也许有更有效的方法，但我认为这能满足你的要求：

WITH x AS (
    SELECT *
    FROM (VALUES 
          ('xys','txq','2020-03-01'::DATE,1),
          ('xys','txs','2020-03-08'::DATE,0),
          ('xys','txt','2020-03-12'::DATE,0),
          ('xyb','txl','2020-03-13'::DATE,0),
          ('xyb','txp','2020-03-14'::DATE,1)) as y (card_id, txn_id, date_, legit)
    )
SELECT card_id,
    CASE WHEN legit = 0 THEN txn_id
         WHEN DATEDIFF('d',date_next,date_) <= 10 THEN txn_id_next
         ELSE txn_id END as txn_id,
    CASE WHEN legit = 0 THEN date_
         WHEN DATEDIFF('d',date_next,date_) <= 10 THEN date_next
         ELSE date_ END as date_,   
    CASE WHEN legit = 0 THEN legit
         WHEN DATEDIFF('d',date_next,date_) <= 10 THEN legit_next
         ELSE legit END as legit
FROM (
  SELECT *,
      ROW_NUMBER() OVER (PARTITION BY card_id ORDER BY date_) as row_num,
      LEAD(txn_id) OVER (PARTITION BY card_id ORDER BY date_) as txn_id_next,
      LEAD(date_) OVER (PARTITION BY card_id ORDER BY date_) as date_next,
      LEAD(legit) OVER (PARTITION BY card_id ORDER BY date_) as legit_next
  FROM x
  QUALIFY row_num = 1
  );

首先获取第一条记录和“下一条记录”，并将它们移动到一行，然后 CASE 语句根据我认为您指定的规则确定是使用第一条记录还是“下一条”记录。

我希望能够比较该卡id的第一个记录和任何未来记录之间的日期，其中该未来记录的合法性为0，如果第一个记录在10天内，则显示该记录。
如果我理解正确，您可以使用窗口函数：

select t.*
from (select t.*, min(date) over (partition by cardid) as min_date
      from t
     ) t
where legit = 0 and date < min_date + interval '10 day';