sastiger
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If you would like to shair that with me chronister then please do also how do I read date from it? Like lets say the petname ? Thanks Everyone for your help :>
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Thanks for that andy but on the database it just shows up as blank input ???
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Still am getting the same error here is my page which is posting to the php file <html> <head> <title>Pet creation page</title> </head> <body> <form method="post" action="post.php"> <td>Pet name:</td> <input type="text" name="$petname" value="" size="12" /> <br> <td>Type:</td> <input type="text" name="$pettype" value="" size="12" /> <br> <td>Price:</td> <input type="text" name="$petprice" value="" size="12" /> <br> <input type="submit" value="Update Database" /> </form> </body> </html> and here is my php post page <?php $petname = $_POST['name']; $pettype = $_POST['type']; $petprice = $_POST['250']; $host="localhost"; $user="root"; $pass="majick"; mysql_connect ("$host", "$user", "$pass") or die ( mysql_select_db(pet) $result = mysql_query("INSERT INTO animal (name, type, price) VALUES ('$petname', '$pettype', '$petprice'); mysql_query or die (’Error updating database’); ?> PLEASE HELP
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I keep getting this Parse error: syntax error, unexpected T_VARIABLE in C:\xampp\htdocs\home\post.php on line 8 Here is my actual code for the post.php page <?php $loginname = $_POST['loginname']; $password = $_POST['password']; mysql_connect ("localhost", "root", "majick") or die ( mysql_select_db ("animal1") $result = mysql_query("INSERT INTO login ($username, $password) VALUES ('$username', $password')"); mysql_query($query) or die (’Error updating database’); echo “database was updated with: ” ".$username.” “.$password.”; ?> Please help me any help would be greatly appreciated!