I have the following table:

location eventdate value
loc1 2021-11-28 1
loc2 2021-11-28 2
loc1 2021-11-29 1
loc2 2021-11-29 3
loc1 2021-11-30 2
loc2 2021-11-30 5

And I'd like to compare the values from the oldest availavle day and the day before that to get this result:

location diff
loc1 1
loc2 2

What's the SQL statement to get this?

  • What is your MySQL version? Please tag your question accordingly. You may also want to consider following these suggestions.
    – mustaccio
    Commented Nov 29, 2021 at 20:43
  • Hi, and welcome to dba.se! Please go to dbfiddle.uk and create your table(s) and input your data. This reduces duplication of effort, provides a single source of truth and saves on duplication of effort for those who might wish to answer. Help us to help you!
    – Vérace
    Commented Nov 29, 2021 at 21:48
  • Is this only today and yesterday? Or any day and its previous day? What have you tried. Hint, use a self-join, join the table to itself with the second table aliased, and make the join criteria suit your question.
    – danblack
    Commented Nov 30, 2021 at 0:16
  • Use 2 copies of your table.
    – Akina
    Commented Nov 30, 2021 at 4:58
  • Here is a fiddle with a socultion joining the table to itself: dbfiddle.uk/… But I have to extend this because I don't know if there is data available for today when I run the select. So I have to compare the max. available the to the day before. Any hint for that?
    – jlai
    Commented Nov 30, 2021 at 6:43

1 Answer 1


As others have pointed out in the comments, you can JOIN your table to itself, also known as a self join. You can achieve this like so:

    CurrentDay.value - PreviousDay.value AS diff
FROM YourTable CurrentDay
INNER JOIN YourTable PreviousDay
    ON CurrentDay.location = PreviousDay.location
    AND CurrentDay.eventdate = DATE_SUB(PreviousDay.eventdate, INTERVAL 1 DAY);

This self join takes your table and joins it to itself on the same location but where the eventdate differs by 1 day.

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