[Banditul18]

PHP код:

mysql_format(SQL, query, sizeof query, "UPDATE `%s` SET `%s`='%d' WHERE `%s`=%d", table, column, value, sqlID, sqlid); 


You forgot to pass the value that you wanted to be updated

Quote:

Originally Posted by StRaphael Try: PHP код: mysql_format(SQL, query, sizeof query, "UPDATE `%s` SET `%s`='%d' WHERE `%s`='%d'", table, column, sqlID, sqlid);

That does same thing but worst, it will not gonna use the index for the sqlID so it makes it a bit slower