Sql 选择符合条件的所有记录加上不符合条件的下一条记录

我有一个表，它有如下列：

ID, Description, PC, Account, Amount

我将在“PC”列中寻找一个009值，每个数据在“ID”列中都有一个顺序编号的唯一ID。具有009“PC”值的每个数据后面至少有一个项目没有将“009”作为PC值

我有没有办法写一个真实的、类似于下面伪代码的查询？这样的事情真的可能吗

SELECT  
table.ID AS ID,  
table.Description AS Description,  
table.PC AS PC
WHERE ((table.PC = 009) OR (whatever, as long as table.ID-1's "PC" value = 009))

---

如果您知道

id

s实际上是顺序的，那么可以将逻辑放在

WHERE

子句中：

SELECT t.ID, t.Description, t.PC AS PC
FROM table as t
WHERE t.PC = 009 OR
      EXISTS (SELECT 1
              FROM table as t2
              WHERE t2.id = t.id - 1 AND t2.PC = 009
             )
ORDER BY t.ID;

---

样本数据和预期输出将非常好，并且易于理解。问题的格式不正确，没有任何意义，因为它的表述如下：

选择所有符合条件的记录，再加上下一条不符合条件的记录

SELECT  
table.ID AS ID,  
table.Description AS Description,  
table.PC AS PC
FROM table
WHERE ((table.PC = '009') OR 
table.ID - 1 = 
  (
  SELECT table.ID
  FROM table
  WHERE table.PC = '009'
  )
)

SELECT  
    t.ID AS ID,  
    t.Description AS Description,  
    t.PC AS PC
FROM tbl t
WHERE 
   (t.PC = 009) OR 
    (t.id - 1 = (SELECT t2.ID FROM tbl t2 where t.PC = 009))