计算与MySQL中每个唯一ID的前一行的日期差

Question

I am a SQL beginner and am learning the ropes of querying. I'm trying to find the date difference between purchases by the same customer. I have a dataset that looks like this:

ID | Purchase_Date
==================
1  | 08/10/2017
------------------
1  | 08/11/2017
------------------
1  | 08/17/2017
------------------
2  | 08/09/2017
------------------
3  | 08/08/2017
------------------
3  | 08/10/2017

I want to have a column that shows the difference in days for each unique customer purchase, so that the output will look like this:

ID | Purchase_Date | Difference
===============================
1  | 08/10/2017    | NULL
-------------------------------
1  | 08/11/2017    | 1
-------------------------------
1  | 08/17/2017    | 6
-------------------------------
2  | 08/09/2017    | NULL
-------------------------------
3  | 08/08/2017    | NULL
-------------------------------
3  | 08/10/2017    | 2

What would be the best way to go about this using a MySQL query?

Answer 1

This is rather tricky in MySQL. Probably the best way to learn if you are a beginning is the correlated subquery method:

select t.*, datediff(purchase_date, prev_purchase_date) as diff
from (select t.*,
             (select t2.purchase_date
              from t t2
              where t2.id = t.id and
                    t2.purchase_date < t.purchase_date
              order by t2.purchase_date desc
              limit 1
             ) as prev_purchase_date
      from t
     ) t;

Performance should be okay if you have an index on (id, purchase_date) .

Answer 2

Not so hard, just use a subquery to find previous purchase for each existing purchase for the customer, and self-join to that record.

Select t.id, t.PurchaseDate, p.Purchase_date,
    DATEDIFF(t.PurchaseDate, p.Purchase_date) Difference
From myTable t          -- t for This purchase record
    left join myTable p -- p for Previous purchase record
       on p.id = t.Id 
         and p.purchase_date =
           (Select Max(purchase_date)
            from mytable
            where id = t.id
               and purchase_date < 
                   t.purchaseDate) 

Answer 3

It is possible to solve it not using dependent subquery

SELECT yt.id, create_date, NULLIF(yt.create_date - tm.min_create_date, 0)
FROM your_table yt
JOIN
(
  SELECT id, MIN(create_date) min_create_date
  FROM your_table
  GROUP BY id
) tm ON tm.id = yt.id

sqlfiddle demo