SQL Find Customers who have not paid 100

Hi
I have a Customers table and a Payments table, shown in that order below
What joined query would I use to find the CustID of those Customers who have
not paid an AMOUNT of 100. The [LINK ID] is the same as the CustID in CUSTOMERS

Customers
Payments
Murray BrownMicrosoft Cloud Azure/Excel Solution DeveloperAsked:
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Scott PletcherSenior DBACommented:
SELECT c.*
FROM (
    SELECT [Link Id]
    FROM Payments
    GROUP BY p.[Link Id]
    HAVING MAX(CASE WHEN Amount = 100.00 THEN 1 ELSE 0 END) = 0
) AS p
INNER JOIN Customers c ON
    c.CustID = p.[Link Id]
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Jim HornMicrosoft SQL Server Developer, Architect, and AuthorCommented:
btw, you might have the image labels off, as the set labeled 'Customers' looks like a payments table, and vise versa

Looks like Scott answered it above.

Also, do yourself a favor and lose the spaces in the First Name, Last Name columns, as it forces T-SQL to delineate it using square brackets [ ], which is one thing to potentially forget and throw an error.
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Phillip BurtonDirector, Practice Manager and Computing ConsultantCommented:
I assume that these two tables are the wrong way round - the first one should be payments and the customers.

Select C.CustID, C.[First Name], C.[Last Name], Sum(Amount) as TotalAmount
From Customers C
LEFT JOIN Payments P
ON C.CustID = P.[LINK ID]
GROUP BY C.CustID, C.[First Name], C.[Last Name]
HAVING Sum(Amount) < 100 OR P.[LINK ID] IS NULL

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Murray BrownMicrosoft Cloud Azure/Excel Solution DeveloperAuthor Commented:
Hi
Yes the two tables are the wrong way around. I was looking for Customers who have not paid the amount of 100 once(not in total).
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Phillip BurtonDirector, Practice Manager and Computing ConsultantCommented:
Select C.CustID, C.[First Name], C.[Last Name], Max(Amount) as MaxAmount
From Customers C
LEFT JOIN Payments P
ON C.CustID = P.[LINK ID]
WHERE Max(Amount) < 100 OR P.[LINK ID] IS NULL
GROUP BY C.CustID, C.[First Name], C.[Last Name]

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Murray BrownMicrosoft Cloud Azure/Excel Solution DeveloperAuthor Commented:
Thanks for the help
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