[Resolved] Running a query in PHP!!!

GaryAC

I may be making a silly mistake, but I've been looking at it for ages and i cannot figure out what it is. So i was wondering if anyone can help!!! Please...

I have some data stored in PHP variables that are used to insert data to a mySQL db.

I have a customer table which automatically generates a custid number. So when I enter a new customer it is automatically given an id.

Now I need this custid, to be able to insert it into the next table. However for some reason when I query the customer table to obtain this custid, it returns nothing.

Here is my code:

if ( $referer[0] == URLLIST && $yesButton == 'Yes' ) {
if ( !($db = mysql_connect($host, $user, $passwd)) ) {
printError(sprintf("Error connecting to database from %s, by user %s", $host, $user));
} else {
mysql_select_db($dbName, $db);

  $query1 = "INSERT INTO customer (Name, addr1, addr2, addr3, postcode, email, passwd, cardnum, expiry)" .

"VALUES('$Name', '$Add_1', '$Add_2', '$Add_3', '$P_code', '$Email', '$pword', '$Card', '$expiry')";

  if ( !mysql_query($query1,$db) ) {
     printError(sprintf("Error %s : %s", mysql_errno(), mysql_error()));
  } else {
     echo ("Thank you for completing the registration form $Name <br /><br />\n");
  }

//Now I want to find the custid that has been generated to put in the next table

//Query to find custid
  $find = "select custid from customer where Name = $Name";
  $answer = mysql_query($find,$db);

//Put custid into PHP variable $custid
$custid = $answer["custid"];
echo "Customer id is " . $custid;

  $query2 = "INSERT INTO custord (custid, ortime, numdvd, carriage, orvalue)" .
			"VALUES ('$custid', '', '$NoDVD', '$Tx', '$Total')";

  if ( !mysql_query($query2,$db) ) {
     printError(sprintf("Error %s : %s", mysql_errno(), mysql_error()));
  } else {
     echo ("Thank you for making an order $Name <br /><br />\n");
  }
  mysql_close($db);

}
exit('</head><body></body></html>');
}

I can't figure out why $custid is empty!!!!!
Any solutions??

Thanks for your help

Reg

Gary

cahva

Well, you have forgotten to add simple quotes around the $Name.

$find = "select custid from customer where Name = $Name";
should be:
$find = "select custid from customer where Name = '$Name'";

But actually you dont need that query at all! You can use mysql_insert_id()-function which returns the last inserted id-number.

...
//Put custid into PHP variable $custid 
$custid = mysql_insert_id(); 
echo "Customer id is " . $custid;
...

GaryAC

You are a diamond!!!!!

Thanks alot, it works now, and mysql_insert_id() function is pretty handy hey!!!

Cheers I aprreciate your help

regards

Gary