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About FUNKAM35

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  1. $first1 = $_POST['first1']; $last1 = $_POST['last1']; $email1 = $_POST['email1']; $register = $_POST['register']; ok I think the error is on some different lines as posted here how can I fix this please thanks
  2. <form name=\"form1\" ACTION=\"$_SERVER[REQUEST_URI]\" METHOD=\"POST\">Your First Name<br><input name=\"first\" value = \"$first\" type=\"text\" id=\"first\" size=\"25\"><br><br>Last Name<br><input name=\"last\" value = \"$last\" type=\"text\" id=\"last\" size=\"25\"><br><br>Email<br><input name=\"email\" value = \"$email\" type=\"text\" id=\"email\" size=\"25\"><br><br>Telephone Number<br><input name=\"phone\" value = \"$phone\" type=\"text\" id=\"phone\" size=\"25\"><br><br><label for=\"comments\">Comments</label><textarea name=\"comments\" cols=\"18\" rows=\"4\" class=\"txt\" id=\"comments\"></textarea><div style=\"visibility:hidden\"><input name=\"email2\" type=\"text\" size=\"45\" id=\"email2\" > </div><input type=\"submit\" name=\"Submit\" value=\"Submit\" class=\"tpc4\"></form>\n"; Hi I am getting the error message Notice: Undefined Index on the following form before submission I would like to fix this rather than suppress errors Any help would be much appreciated
  3. $total_results = mysqli_result(mysqli_query("SELECT COUNT(*) as Num FROM property WHERE $where "),0) ; $total_results = mysql_result(mysql_query("SELECT COUNT(*) as Num FROM property WHERE $where "),0) ; Hi, Please can anyone help I have changed MySQL_ to mysqli_ but I am getting fatal error on above line, please can anyone help Thanks
  4. Thanks Ch0cu3r tried that and it says Fatal error: Call to a member function query() on a non-object
  5. $results = mysql_query("$select", $link_id); while ($query_data = mysql_fetch_row($results)) { $link_id = mysqli_connect("$db_host","$db_user","$db_password","$db_database"); if ($mysqli->connect_errno) { echo "Failed to connect to MySQL: (" . $mysqli->connect_errno . ") " . $mysqli->connect_error; } $link_id = mysql_connect("$db_host","$db_user","$db_password"); if (mysql_select_db("$db_database", $link_id)); else { echo "connection failed."; } I am having to update MySQL to mysqli as my host is upgrading to PHP5.6 I have managed to convert and connect to the database converted to I cannot get the fetch results to work, can anyone help me convert the following code Many Thanks
  6. Hi, I need to do a multiple select eg select * FROM stock WHERE and then it will be where (width=100 AND size=50 AND profile=20) OR (width=50 AND size=120 AND profile =30) OR (width=30 AND size=40 AND profile=70) How would I do this? many thanks
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