Help with database query and records

[destiny]

1. I am trying to have data submitted by a form checked against my database. If a record exists based upon name and email, then an update happens. If no record is there the data is entered.
   I need to know how to identify the CustomerId number already in existence in the table for the UPDATE query.
2. Next the second part of the informatiion gets put into a second table calling upon the lastId generated to put in the second field as a cross-reference. Will this code work for what has transpired before it?
   I'm a little confused by it right now so any help to get this working would be greatly appreciated.

mysql_query("select * from customers where Email='$Email' and FirstName='$FirstName'");
if (mysql_num_results == 0) // then not already there
$query = "INSERT INTO customers VALUES ('NULL','$FirstName','$LastName','$PartnerName','$Number','$Mobile',
'$Fax','$Email','$Gender','$Age','$Bought_Cd','$CustRemarks','$CoName','$CustAddress','$CustCity','$CustState','$CustZip',
'$CustCountry','$Newsletter','$ClientVname','$ClientContact','$ClientAddress','$ClientCity','$ClientCountry','$ClientPhone'
,'$ClientFax','$Clientmobile','$ClientEmail','$ClientRemarks','$mysqlDate')";
$result = mysql_query($query);

else
$query = "UPDATE customers SET FirstName='$FirstName',LastName='$LastName',PartnerNames='$PartnerNames',Number='$Number'
,Mobile='$Mobile',Fax='$Fax',Email='$Email',Gender='$Gender',Age='$Age',Bought_Cd='$Bought_Cd',CustRemarks='$CustRemarks'
,CoName='$CoName',CustAddress='$CustAddress',CustCity='$CustCity',CustState='$CustState',CustZip='$CustZip',CustCountry='$CustCountry'
,Newsletter='$Newsletter',ClientVname='$ClientVname',ClientContact='$ClientContact',ClientAddress='$ClientAddress'
,ClientCity='$ClientCity',ClientCountry='$ClientCountry',ClientPhone='$ClientPhone'
,ClientFax='$ClientFax',Clientmobile='$Clientmobile',ClientEmail='$ClientEmail',ClientRemarks='$ClientRemarks'
,mysqlDate='$mysqlDate'WHERE CustomerId=$lastId";

//not sure about this:
$bookingId = mysql_result /// get the id result from the prev query

$lastId = mysql_insert_id($databaseLink)
or die("Invalid query: " . mysql_error());

$gigDate = "$ShowYear-$ShowMonth-$ShowDate";
$query = "INSERT INTO booking VALUES ('NULL','$lastId','$Vname','$Vaddress','$Vcity','$Vphone',
'$Bookas','$gigDate','$ShowDay','$ShowMonth','$ShowDate','$ShowYear','$ShowLength','$SoundCheck','$ShowTime','$CheckTime','$MultipleShow',
'$EventType',
'$Environment','$Capacity','$TicketPrice','$OpenPublic','$OtherDescription','$Guarantee','$BuyerProvide','$ArtistProvide'
,'$PaysTechnician','$Accommodations',NOW(),'')";
$queryresult = mysql_query($query, $databaseLink);
$bookingId = mysql_insert_id();
printf("$FirstName your unique booking id number is:%d\n", $bookingId);

[rachel2004]

It's a bit hard to read your code without any indents, curly braces, or color coding, but I can see you have some problems you need to address if you want to get anywhere.

My suggestions are just some that I see from a quick scan of your code, so other problems may exists. Make these changes, clean up the code so people can read it more easily and then post again so others or I will continue helping you out:

1. if (mysql_num_results == 0) // then not already there
   (you are incorrectly using the mysql_num_rows identifier, so that will fail. Look at the PHP manual... )

2. $bookingId = mysql_result /// get the id result from the prev query

(You can use the function mysql_insert_id() to get the id of the last database INSERT performed in your script)