Question

我有这个简单的表格设置see fiddle

    CREATE TABLE mytable
    (`id` int, `itemid` int);

    INSERT INTO mytable
    (`id`, `itemid`)
    VALUES
    (1, 111),
    (2, 222),
    (3, 333),
    (4, 444),
    (5, 111),
    (6, 222),
    (7, 333),
    (8, 564),
    (9, 111),
    (10, 121),
    (11, 131),
    (12, 111),
    (13, 353),
    (14, 373);

我想显示最后一个itemid在当前行itemid旁边的行中的内容。我已经用下面的

完成了

    SELECT 
    mt.id,
    mt.itemid,
    (
        select mt2.itemid
        from mytable mt2
        where mt2.id < mt.id
        ORDER BY mt2.id DESC
        LIMIT 1
    ) as lastitemid
     FROM mytable mt 
     ORDER BY id DESC
     LIMIT 5

这可以按预期返回

    ID  ITEMID  LASTITEMID
    14  373     353
    13  353     111
    12  111     131
    11  131     121
    10  121     111

但是我只想显示lastitemid = 111的行。

我尝试过做

    SELECT 
    mt.id,
    mt.itemid,
    (
        select mt2.itemid
        from mytable mt2
        where mt2.id < mt.id
        ORDER BY mt2.id DESC
        LIMIT 1
    ) as lastitemid
    FROM mytable mt 
    WHERE lastitemid = 111
    ORDER BY id DESC
    LIMIT 5

在'where子句'

中获取未知列'lastitemid'

我也尝试添加

    AND mt2.itemid = 111

到内部查询

这不会得到任何错误，但是对于所有行都返回111，这不是我想要的，因为它是无效的，例如对于id = 12，lastitemid是131但是它说111

    ID  ITEMID  LASTITEMID
    14  373     111
    13  353     111
    12  111     111
    11  131     111
    10  121     111

使用我的示例数据集，如果我有正确的查询，我应该得到以下结果

    ID  ITEMID  LASTITEMID
    13  353     111
    10  121     111
    6   222     111
    2   222     111

我怎样才能做到这一点？

Answer 1

试试这个：

SELECT mt.id, mt.itemid,
      (SELECT mt2.itemid FROM mytable mt2 WHERE mt2.id < mt.id ORDER BY mt2.id DESC LIMIT 1) AS lastitemid
FROM mytable mt 
HAVING lastitemid = 111
ORDER BY id DESC
LIMIT 5

检查SQL FIDDLE DEMO

<强>输出

| ID | ITEMID | LASTITEMID |
|----|--------|------------|
| 13 |    353 |        111 |
| 10 |    121 |        111 |
|  6 |    222 |        111 |
|  2 |    222 |        111 |

Answer 2

如果保证id是连续的，你可以这样做。

SELECT curr.id, curr.itemid, previous.itemid AS lastitemid
FROM   mytable curr JOIN mytable previous ON previous.id = curr.id - 1
WHERE  previous.itemid = 111

否则，你需要像

这样的东西

SELECT curr.id, curr.itemid, previous.itemid AS lastitemid
FROM   mytable curr, mytable previous
WHERE  previous.id < curr.id 
  AND  previous.itemid = 111
  AND NOT EXISTS ( 
      SELECT 1 FROM mytable interloper
      WHERE  interloper.id < curr.id
        AND  previous.id < interloper.id )

Answer 3

您只需将当前查询用作表格，然后使用where子句从中进行选择。

SELECT 
l.id,
l.itemid
FROM
(
    SELECT 
    mt.id,
    mt.itemid,
    (
        select mt2.itemid
        from mytable mt2
        where mt2.id < mt.id
        ORDER BY mt2.id DESC
        LIMIT 1
    ) as lastitemid
    FROM mytable mt 
) As l
WHERE l.lastitemid = 111
ORDER BY l.id DESC
LIMIT 5

根据最后一行mysql过滤数据

3 个答案: