Hi I have two mysql problems
a) I'm confused on why this code won't add data to my database:
$con = mysql_connect("localhost","username","password");
mysql_select_db("database", $con);
$sql="INSERT INTO Table ('adder', 'username')
VALUES('$myusername','$username1')";
mysql_query( $sql, $con ) or trigger_error( mysql_error( $con ), E_USER_ERROR );
here is the full error: "you have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'Add (username
, name
, number
) VALUES('56', 'Name', 'number')' at line 1"
and before you ask, yes the data is there
b) I'm having problems with mysql_fetch_array
This is the code:
mysql_connect("$host", "$mysqlusername", "$password")or die("cannot connect");
mysql_select_db("$db_name")or die("cannot select DB");
$sql = "SELECT * FROM Add WHERE adder=\"$username\"";
$result=mysql_query($sql);
$row=mysql_fetch_array($result);
if (empty($row)) {
echo "$username hasn\'t added anyone";
}
else {
while($row = mysql_fetch_array($result)){
echo "<a href=\"/".$row[username]."><".$row[username]."</a>";
}
}
The error here is: Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in....
And yes the data is there on the database aswell..
Any ideas??
Thanks for the help :) Niall
You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near...
You omitted the important part that comes afterwards. – Richard86 May 25 '11 at 19:39