mysql find the most recent DISTINCT customers from multiple orders

Why not:

SELECT  CustID, Name FROM tblCustomers ORDER BY CustID DESC LIMIT 5

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You only need the five most recent customers, right? Why involve the Orders table?

HTH,
Dan

sorry i was indeed imprecise,
the five customers with the most recent orders.

OK, try this:

SELECT  CustID, Name FROM tblCustomers
    WHERE CustID IN (
        SELECT DISTINCT tblOrders.CustID FROM tblOrders 
        ORDER BY tblOrders.CustID DESC LIMIT 5)

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This will first select the customer ids for the most recent orders. Distinct is there so if a customer has multiple orders it will still be counted as 1.
Then it will give you the names of those customers.

HTH,
Dan

close, very close -
any way to do this in version 5.3?

i get the error    (1235)  THis version of MySQL doesn't yet support LIMIT & IN subquery

thanks for your patience!

OK, let's try it with a Join

SELECT DISTINCT tblOrders.CustID, tblCustomers.Name, tblOrders.ID
    FROM tblOrders  JOIN tblCustomers 
        ON tblOrders.CustID = tblCustomers.ID
    ORDER BY tblOrders.ID DESC LIMIT 5

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thank you for your continued help!
i'm not that familiar with JOIN, but have tried it.

my table Orders is actually called Chunk (an inherited name!)
as a check, the following works:  SELECT tblChunk.CustID FROM tblChunk

but  this error comes about when i execute the following query.
error:
SQL Error (1054): Unknown column 'tblChunk.CustID' in 'on clause'

query (based on your suggestion, which seems like it should work) :
SELECT  DISTINCT CUS.CustID, CUS.Name, CH.ChunkID
   FROM  tblChunk CH JOIN tblCustomers CUS
   ON tblChunk.CustID = tblCustomers.CustID
   ORDER BY tblChunk.ChunkID DESC LIMIT 5

OK, if you insist on aliasing your tables:

SELECT DISTINCT CH.CustID, CUS.Name, CH.ChunkID
    FROM tblChunk CH JOIN tblCustomers CUS
        ON CH.CustID = CUS.CustID
    ORDER BY CH.ChunkID DESC LIMIT 5

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If it gives the same error (unknown CH.CustID) it simply means you do not have a CustID field on your chunk table, so post here the create statement for your two tables so we can do a correct query.

this one works (also see that i confirmed the column field existence in previous post)
however it is not distinct for Customers
result:
CustID      Name      ChunkID
229              Casm      4015
251              Thor      4014
251              Thor      4013
240              Sher       4012
240              Sher       4011

suggestions?  (again thanks)

I've been following this with some interest because I can't think of a way that will work. I'd be surprised is Dan's last response handles a situation where the last 5 orders were placed by only two customers for example.

What I am struggling with is that we may need only 5 orders (each customer placed one) or 25 (each customer placed 5). I think the solution may need a combination of group by and having to identify those that placed multiple orders with a union to the last 25 orders placed but my sql is not good enough to figure it out - Dan?

on other hand maybe the above works and I'll get back in my box

Tested using this:

CREATE TABLE IF NOT EXISTS `tblchunk` (
  `ChunkId` int(10) unsigned NOT NULL,
  `CustID` int(10) unsigned NOT NULL,
  `Order` varchar(10) COLLATE latin1_general_ci NOT NULL,
  PRIMARY KEY (`ChunkId`)
) ENGINE=MyISAM DEFAULT CHARSET=latin1 COLLATE=latin1_general_ci;

--
-- Dumping data for table `tblchunk`
--

INSERT INTO `tblchunk` (`ChunkId`, `CustID`, `Order`) VALUES
(4011, 240, 'aaa'),
(4012, 240, 'gagsa'),
(4013, 251, 'hasha'),
(4014, 251, 'hashg'),
(4015, 229, 'gsdhas'),
(5, 0, ''),
(6, 1, '');


CREATE TABLE IF NOT EXISTS `tblcustomers` (
  `CustID` int(10) unsigned NOT NULL,
  `Name` varchar(50) COLLATE latin1_general_ci NOT NULL,
  PRIMARY KEY (`CustID`)
) ENGINE=MyISAM DEFAULT CHARSET=latin1 COLLATE=latin1_general_ci;

--
-- Dumping data for table `tblcustomers`
--

INSERT INTO `tblcustomers` (`CustID`, `Name`) VALUES
(229, 'Casm'),
(251, 'Thor'),
(240, 'Sher'),
(0, 'fdhgasdg'),
(1, 'tyo'),
(2, 'sagsd'),
(3, 'kfg');

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(Sorry i was out and didn't respond immediately.)

YES Dan's last one works!   Thank you all for your VERY PATIENT help.
this is inspiration for me to dig in more to learn the subtle details of SQL
again MUCH APPRECIATED!

many thanks!

Glad I could help!