从mysql中的前一行减去值

Question

I have table with the following structure

ID  VALUE
1   100
2   200
3   300
4   400
5   500

I want output like this

ID  VALUE  DIFF_to_Prev
1   100     0
2   200    100 
3   300    100 
4   400    100
5   500    100

This is the query I've tried so far

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

This is the results I get from it:

ID      VALUE   DIFF_to_Prev
1       100     0
2       200     0
3       300     0
4       400     0
5       500     0

I want to know what I'm doing wrong. Please tell me how can I fix it with suggestions.

Thanks!!

Answer 1

Of course you a getting 0 in this column, you are not giving the parameter any value..

If the ID's are continuously and 3 will always be previous of 4 , and 4 of 5 and ETC... then it can be done with a join:

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)

Answer 2

Eg:

SELECT x.*
     , COALESCE(x.value-@prev,0) diff_to_prev
     , @prev:=value  
  FROM my_table x
     , (SELECT @prev:=null) vars 
 ORDER 
    BY id;

Answer 3

SET @LastVALUE:= 0;
SELECT dtr.ID, dtr.VALUE, 
    0 - @LastVALUE + (@LastVALUE := dtr.VALUE) DIFF_to_Prev     
FROM difftworows as dtr
ORDER BY dtr.ID

If first Diff really needs to be 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

Answer 4

Without using user variables

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

EDIT - avoiding the sub query:-

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

Answer 5

I'll revive it, as now there is way better way to do it - in mysql 8+ there are window functions for such a problems. For this situation you could use lag, like here: https://learnsql.com/blog/difference-between-two-rows-in-sql/

For you itd be:

VALUE - LAG(VALUE) OVER (ORDER BY ID ) AS DIFF_to_Prev