larrypelican
New Member
I have never encountered an error like this before. I was using prepared statements when it stared and have tried everything to correct the problem including stripping the form down to it's bare components.The following code, when executed, is creating a duplicate row.\[code\] $sql = "INSERT INTO inventory VALUES ('','$stocknum','$year','$make','$model','$price','$body','$mile','$engine','$trans', '$drive','$doors','$fuel','$vin','$des','$featured','$sale','$features','$safety','$airbags','$power', '$exterior','$interior','','','','','')";$insert = mysql_query($sql,$connection) or die(mysql_error());$name = mysql_insert_id();\[/code\]I can't wrap my head around why it would do this.