I want to hit a URL in php from java servlet.I just want to send a information to that url.I didn't need to go to that url.i want to stay in my page.Anyway to do that.?
You could simply make a post on that url like in the http components doc
PostMethod post = new PostMethod("http://jakarata.apache.org/");
NameValuePair[] data = {
new NameValuePair("user", "joe"),
new NameValuePair("password", "bloggs")
};
post.setRequestBody(data);
// execute method and handle any error responses.
...
InputStream in = post.getResponseBodyAsStream();
// handle response.
Try following code
URL url = new URL("your url");
URLConnection connection = url.openConnection();
connection.setConnectTimeout(5000); // time out
connection.setDoOutput(true);
PrintWriter out = new PrintWriter(new OutputStreamWriter(connection.getOutputStream()));
String postData = "your post data";
out.print(postData);
out.close();
String response connection.getInputStream();
or you can use
Request.Post("your url").bodyForm(
Form.form().add("parameter name", "value")
.add("parameter name1", "value").build()).execute();
Related
NOTICE UPDATE!!
The problem got solved and i added my own answer in the thread
In short, I have attempted to add the parameter "scan_id" value but since it is a POST i can't add the value directly in the url path.
using the code i already have, how would i go about modifying or adding so that the url is correct, that is, so that it accepts my POST?.
somehow i have been unable to find any examples that have helped me in figuring out how i would go about doing this..
I know how to do a POST with a payload, a GET with params. but a post with Params is very confusing to me.
Appreciate any help. (i'd like to continue using HttpUrlConnection unless an other example is provided that also tells me how to send the request and not only configuring the path.
I've tried adding it to the payload.
I've tried UriBuilder but found it confusing and in contrast with the rest of my code, so wanted to ask for help with HttpUrlConnection.
URL url = new URL("http://localhost/scans/{scan_id}/launch");
HttpURLConnection con = (HttpURLConnection) url.openConnection();
con.setRequestMethod("POST");
con.setRequestProperty("tmp_value_dont_mind_this", "432432");
con.setRequestProperty("X-Cookie", "token=" + "43432");
con.setRequestProperty("X-ApiKeys", "accessKey="+"43234;" + " secretKey="+"43234;");
con.setDoInput(true);
con.setDoOutput(true); //NOT NEEDED FOR GETS
con.setRequestMethod("POST");
con.setRequestProperty("Accept", "application/json");
con.setRequestProperty("Content-Type", "application/json; charset=UTF-8");
//First example of writing (works when writing a payload)
OutputStreamWriter writer = new OutputStreamWriter(con.getOutputStream(), "UTF-8");
writer.write(payload);
writer.close();
//second attemp at writing, doens't work (wanted to replace {scan_id} in the url)
DataOutputStream writer = new DataOutputStream(con.getOutputStream());
writer.writeChars("scan_id=42324"); //tried writing directly
//writer.write(payload);
writer.close();
Exception:
Exception in thread "main" java.io.IOException: Server returned HTTP response code: 400 for URL: http://localhost/scans/launch
I'd like one of the three response codes because then i know the Url is correct:
200 Returned if the scan was successfully launched.
403 Returned if the scan is disabled.
404 Returned if the scan does not exist.
I've tried several urls
localhost/scans/launch,
localhost/scans//launch,
localhost/scans/?/launch,
localhost/scans/{scan_id}/launch,
So with the help of a friend and everyone here i solved my problem.
The below code is all the code in an entire class explained bit by bit. at the bottom you have the full class with all its syntax etc, that takes parameters and returns a string.
in a HTTP request there are certain sections.
Such sections include in my case, Request headers, parameters in the Url and a Payload.
depending on the API certain variables required by the API need to go into their respective category.
My ORIGINAL URL looked like this: "http://host:port/scans/{scan_id}/export?{history_id}"
I CHANGED to: "https://host:port/scans/" + scan_Id + "/export?history_id=" + ID;
and the API i am calling required an argument in the payload called "format" with a value.
String payload = "{\"format\" : \"csv\"}";
So with my new URL i opened a connection and set the request headers i needed to set.
HttpsURLConnection con = (HttpsURLConnection) url.openConnection();
The setDoOutput should be commented out when making a GET request.
con.setDoInput(true);
con.setDoOutput(true);
con.setRequestMethod("POST");
con.setRequestProperty("Accept", "application/json");
con.setRequestProperty("Content-Type", "application/json; charset=UTF-8");
con.setRequestProperty("X-Cookie", "token=" + token);
con.setRequestProperty("X-ApiKeys", "accessKey="+"23243;" +"secretKey="+"45543;");
Here i write to the payload.
