Group consecutively values in MySQL and add an id to such groups

问题

I have a simple table and I need to identified groups of four rows (the groups aren't consecutives), but each rows of each row has a +1 in the value. For example:

----------------------
| language     | id  |
----------------------
| C            |  16 |
| C++          |  17 |
| Java         |  18 |
| Python       |  19 |
| HTML         |  65 |
| JavaScript   |  66 |
| PHP          |  67 |
| Perl         |  68 |
----------------------

I want to add a column that indicates the group or set, how is possible to get this output using MySQL?:

----------------------------
| language     | id  | set |
----------------------------
| C            |  16 |  1  |
| C++          |  17 |  1  |
| Java         |  18 |  1  |
| Python       |  19 |  1  |
| HTML         |  65 |  2  |
| JavaScript   |  66 |  2  |
| PHP          |  67 |  2  |
| Perl         |  68 |  2  |
----------------------------

Note that in this examples is only 2 sets (it could be 1 or more sets) and they didn't start in 16 (such values are not knowledged, but the restriction is that each id value of each row has this form n, n+1, n+2 and n+3).

I've been investigating about Gaps & Islands problem but didn't figure how to solve it by using their solutions. Also I search on stackoverflow but the closest question that I found was How to find gaps in sequential numbering in mysql?

Thanks

回答1:

select language, 
       @n:=if(@m+1=id, @n, @n+1) `set`, 
       (@m:=id) id 
   from t1, 
        (select @n:=0) n, 
        (select @m:=0) m

Demo on sqlfiddle

回答2:

SELECT language,id,g
FROM (
  SELECT language,id,
    CASE WHEN id=@lastid+1 THEN @n ELSE @n:=@n+1 END AS g,
    @lastid := id As b
  FROM
    t, (SELECT @n:=0) r
  ORDER BY
    id
) s

EDIT

In case you want just 4 per group add a row number variable:

SELECT language,id,g,rn
FROM (
  SELECT language,id,
    CASE WHEN id=@lastid+1 THEN @n ELSE @n:=@n+1 END AS g,
   @rn := IF(@lastid+1 = id, @rn + 1, 1) AS rn,
    @lastid := id As dt
  FROM
    t, (SELECT @n:=0) r
  ORDER BY
    id
) s
Where rn <=4

FIDDLE

回答3:

You can use the following query:

SELECT l.*, s.rn
FROM languages AS l
INNER JOIN (
  SELECT minID, @rn2:=@rn2+1 AS rn
  FROM (
    SELECT MIN(id) AS minID
    FROM (
      SELECT id,
             id - IF (true, @rn1:=@rn1+1, 0) AS grp
      FROM languages
      CROSS JOIN (SELECT @rn1:=0) AS var1 
      ORDER BY id) t    
    GROUP BY grp
    HAVING COUNT(grp) = 4 ) u
  CROSS JOIN (SELECT @rn2:=0) AS var2  
) s ON l.id BETWEEN minID AND minID + 3

The above query identifies islands of exactly 4 consecutive records and returns there records only. It is easily modifiable to account for a different number of consecutive records.

Please also note the usage of IF conditional: it guarantees that @rn1 is first initialized and then used in order to calculate grp field.

Demo here

来源：https://stackoverflow.com/questions/31814492/group-consecutively-values-in-mysql-and-add-an-id-to-such-groups

标签

mysql

sql

database

gaps-and-islands