仅当至少有一行匹配时才选择条件

Question

我有一个select语句，并且只有当其中至少有一个与今天有60天不同的日期时才返回所有值。 问题是我有一个外部应用程序，它返回我想要比较的列，它们来自不同的表（一个属于现金项目，另一个属于卡项目）。
考虑到我有以下几点：

OUTER APPLY (
    SELECT COUNT(*) AS quantity, MIN(date) AS item_date 
     FROM dbo.get_cash_items(loans.id_cash) AS cash_item 
     HAVING loans.id_product_type = 1 --Cash 
    UNION 
    SELECT COUNT(*) AS quantity, MIN(date) AS item_date 
     FROM dbo.get_card_items(loans.id_card) AS card_item 
     HAVING loans.id_product_type = 2 --Card 
) AS items 

我想返回所有的行，只有当DATEDIFF(DAY, MIN(items.item_date), GETDATE()) >= 60，但我想他们都哪怕只有一个匹配这个条件。
这样做的最佳方法是什么？

编辑
使其更清晰，我将解释使用案例：
我需要显示每一笔贷款的项目，只有当客户端是在到期日上的60多天晚任何它

Answer 1

我也不太清楚，你能指望什么，而是怎么样：

WITH items 
    AS (SELECT Count(*) AS quantity, 
       Min(date) AS item_date 
     FROM dbo.Get_cash_items(loans.id_cash) AS cash_item 
     HAVING loans.id_product_type = 1 
     UNION 
     SELECT Count(*) AS quantity, 
       Min(date) AS item_date 
     FROM dbo.Get_card_items(loans.id_card) AS card_item 
     HAVING loans.id_product_type = 2) 
SELECT a.* 
FROM items AS a, 
     (SELECT TOP 1 * 
     FROM items AS b 
     WHERE Datediff(day, b.item_date, Getdate()) >= 60) AS c 

它是一种CROSS JOIN的，在表C都会有一个或零行取决于是否条件满足 - 它会加入到其他表中的每一行。

Answer 2

你有没有尝试过这样的事情？

SELECT a.quantity, a.item_date 
FROM 
    (SELECT COUNT(*) AS quantity, MIN(date) AS item_date 
     FROM dbo.get_cash_items(loans.id_cash) AS cash_item 
     HAVING loans.id_product_type = 1 
    UNION 
    SELECT COUNT(*) AS quantity, MIN(date) AS item_date 
     FROM dbo.get_card_items(loans.id_card) AS card_item 
     HAVING loans.id_product_type = 2) a 
WHERE DATEDIFF(day, a.item_date, GETDATE()) >= 60 

Answer 3

通常我使用CTE来选择我想要选择的记录的关键字，然后加入。下面是一个例子尝试：

with LateClients as 
(
    SELECT LoadId FROM Payment Where /*payment date later than 60 days*/ 
) 

SELECT  p.LoanId, 
      p.UserId 
FROM  Payment as p 
INNER JOIN LateClients as LC 
ON   p.LoanId = lc.LoanId 
OrderBy p.LoanId, p.UserId 

我知道这是从你们发布的代码有点不同，但是这是要解释这个概念一个简单的例子。祝你好运！