RE: [PHP] what is wrong about this sniplet?

HI Jack, What is the variable $HotelID ? mysql_insert_id - has an optional parameter: link_identifier (the database handle) So in this case, is $HotelID your database handle? If not, then just call mysql_insert_id() without any parameters. Otherwise, it is always best to pass around the database handle so that you know which connection you are using. Sam > -----Original Message----- > From: Jacky@lilst [mailto:jacky@activelifestyle.com] > Sent: Thursday, 25 January 2001 03:17 > To: php-general@lists.php.net > Subject: [PHP] what is wrong about this sniplet? > > > I have the sniplet to run teh query at the page to insert data as > shown below. After it is run, there was en error said something like > "Mysql warning, 0 ( zero) is not Mysql index" and the error > point to the line using mysql_insert_id($HotelID); > My limited experience cannot tell me what should I be doing in > order to get what I need. Any thought? > ****************************************************** > //insert Hotel detail > $insertHotel = "INSERT INTO Hoteldetail > (HotelName,HotelLocation, HotelCountry,HotelPostcode, > HotelTelephone,Hotelfax,HotelURL, > HotelContactFirstName,HotelContactLastName, HotelRoomProvided, > HotelEmail) VALUES ('$HotelName', '$HotelLocation', > '$HotelCountry', '$HotelPostcode', '$Hoteltelephone', > '$Hotelfax', '$HotelURL', '$HotelContactFirstName', > '$HotelContactLastName', '$HotelRoomProvided', > '$HotelEmail')"; > $resultHotel = mysql_query($insertHotel); > // retrive latest HotelID > $latestHotelID = mysql_insert_id($HotelID); > ****************************************************** > cheers > Jack > jacky@activelifestyle.com > "There is nothing more rewarding than reaching the goal you set > for yourself" >