I have a simple section of code that attempts to UPDATE a row in a MySQL table.  If mysql_affected_rows() = 0 I assume the row didn't exist so I go ahead and INSERT it.  This works about 95% of the time.  

The other 5% of the time, the row does exist and IS updated, but mysql_affected_rows() returns 0--so I go ahead and INSERT it.  I then end up with a duplicate entry.  I could define the appropriate columns as "unique" so that the INSERT would fail, but I feel that something else is wrong here and doing this would be a work-around for what might be a real problem.

My table is entirely numeric, no strings.  It's not a case of updating a row to its same values and so it's not counted as an affected row.  The value DOES change.

My code is:

$SQL = "UPDATE UserStats SET MS=MS+$Var,MC=MC+1 
       WHERE idUser=$idUser AND Month=$Month";
$updateStats = mysql_db_query ($sDB, $SQL, $nConnection);
if(mysql_affected_rows($nConnection) == 0)
 {
 $SQL = "INSERT INTO UserStats SET idUser=$idUser,
   Month=$Month, MS=$Var,MC=1";
 $updateStats = mysql_db_query ($sDB, $SQL, $nConnection);
 }

MS, MC, and idUser are all BigInts.  Month is MediumInt. The $Var variable is always non-zero.  I've dumped the SQL queries to a logfile--they are always valid queries, so it's not an issue of one of my variables not being defined and producing an improper SQL.  I've also tried the mysql_affected_rows without the $nConnection paramter.

In summary, the above UPDATE *ALWAYS* works in that the actual row in question is always updated correctly in the MySQL database.  However, sometimes the mysql_affected_rows() returns 0 instead of 1; so my code continues to INSERT a new row and I end up with a duplicate.

MySQL version is 3.23.49.  Same UPDATE instruction works fine when executed manually multiple times in MySQL command-line, etc.  Always returns the correct number of rows having been updated.

I am not sure if this is a PHP problem or a MySQL problem, but I lean towards PHP since MySQL *IS* updating the row as requested and I can't duplicate the problem outside of PHP.  I'm also open to it being a code problem on my end, but at this point I don't see how.