根据条件postgresql查找以前的值

find previous value based on condition postgresql 

id    account   amount  
123     abc      10        
1234    abc      12        
4568    abc      16        
456     def      20        
458     def      30

我想过滤差值（金额）小于 5 的唯一 ID（按帐户分组）。这里的答案是这样的：

id:
123
1234
4568

您可以结合使用 LEAD 和 LAG 以获得所需的结果。

SELECT id
FROM (
    SELECT t.*
        ,amount - LAG(amount, 1, 0) OVER (
            PARTITION BY account ORDER BY amount
            ) AS diff1
        ,amount - LEAD(amount, 1, 0) OVER (
            PARTITION BY account ORDER BY amount
            ) AS diff2
    FROM t
    ) s
WHERE abs(diff1) < 5
    OR abs(diff2) < 5;

Demo