avoiding writing duplicate records

[adavis]

Okay, start over here. What I want to do is query the database to see if the id exists. If it doesn't then I want to add a row. If it does, then I want to skip that part. The trouble I'm having is that if the row is not out there, then I get an error message on my query. I have tried numerous combinations, but haven't been able to find what works. Current code:

$sql = "SELECT COUNT(*) FROM speaker WHERE sid = '".$_POST['sid']."' LIMIT 1";
$result = mysql_query($sql,$dblink) || die("QUERY FAILED: $sql");
$numrows = 0;
if (mysql_num_rows($result)) { $numrows++; }
echo "numrows:".$numrows."<br>sql: ".$sql."<br>";

I get this error:

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in /web/apache/htdocs/x/x/input.php on line 24
numrows:0
sql: SELECT COUNT(*) FROM speaker WHERE id='123456' LIMIT 1

Line 24 is the if statement. I've tried a check $result, but on that, I get a return of 1, regardless of whether there's a match or not.

Please, any ideas, I'm quickly running out of time. I have a 5/1 deadline for customer delivery.

[ir4z0r]

$sql = "SELECT * FROM speaker WHERE sid = '".$_POST['sid']."'";
$result = mysql_query($sql,$dblink) or die("QUERY FAILED: $sql");
if (mysql_affected_rows() > 0)
{
  echo "found";
}
else
{
  echo "numrows:". mysql_num_rows() ."<br>sql: ".$sql."<br>";
}

you have used COUNT(*) before, this ALWAYS returns a row containing the count

[adavis]

Hey ir4z0r, Thanks! Finally. that worked. I just had to figure out how to ask my question right. Take care.

[adavis]

All this trouble I've been having had been due to an error in my die statement. Oops...