//WRITING THE PAYLOAD to the http call
OutputStreamWriter writer = new OutputStreamWriter(con.getOutputStream(), "UTF-8");
writer.write(payload);
writer.close();
After i've written the payload i read whatever response i get back (this depends on the call, when i do a file download (GET Request) i don't have a response to read as i've already read the response through another piece of code).
I hope this helps anyone who might encounter this thread.
public String requestScan(int scan_Id, String token, String ID) throws MalformedInputException, ProtocolException, IOException {
try {
String endpoint = "https://host:port/scans/" + scan_Id + "/export?history_id=" ID;
URL url = new URL(endpoint);
String payload= "{\"format\" : \"csv\"}";
HttpsURLConnection con = (HttpsURLConnection) url.openConnection();
con.setDoInput(true);
con.setDoOutput(true);
con.setRequestMethod("POST");
con.setRequestProperty("Accept", "application/json");
con.setRequestProperty("Content-Type", "application/json; charset=UTF-8");
con.setRequestProperty("X-Cookie", "token=" + token);
con.setRequestProperty("X-ApiKeys", "accessKey="+"324324;" +
"secretKey="+"43242;");
//WRITING THE PAYLOAD to the http call
OutputStreamWriter writer = new OutputStreamWriter(con.getOutputStream(), "UTF-8");
writer.write(payload);
writer.close();
//READING RESPONSE
BufferedReader br = new BufferedReader(new InputStreamReader(con.getInputStream()));
StringBuffer jsonString = new StringBuffer();
String line;
while ((line = br.readLine()) != null) {
jsonString.append(line);
}
br.close();
con.disconnect();
return jsonString.toString();
} catch (Exception e) {
throw new RuntimeException(e.getMessage());
}
}
As discussed here the solution would be to change the content type to application/x-www-form-urlencoded, but since you are already using application/json; charset=UTF-8 (which I am assuming is a requirement of your project) you have no choise to redesign the whole thing. I suggest you one of the following:
Add another GET service;
Add another POST service with content type application/x-www-form-urlencoded;
Replace this service with one of the above.
Do not specify the content type at all so the client will accept anything. (Don't know if possible in java)
If there are another solutions I'm not aware of, I don't know how much they would be compliant to HTTP protocol.
(More info)
Hope I helped!
Why you are not using like this. Since you need to do a POST with HttpURLConnection, you need to write the parameters to the connection after you have opened the connection.
String urlParameters = "scan_id=42324";
byte[] postData = urlParameters.getBytes(StandardCharsets.UTF_8);
DataOutputStream dataOutputStream = new DataOutputStream(conn.getOutputStream());
dataOutputStream.write(postData);
Or if you have launch in the end, just change the above code to the following,
String urlParameters = "42324/launch";
byte[] postData = urlParameters.getBytes(StandardCharsets.UTF_8);
DataOutputStream dataOutputStream = new DataOutputStream(conn.getOutputStream());
dataOutputStream.write(postData);
URL url = new URL("http://localhost/scans/{scan_id}/launch");
That line looks odd to me; it seems you are trying to use a URL where you are intending the behavior of a URI Template.
The exact syntax will depend on which template implementation you choose; an implementation using the Spring libraries might look like:
import org.springframework.web.util.UriTemplate;
import java.net.url;
// Warning - UNTESTED code ahead
UriTemplate template = new UriTemplate("http://localhost/scans/{scan_id}/launch");
Map<String,String> uriVariables = Collections.singletonMap("scan_id", "42324");
URI uri = template.expand(uriVariables);
URL url = uri.toURL();
I'm trying to login to a portal. It works using Postman. When I try the same request using plain Java or OkHttp the login fails and I will be redirected to the login page.
HttpUrl.Builder httpBuilder = HttpUrl.parse("https://test58.cashctrl.com/auth/login.html").newBuilder();
httpBuilder.addQueryParameter("JMCF_AUTH_EMAIL", "email");
httpBuilder.addQueryParameter("JMCF_AUTH_PASSWORD", "password");
Request request = new Request.Builder()
.url(httpBuilder.build())
.get()
.build();
I know the Url looks weird but it works this way using Postman or even simply use a browser.
