如何基于前一行更新MySQL行？

Question

I have a table like this:

ACTUAL TABLE

ID | Type  |   Number
----------+----------
1  |   x   |   2000
2  |   y   |   4500
3  |   y   |   4500
4  |   x   |   3000
5  |   y   |   5000
6  |   x   |   4000
7  |   y   |   7500
8  |   y   |   7500

And I want to update the y to be exact as the x before it:

EXPECTED TABLE

ID | Type  |   Number
----------+----------
1  |   x   |   2000
2  |   y   |   2000
3  |   y   |   2000
4  |   x   |   3000
5  |   y   |   3000
6  |   x   |   4000
7  |   y   |   4000
8  |   y   |   4000

How should I do that in MySQL?

Answer 1

Try Like This

t is tablename

a is type

va is number

update t join (
      select curr.id, max(prior.id) prior_id
        from t curr
        join t prior
          on prior.a = 'x'
             and prior.id < curr.id
       where curr.a = 'y'
    group by curr.id) g
          on g.id = t.id
        join t v on v.id = g.prior_id
   set t.va = v.va;

Sql Fiddle

Answer 2

只需根据您的ID进行更新： UPDATE <tablename> SET <fieldname> = (SELECT <fieldname> FROM <tablename> WHERE id = 1) WHERE id = 2这可能有用。

Answer 3

START TRANSACTION;

SELECT id, va into @max_id, @value_x FROM t 
WHERE a = 'x'
ORDER BY id DESC
LIMIT 1;


UPDATE t
set va = @value_x
where ID > @max_id;

COMMIT;

Try to get the value of x and id from the table first, into sql variables. Then do a simple update using those variabes.

Answer 4

UPDATE my_table a
  JOIN
     ( 

       SELECT y.*
            , MAX(x.number) max_number 
         FROM my_table y 
         JOIN my_table x
           ON x.type = 'x' 
          AND x.id <= y.id 
        GROUP 
           BY y.id

     ) b
    ON b.id = a.id
   SET a.number = b.max_number;