Question

我有非连续但有序的数字标识符。我想得到连续的值。

当前表：

original_value

1
1
1
3
3
29
29
29
29
1203
1203
5230304
5230304
5230304
5230304

所需表：

original_value   desired_value

1                1
1                1
1                1
3                2
3                2
29               3
29               3
29               3
29               3
1203             4
1203             4
5230304          5
5230304          5
5230304          5
5230304          5

Answer 1

另一种方法，无法加入：

select

  original_value,
  case when @original = original_value then
    @group_number
  else
    @group_number := @group_number + 1
  end,
  (@original := original_value) as x

from tbl,(select @original := null, @group_number := 0) z
order by original_value

实时测试：http://www.sqlfiddle.com/#!2/b82d6/6

如果要删除result中的计算，table-派生查询：

select w.original_value, w.group_number
from
(
  select

    original_value,
    case when @original = original_value then
      @group_number
    else
      @group_number := @group_number + 1      
    end as group_number,
    (@original := original_value) as x

  from tbl,(select @original := null, @group_number := 0) z
  order by original_value
) w

实时测试：http://www.sqlfiddle.com/#!2/b82d6/4

Answer 2

汇总您的值，使用行号来获得您想要的值，然后再加入原始值：

SELECT t.original, sq.desired
FROM Table t
INNER JOIN(
  SELECT ssq.original, @rownum:=@rownum+1 ‘desired’
  FROM (
    SELECT t.original
    FROM Table t
    GROUP BY t.original
    ORDER BY t.original
  ) ssq,
  (SELECT @rownum:=0) r 
) sq ON sq.original = t.original

Answer 3

另一种方法，这个模拟带有连接的DENSE_RANK：

    SELECT t.original, d.derived
      FROM t
INNER JOIN (    SELECT t.original, COUNT(DISTINCT t1.original) AS "derived"
                  FROM t
            INNER JOIN t t1
                       ON t.original >= t1.original
              GROUP BY 1) d
           ON t.original = d.original
  ORDER BY t.original;

Answer 4

如果我理解正确的话：

select 
    original_value, 
    dense_rank() over (order by original_value) as desired_value

SQL：用连续值替换非连续值

4 个答案: