mySql - updating by comparing rows in same table - mysql

Question

I want to update a column by comparing each row to all other rows in the table but I cant figure out how to distinguish the column names in the row being updated with the rows being searched through.
Here's a simplified example...
people:
+--------+-----+----------------+
| name | age | nameClosestAge |
+--------+-----+----------------+
| alice | 20 | |
| bob | 30 | |
| clive | 22 | |
| duncan | 24 | |
+--------+-----+----------------+
To fill in the 'nameClosestAge' column with the name of the person that is closest in age to each person, you could do this...
create temporary table peopleTemp like people;
insert into peopleTemp select * from people;
update people set nameClosestAge =
(select name from peopleTemp where people.name != peopleTemp.name
order by abs(people.age - peopleTemp.age) asc limit 1);
Which produces this....
+--------+-----+----------------+
| name | age | nameClosestAge |
+--------+-----+----------------+
| alice | 20 | clive |
| bob | 30 | duncan |
| clive | 22 | alice |
| duncan | 25 | clive |
+--------+-----+----------------+
Surely there is a way to do this without creating a duplicate table.
I'm looking for the most efficient method here as I have a very large table and its taking too long to update.
I'm using mySql with PHP.

Answer 1

You could perform this with just one sub-query and no temp table.
SELECT name, age, (
SELECT name
FROM people
WHERE name != ppl.name
ORDER BY ABS( people.age - ppl.age )
LIMIT 1
) AS nameClosestAge
FROM people AS ppl;
Checked and works :)
EDIT: If you want to be able to work with the calc'ed row, you can use a view;
CREATE VIEW people_close AS
SELECT name, age, (
SELECT name
FROM people
WHERE name != ppl.name
ORDER BY ABS( people.age - ppl.age )
LIMIT 1
) AS nameClosestAge
FROM people AS ppl;
You can't update the calculated field but can query against it easily.