Can anyone tell me why this does not work? (here is a very simplified version). There are two pages, the first one (test.html) contents a HTML form, the second one (test.php) should evaluate the parameters. I use the POST method. The parameters don't appear in the second page. Can somebody write a simple example that do it? Very thankful, Sam doGet: ------------------------ OutputStream out = response.getOutputStream(); try { URL u = new URL("http://someadr/test.html"); URLConnection uco = u.openConnection (); response.setContentType(uco.getContentType()); InputStream in = uco.getInputStream(); DataInputStream data = new DataInputStream (new BufferedInputStream(in)); int b; while ((b = data.read()) != -1) out.write(b); out.close(); }
doPost: ------------------------ OutputStream out = response.getOutputStream(); try { URL u = new URL("http://someadr/test.php"); URLConnection uco = u.openConnection (); response.setContentType(uco.getContentType()); InputStream in = uco.getInputStream(); DataInputStream data = new DataInputStream (new BufferedInputStream(in)); int b; while ((b = data.read()) != -1) out.write(b); out.close();
If you use the phpinfo() function, you will see the parameters at the bottom of the page... This probably wasn't what you were looking for though... /Peter
I don't even see where you're SENDING parameters! For POST you should be writing out a stream to the PHP server. I believe the format is a line with the parameter name followed by a line with the parameter value (repeat as needed). Also you should set the content-type of the output stream.
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Thanks for the response. Could you tell me, how I can tell to the server, that the new request is POST. Normally, the server gets a new request as GET. thanks, Sam
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