我有以下结构的表格
ID VALUE
1 100
2 200
3 300
4 400
5 500
我想要这样的输出
ID VALUE DIFF_to_Prev
1 100 0
2 200 100
3 300 100
4 400 100
5 500 100
这是我到目前为止尝试过的查询
SET @LastVALUE:= 0;
SET @LastSN:= 0;
SELECT dtr.SN, dtr.VALUE,
IF(@LastSN = dtr.SN, dtr.Value - @LastVALUE, 0) DIFF_to_Prev
FROM difftworows as dtr
这是我从中得到的结果:
ID VALUE DIFF_to_Prev
1 100 0
2 200 0
3 300 0
4 400 0
5 500 0
我做错了什么?我该如何解决?
当然你在这一列中得到0,你没有给参数任何值..
如果 ID 是连续的,并且 3 始终是 4 的前一个,以及 5 中的 4 等等......那么可以通过连接来完成:
SELECT t.id,t.value,t.value-coalese(s.value,0) as DIFF_to_Prev
FROM YourTable t
LEFT OUTER JOIN YourTable s ON(t.id = s.id + 1)
例如:
SELECT x.*
, COALESCE(x.value-@prev,0) diff_to_prev
, @prev:=value
FROM my_table x
, (SELECT @prev:=null) vars
ORDER
BY id;
SET @LastVALUE:= 0;
SELECT dtr.ID, dtr.VALUE,
0 - @LastVALUE + (@LastVALUE := dtr.VALUE) DIFF_to_Prev
FROM difftworows as dtr
ORDER BY dtr.ID
如果第一个 Diff 确实需要为 0:
SET @LastVALUE:= NULL;
SELECT dtr.ID, dtr.VALUE,
CASE
WHEN @LastVALUE IS NULL THEN 0 * (@LastVALUE := dtr.VALUE)
ELSE 0 - @LastVALUE + (@LastVALUE := dtr.VALUE)
END DIFF_to_Prev
FROM difftworows as dtr
ORDER BY dtr.ID
我将恢复它,因为现在有更好的方法来做到这一点 - 在 mysql 8+ 中有针对此类问题的窗口函数。对于这种情况,您可以使用滞后,如下所示: https://learnsql.com/blog/difference- Between-two-rows-in-sql/
对你来说是:
VALUE - LAG(VALUE) OVER (ORDER BY ID ) AS DIFF_to_Prev
不使用用户变量
SELECT sub0.ID,
sub0.VALUE,
sub0.VALUE - (COALESCE(d2a.VALUE, sub0.VALUE)) AS DIFF_to_Prev
FROM
(
SELECT d1.ID, d1.VALUE, MAX(d2.ID) AS d2_max_id
FROM difftworows d1
LEFT OUTER JOIN difftworows d2
ON d1.ID > d2.ID
GROUP BY d1.ID, d1.VALUE
) sub0
LEFT OUTER JOIN difftworows d2a
ON sub0.d2_max_id = d2a.ID
ORDER BY sub0.ID
编辑 - 避免子查询:-
SELECT d1.ID,
d1.VALUE,
d1.VALUE - IF(COUNT(d2.ID) = 0, d1.VALUE, SUBSTRING_INDEX(GROUP_CONCAT(d2.VALUE ORDER BY d2.ID DESC), ',', 1)) AS DIFF_to_Prev
FROM difftworows d1
LEFT OUTER JOIN difftworows d2
ON d1.ID > d2.ID
GROUP BY d1.ID,
d1.VALUE