Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
586 views
in Technique[技术] by (71.8m points)

mysql - Subtracting one row of data from another in SQL

I've been stumped with some SQL where I've got several rows of data, and I want to subtract a row from the previous row and have it repeat all the way down.

So here is the table:

CREATE TABLE foo (
  id,
  length
)
INSERT INTO foo (id,length) VALUES(1,1090)
INSERT INTO foo (id,length) VALUES(2,888)
INSERT INTO foo (id,length) VALUES(3,545)
INSERT INTO foo (id,length) VALUES(4,434)
INSERT INTO foo (id,length) VALUES(5,45)

I want the results to show a third column called difference which is one row subtracting from the one below with the final row subtracting from zero.

+------+------------------------+
| id   |length |  difference  |
+------+------------------------+
|    1 | 1090  |  202         |
|    2 |  888  |  343         |
|    3 |  545  |  111         |
|    4 |  434  |  389         |
|    5 |   45  |   45         |

I've tried a self join but I'm not exactly sure how to limit the results instead of having it cycle through itself. I can't depend that the id value will be sequential for a given result set so I'm not using that value. I could extend the schema to include some kind of sequential value.

This is what I've tried:

SELECT id, f.length, f2.length, (f.length - f2.length) AS difference
FROM foo f, foo f2

Thank you for the assist.

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

This might help you (somewhat).


select a.id, a.length, 
coalesce(a.length - 
    (select b.length from foo b where b.id = a.id + 1), a.length) as diff
from foo a


与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...