2=MyContent2
And in PHP:
$content1 = $_GET['myVariable1'];
$content2 = $_GET['myVariable2'];
Also the content of "MyContent" needs to be an string encoded. using any UrlEncoder.
To pass a byte[] array using this method you need to convert the byte array to an string encoded in some printable encoding like base64
The GET method also has a sort limit of data that can be passed safely (usually 2048 bytes)
The other method "POST" is more complex (but not a lot), way to add more data.
You need to prepare the HttpConnection to pass the data as POST.
Also the data stored in urlParamenters need to be according to the url enconding.
Passing the data using post is similar to GET but instead of adding all the variables next to the url the varuiables are added in the Stream of the httpConnection request.
example of the java code:
String urlParameters = "myVariable1=myValue1&myVariable2=myValue2";
HttpURLConnection connection = null;
try {
url = new URL(targetURL);
connection = (HttpURLConnection)url.openConnection();
// Use post and add the type of post data as URLENCODED
connection.setRequestMethod("POST");
connection.setRequestProperty("Content-Type","application/x-www-form-urlencoded");
// Optinally add the language and the data content
connection.setRequestProperty("Content-Length", "" + Integer.toString(urlParameters.getBytes().length));
connection.setRequestProperty("Content-Language", "en-US");
// Set the mode as output and disable cache.
connection.setUseCaches (false);
connection.setDoInput(true);
connection.setDoOutput(true);
//Send request
DataOutputStream wr = new DataOutputStream (connection.getOutputStream ());
wr.writeBytes (urlParameters);
wr.flush ();
wr.close ();
// Get Response
// Optionally you can get the response of php call.
InputStream is = connection.getInputStream();
BufferedReader rd = new BufferedReader(new InputStreamReader(is));
String line;
StringBuffer response = new StringBuffer();
while((line = rd.readLine()) != null) {
response.append(line);
response.append('\r');
}
rd.close();
return response.toString();
The php is similar, you only need to replace $_GET by $_POST:
$content1 = $_POST['myVariable1'];
$content2 = $_POST['myVariable2'];