Question

我的数据库中有此表：

 id   name             source
------------------------------
  1    John Doe          fbs
  2    Clara Wayne       fbs
  3    John Doe          dem
  4    Markus Clark      lp
  5    Markus Clark      fbs
  6    John Doe          sms

我需要提取这样的内容，其中列source既不是dem也不是sms：

 id   name             source
------------------------------
  2    Clara Wayne       fbs
  4    Markus Clark      lp

Answer 1

我假设未选择John Doe，因为他的记录之一属于sms来源，您不想在结果中包括该信息。

SELECT * 
FROM YOUR_TABLE 
WHERE name NOT IN (
  SELECT name
  FROM YOUR_TABLE
  WHERE source IN ('dem','sms')
);

Answer 2

您可以选择喜欢的数据。...

    SELECT * FROM YOUR_TABLE_NAME WHERE SOURCE NOT IN ('DEM', 'SMS')

或者您可以这样做...

   SELECT DISTINCT SOURCE, NAME, ID FROM TABLE_NAME T WHERE T.NAME NOT IN (SELECT TOP 1 NAME FROM TABLE_NAME WHERE SOURCE IN('DEM', 'SMS'))

Answer 3

select * from your_table where source != 'dem' and source != 'sms'

另一种解决方法

Answer 4

您可以像这样列出“不想要的值”：

SELECT * FROM yourtable WHERE source NOT IN ('fbs','lp');

Answer 5

您可以使用：

CANEraseResponse.data[8] = {0x00, 0xFF, 0x00, 0x04, 0x02, 0x00, 0x00, 0x00};

Answer 6

尝试一下

SELECT *
FROM yourtable t
WHERE NOT EXISTS (
    SELECT 1
    FROM yourtable
    WHERE id = t.id 
        AND name = t.name 
        AND source IN ('dem', 'sms')
)

查询以提取没有特定值的行

6 个答案: