How to Call SMS API Using java

Question

1.00/5 (1 vote)

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Following is the java code but I get response only "OK"
while I am using Vb.net with same API I got response "SMS Sent Successfully"

JAVA CODE

String requestUrl = "SOME API";

URL url = new URL(requestUrl);
HttpURLConnection uc = (HttpURLConnection)url.openConnection();

System.out.println(uc.getResponseMessage());

uc.disconnect();

VB.net COde
Imports System.Net
Imports System.IO

Dim sURL As String
Dim objReader As StreamReader
Dim objStream As Stream
Dim sResponse As WebRequest

sResponse = WebRequest.Create(sURL)
objStream = sResponse.GetResponse.GetResponseStream()

Please Please Answer me with java code stuck in this...

thank you

Posted 6-Dec-15 21:22pm

hareshdgr8

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Richard MacCutchan · Answer 1 · 2015-12-06T21:56:00

Solution 1

See https://www.google.com/search?q=java%20sms[^].

Posted 6-Dec-15 21:56pm

Richard MacCutchan

Comments

hareshdgr8 7-Dec-15 4:54am

sir I refer that link but didn't work for me and first link is in php not in java I need in java please help me

and
sir other link is not helpfull

Richard MacCutchan 7-Dec-15 6:24am

Then you will need to refine the search terms to meet your specific needs.

Richard MacCutchan 7-Dec-15 6:26am

You should also read http://docs.oracle.com/javase/7/docs/api/java/net/HttpURLConnection.html#getResponseMessage().

hareshdgr8 7-Dec-15 8:00am

Not Usefull sorry sir :-(

Richard MacCutchan 7-Dec-15 12:32pm

In what way not useful? Did you follow that link and read the description? Do you now understand why you get the response "OK" and nothing else? Do you know what other calls you need to make to get the full details from the HTML response?

hareshdgr8 8-Dec-15 0:58am

but while I am running this api via script I got response then what is problem ...

Richard MacCutchan 8-Dec-15 4:02am

There is no problem, you got a response saying tyour request has been received sussessfully at the server. If you want more detail then you need to read the full web stream.

hareshdgr8 9-Dec-15 8:28am

thank you so much

Adil_J · Answer 2 · 2015-12-10T19:40:00

In your code use the method below after changing your code as follow:

From

C#

System.out.println(uc.getResponseMessage());

To

C#

System.out.println(doHttpUrlConnectionAction(uc));

C#

/**
   * Returns the output from the given URL.
   * 
   * I tried to hide exception-handling in this method, and just return a high
   * level Exception from here.
   * Modify this behavior as desired.
   * 
   * @param connection
   * @return
   * @throws Exception
   */
  private String doHttpUrlConnectionAction(HttpURLConnection connection)
  throws Exception
  {
    BufferedReader reader = null;
    StringBuilder stringBuilder;
 
    try
    {
      connection.connect();
 
      // read the output from the server
      reader = new BufferedReader(new   
     InputStreamReader(connection.getInputStream()));
      stringBuilder = new StringBuilder();
 
      String line = null;
      while ((line = reader.readLine()) != null)
      {
        stringBuilder.append(line + "\n");
      }
      return stringBuilder.toString();
    }
    catch (Exception e)
    {
      e.printStackTrace();
      throw e;
    }
    finally
    {
      // close the reader; this can throw an exception too, so
      // wrap it in another try/catch block.
      if (reader != null)
      {
        try
        {
          reader.close();
        }
        catch (IOException ioe)
        {
          ioe.printStackTrace();
        }
      }
    }
  }