Alternative with plain Java, which I tried:
Map<String, String> parameters = new HashMap<>();
parameters.put(PARAM_EMAIL, EMAIL);
parameters.put(PARAM_PASSWORD, PASSWORD);
URL url = new URL(LOGIN_URL + "?" + ParameterStringBuilder.getParamsString(parameters));
HttpURLConnection con = (HttpURLConnection) url.openConnection();
con.setRequestMethod("POST");
con.setInstanceFollowRedirects(true);
con.setDoOutput(true);
DataOutputStream out = new DataOutputStream(con.getOutputStream());
out.writeBytes(ParameterStringBuilder.getParamsString(parameters));
out.flush();
out.close();
int status = con.getResponseCode();
BufferedReader in = new BufferedReader(
new InputStreamReader(con.getInputStream()));
String inputLine;
StringBuffer content = new StringBuffer();
while ((inputLine = in.readLine()) != null) {
content.append(inputLine + "\n");
}
in.close();
con.disconnect();
System.out.println(status);
System.out.println(content.toString());
Postman must be doing something special or also a browser which I don't see.
I had the same issue, I got to know that Postman has "code" feature. Below the send button you can see the code option it will generate the code for you. There is a list of language to choose from and java is one of them. Do check that out. Also you must be missing the cookie, see the temporary headers in Postman add all in your code and do include the cookie one.
Thanks I hope it helps.
I have a problem on HttpURLConnection in post method. Everything is working fine on get method however, when I try to use Post method. I'm getting this error message.
Exception in thread "main" java.io.IOException: Server returned HTTP response code: 403 for URL
Here's my code snippet. I hope you could help me about this.
URL url = new URL(my url/userInfo);
String encoding = Base64.getEncoder().encodeToString(("username:password").getBytes("UTF-8"));
HttpURLConnection connection = (HttpURLConnection) url.openConnection();
connection.setRequestMethod("GET");
connection.setDoOutput(true);
connection.setRequestProperty("Content-Type", "text/plain");
connection.setRequestProperty("Authorization", "Basic " + encoding);
connection.setRequestProperty("x-csrf-token", "fetch");
String csrfToken = connection.getHeaderField("x-csrf-token");
BufferedReader in = new BufferedReader(new InputStreamReader(connection.getInputStream()));
String inputLine;
String output = in.readLine();
in.close();
String content = data // expected data to retrieve
URL url2 = new URL(my URL);//another url to push the data retrieve
HttpURLConnection connection2 = (HttpsURLConnection) url2.openConnection();
connection2.setDoInput(true);
connection2.setDoOutput(true);
connection2.setRequestMethod("POST");
connection2.setRequestProperty("Authorization", "Basic " + encoding);
connection2.setRequestProperty("Accept", "application/json");
connection2.setRequestProperty("x-CSRFToken", csrfToken);
connection2.setRequestProperty("cache-control", "no-cache");
OutputStream os = connection2.getOutputStream();
OutputStreamWriter osw = new OutputStreamWriter(os, "UTF-8");
osw.write(data);//this is where the data will be pushed
osw.flush();
osw.close();
os.close();
the idea is, we need first to get the x-csrf-token and data from the first link, which is okay. After GET Method execution, the POST method will occur. unfortunately, the post method is not working. I'm getting the error message shown above. By the way, we tried to do a post method in POSTMAN and it' working fine.
Hoping you could help me about this.
So I have a problem where if I type this link on the browser and hit enter, an activation happens. I just want to do the same through Java. I don't need any kind of response from the URL. It should just do the same as entering the URL on a browser. Currently my code doesn't throw an error, but I don't think its working because the activation is not happening. My code:
public static void enableMachine(String dns){
try {
String req= "http://"+dns+"/username?username=sputtasw";
URL url = new URL(req);
URLConnection connection = url.openConnection();
connection.connect();
/*BufferedReader br = new BufferedReader(new InputStreamReader(url.openStream()));
String strTemp = "";
while (null != (strTemp = br.readLine())) {
System.out.println(strTemp);
}*/
} catch (Exception ex) {
ex.printStackTrace();
}
}
What's the problem?
If you want to do that with an URLConnection, it isn't sufficient to just open the connection with connect, you also have to send e.g. an HTTP request etc.
That said, i think it would be easier, if you use an HTTP client like the one from Apache HttpComponents (http://hc.apache.org/). Just do a GET request with the HTTP client, this would be the same as visiting the page with a browser (those clients usually also supports redirection etc.).
You may use HttpUrlConnectionClass to do the job:
URL url = new URL("http://my.url.com");
HttpURLConnection httpCon = (HttpURLConnection) url.openConnection();
httpCon.setRequestProperty("Content-Type", "application/json");
httpCon.setDoOutput(true);
httpCon.setRequestMethod("POST");
String params = "foo=42&bar=buzz";
DataOutputStream wr = new DataOutputStream(httpCon.getOutputStream());
wr.writeBytes(params);
wr.flush();
wr.close();
httpCon.connect();
int responseCode = httpCon.getResponseCode();
You may as well use "GET" request method and just append parameters to the url.
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