DanPri Posted October 25, 2008 Share Posted October 25, 2008 Hi Guys I know this is a regular question but i have seached everywhere for the answear. I get this error : Parse error: syntax error, unexpected T_VARIABLE on line 13 <?php $first_name = $_POST['first_name']; $last_name = $_POST['last_name']; $nick_name = $_POST['nick_name']; $ass = $_POST['ass']; $height = $_POST['height']; $weight = $_POST['weight']; $from = $_POST['from']; mysql_connect ("localhost", "cwreaves_fd", "fd") or die ('Error: ' . mysql_error()); mysql_select_db ("cwreaves_fd") [b]$query = "INSERT INTO fighter_info(first_name, last_name, nick_name, ass, height, weight, from) VALUES ('$first_name','$last_name','$nick_name','$ass','$height','$weight','$from')";[/b] mysql_query($query) or die ('Error updating Database'); echo "Database updated with: ".$id."".$first_name."".$last_name." ; ?> if any one can help I would be very greatfull Quote Link to comment Share on other sites More sharing options...
kenshintomoe225 Posted October 25, 2008 Share Posted October 25, 2008 mysql_select_db ("cwreaves_fd") you're missing a ; Quote Link to comment Share on other sites More sharing options...
DanPri Posted October 25, 2008 Author Share Posted October 25, 2008 thank you and thanks for such a quick reply Quote Link to comment Share on other sites More sharing options...
DanPri Posted October 25, 2008 Author Share Posted October 25, 2008 I am now getting this error Parse error: syntax error, unexpected $end can anyone help please Quote Link to comment Share on other sites More sharing options...
dezkit Posted October 25, 2008 Share Posted October 25, 2008 Give us your updated code. Quote Link to comment Share on other sites More sharing options...
DanPri Posted October 25, 2008 Author Share Posted October 25, 2008 Here it is <?php $first_name = $_POST['first_name']; $last_name = $_POST['last_name']; $nick_name = $_POST['nick_name']; $ass = $_POST['ass']; $height = $_POST['height']; $weight = $_POST['weight']; $from = $_POST['from']; mysql_connect ("localhost", "cwreaves_fd", "fd") or die ('Error: ' . mysql_error()); mysql_select_db ("cwreaves_fd"); [b]$query = "INSERT INTO fighter_info(first_name, last_name, nick_name, ass, height, weight, from) VALUES ('$first_name','$last_name','$nick_name','$ass','$height','$weight','$from')";[/b] mysql_query($query) or die ('Error updating Database'); echo "Database updated with: ".$id."".$first_name."".$last_name." ; ?> Quote Link to comment Share on other sites More sharing options...
AndyB Posted October 25, 2008 Share Posted October 25, 2008 echo "Database updated with: ".$id."".$first_name."".$last_name." ; s/b echo "Database updated with: ".$id." ".$first_name." ".$last_name; Quote Link to comment Share on other sites More sharing options...
dezkit Posted October 25, 2008 Share Posted October 25, 2008 <?php $first_name = $_POST['first_name']; $last_name = $_POST['last_name']; $nick_name = $_POST['nick_name']; $ass = $_POST['ass']; $height = $_POST['height']; $weight = $_POST['weight']; $from = $_POST['from']; mysql_connect ("localhost", "cwreaves_fd", "fd") or die ('Error: ' . mysql_error()); mysql_select_db ("cwreaves_fd"); $query = "INSERT INTO fighter_info(first_name, last_name, nick_name, ass, height, weight, from) VALUES ('$first_name','$last_name','$nick_name','$ass','$height','$weight','$from')"; mysql_query($query) or die ('Error updating Database'); echo "Database updated with: ".$id." ".$first_name." ".$last_name; ?> Try this, if this doesn't work, then you must have gave us the wrong code. Quote Link to comment Share on other sites More sharing options...
DanPri Posted October 25, 2008 Author Share Posted October 25, 2008 thank you, both worked Quote Link to comment Share on other sites More sharing options...
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