Question

我想要复制的行为就像带有-A和-B标志的grep。例如grep -A 2 -B 2 "hello" myfile.txt将给我所有在其中有“hello”的行，但也包括前面的2行和后面的2行。让我们假设这个表模式：

+--------+-------------------------+
| id     |    message              |
+--------+-------------------------+
| 1      | One tow three           |
| 2      | No error in this        |
| 3      | My testing message      |
| 4      | php module test         |
| 5      | hello world             |
| 6      | team spirit             |
| 7      | puzzle game             |
| 8      | social game             |
| 9      | stackoverflow           |
|10      | stackexchange           |
+------------+---------------------+

现在查询如下： Select * from theTable where message like '%hello%'将导致：

5 | hello world

如何设置另一个选择N行的参数“N”，匹配记录后的N行，即N = 2，结果应为：

    | 3      | My testing message      |
    | 4      | php module test         |
    | 5      | hello world             |
    | 6      | team spirit             |
    | 7      | puzzle game             |

为简单起见，假设'like％TERM％'仅匹配1行。
此处的结果应该按自动递增id字段排序。

Answer 1

是的，这对我有用：

SELECT child.*
FROM stack as child,
(SELECT idstack FROM stack WHERE message LIKE '%hello%') as parent
WHERE child.idstack BETWEEN parent.idstack-2 AND parent.idstack+2;

Answer 2

不知道这是否完全有效，但是如何

SELECT  t.* 
FROM    theTable t
        INNER JOIN (
          SELECT id FROM theTable where message like '%hello%'
        ) id ON id.id <= t.id
ORDER BY
        ID DESC
LIMIT   3                    
UNION ALL 
SELECT  t.* 
FROM    theTable t
        INNER JOIN (
          SELECT id FROM theTable where message like '%hello%'
        ) id ON id.id > t.id
ORDER BY
        ID
LIMIT   2

Answer 3

尝试这个简单的（已编辑） -

CREATE TABLE messages(
  id INT(11) DEFAULT NULL,
  message VARCHAR(255) DEFAULT NULL
);

INSERT INTO messages VALUES 
  (1, 'One tow three'),
  (2, 'No error in this'),
  (3, 'My testing message'),
  (4, 'php module test'),
  (5, 'hello world'),
  (6, 'team spirit'),
  (7, 'puzzle game'),
  (8, 'social game'),
  (9, 'stackoverflow'),
  (10, 'stackexchange');

SET @text = 'hello world';

SELECT id, message FROM (
  SELECT m.*, @n1:=@n1 + 1 num, @n2:=IF(message = @text, @n1, @n2) pos
    FROM messages m, (SELECT @n1:=0, @n2:=0) n ORDER BY m.id
) t
WHERE @n2 >= num - 2 AND @n2 <= num + 2;

+------+--------------------+
| id   | message            |
+------+--------------------+
|    3 | My testing message |
|    4 | php module test    |
|    5 | hello world        |
|    6 | team spirit        |
|    7 | puzzle game        |
+------+--------------------+

N值可以指定为用户变量;目前它是 - '2'。

此查询适用于行号，这可以保证返回最近的记录。

Answer 4

尝试

Select * from theTable
Where id >=
(Select id - variableHere from theTable where message like '%hello%')
Order by id
Limit (variableHere * 2) + 1

Answer 5

（仅限MS SQL Server）

最可靠的方法是使用row_number函数，如果id中存在间隙则无关紧要。如果搜索结果有多次出现并且在每个结果的上方和下方正确返回两个，那么这也将起作用。

WITH

srt AS (
    SELECT ROW_NUMBER() OVER (ORDER BY id) AS int_row, [id]
    FROM theTable
),

result AS (
    SELECT int_row - 2 AS int_bottom, int_row + 2 AS int_top
    FROM theTable
        INNER JOIN srt
            ON theTable.id = srt.id
    WHERE ([message] like '%hello%')
)

SELECT theTable.[id], theTable.[message]
FROM theTable
    INNER JOIN srt
        ON theTable.id = srt.id
    INNER JOIN result
        ON srt.int_row >= result.int_bottom
        AND srt.int_row <= result.int_top
ORDER BY srt.int_row

Answer 6

使用日期而不是ID添加答案。这里的用例是一个随叫随到的轮换表，每周记录一个记录。由于编辑，id可能出于预期目的而无序。任何具有几周记录，pr日期或其他记录的用例当然都必须修补。

+----------+--------------+------+-----+---------+----------------+
| Field    | Type         | Null | Key | Default | Extra          |
+----------+--------------+------+-----+---------+----------------+
| id       | int(11)      | NO   | PRI | NULL    | auto_increment |
| startdate| datetime     | NO   |     | NULL    |                |
| person   | int(11)      | YES  | MUL | NULL    |                |
+----------+--------------+------+-----+---------+----------------+

查询：

SELECT child.*
FROM rota-table as child,
 (SELECT startdate
    FROM rota-table
    WHERE YEARWEEK(startdate, 3) = YEARWEEK(now(), 3) ) as parent
WHERE
    YEARWEEK(child.startdate, 3) >= YEARWEEK(NOW() - INTERVAL 25 WEEK, 3)
    AND YEARWEEK(child.startdate, 3) <= YEARWEEK(NOW() + INTERVAL 25 WEEK, 3)

在符合条件的行之前和之后选择N行？

6 个答案: