Reputation: 203
I am trying to send two string values from my java program into my php page and i seem to be having some difficulty figuring out how this all works.
public static void main (String args[]) throws IOException{
Scanner input = new Scanner (System.in);
String sampleValue = input.next();
String sampleValue1 = input.next();
URL url = new URL("http://woah.x10host.com/randomfact.php");
String result = "";
String data = "fName=" + URLEncoder.encode(sampleValue, "UTF-8");
String id = "lName=" + URLEncoder.encode(sampleValue1, "UTF-8");
HttpURLConnection connection = (HttpURLConnection) url.openConnection();
try {
connection.setDoInput(true);
connection.setDoOutput(true);
connection.setUseCaches(false);
connection.setRequestMethod("POST");
connection.setRequestProperty("Content-Type","application/x-www-form-urlencoded");
// Send the POST data
DataOutputStream dataOut = new DataOutputStream(connection.getOutputStream());
dataOut.writeBytes(id);
dataOut.writeBytes(data);
dataOut.flush();
System.out.println("Data has been posted");
dataOut.close();
BufferedReader in = null;
try {
String line;
in = new BufferedReader(new InputStreamReader(connection.getInputStream()));
while ((line = in.readLine()) != null) {
result += line;
}
} finally {
if (in != null) {
in.close();
}
}
} finally {
connection.disconnect();
System.out.println(result);
}
}
}
my PHP code is
<?php
$conn = mysqli_connect("localhost","woahx10h_funfact","spk","woahx10h_woah");
// CHECK CONNECTION
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$value_data = $_POST['fName'];
$value_id = $_POST['lName'];
echo $value_data;
echo $value_id;
$conn->close();
?>
However, every time i run the java program, values from both data and id seem to be stored in the $value. I want the value from data to be stored in $value_data while the value from x to be stored in $value_id
Upvotes: 3
Views: 73
Reputation: 436
You have to add ampersand (&) between each two parameters. So in this case all you have to do is edit your code like this:
...
dataOut.writeBytes(id);
dataOut.writeBytes("&");
dataOut.writeBytes(data);
...
If you would want to add another parameter to the request you would have to add another ampersand and then the parameter, for example:
...
dataOut.writeBytes(id);
dataOut.writeBytes("&");
dataOut.writeBytes(data);
dataOut.writeBytes("&");
dataOut.writeBytes(parameter3);
...
Upvotes: 1