json_encode returns <!DOCTYPE of type java.lang.String cannot be converted to JSONObject
My JSON Objects has problems with my values by my EditTexts. If i have a URL as Value for example 

> http://www.url.de

Then eclipse throws out the Error in the title. My PHP Script looks like this:

    <?php
	$response = array();
	
	if (isset($_POST['interpret']) && isset($_POST['titel']) && isset($_POST['album']) && isset($_POST['albumcover'])) {
		require_once ('db_connection.php');
		$db = new DB_CONNECT();
		
		$interpret = $_POST['interpret'];
		$titel = $_POST['titel'];
		$album = $_POST['album'];
		$albumcover = $_POST['albumcover'];


		$result = mysql_query("INSERT INTO Charts (interpret, titel, album, albumcover) VALUES ('$interpret', '$titel', '$album', '$albumcover')");
			if ($result) {
				// successfully inserted into database
				$response["success"] = 1;
				$response["message"] = "Charteintrag erfolgreich!";
				
				echo json_encode($response);	
			} else {
				// failed to insert row
				$response["success"] = 0;
				$response["message"] = "Entschuldigung. Ein Fehler ist aufgetreten.";
				echo json_encode($response);	 
		  } 
	} else {
			// required field is missing
			$response["success"] = 0;
			$response["message"] = "Es wurden nich alle Felder ausgef&uuml;llt!";
			echo json_encode($response);
		}
?>

Is it possible that i have to use "stripsplashes" or "mysql_real_escape_string" or something? I am searching for the solution for 2 days...PLS HELP ME!
json_encode returns <!DOCTYPE of type java.lang.String cannot be converted to JSONObject
===

My JSON Objects has problems with my values by my EditTexts. If i have a URL as Value for example 

> http://www.url.de

Then eclipse throws out the Error in the title. My PHP Script looks like this:

    <?php
	$response = array();
	
	if (isset($_POST['interpret']) && isset($_POST['titel']) && isset($_POST['album']) && isset($_POST['albumcover'])) {
		require_once ('db_connection.php');
		$db = new DB_CONNECT();
		
		$interpret = $_POST['interpret'];
		$titel = $_POST['titel'];
		$album = $_POST['album'];
		$albumcover = $_POST['albumcover'];


		$result = mysql_query("INSERT INTO Charts (interpret, titel, album, albumcover) VALUES ('$interpret', '$titel', '$album', '$albumcover')");
			if ($result) {
				// successfully inserted into database
				$response["success"] = 1;
				$response["message"] = "Charteintrag erfolgreich!";
				
				echo json_encode($response);	
			} else {
				// failed to insert row
				$response["success"] = 0;
				$response["message"] = "Entschuldigung. Ein Fehler ist aufgetreten.";
				echo json_encode($response);	 
		  } 
	} else {
			// required field is missing
			$response["success"] = 0;
			$response["message"] = "Es wurden nich alle Felder ausgef&uuml;llt!";
			echo json_encode($response);
		}
?>

Is it possible that i have to use "stripsplashes" or "mysql_real_escape_string" or something? I am searching for the solution for 2 days...PLS HELP ME!