Honestly I have no clue what you trying to say, can you please be divulge more details and especially the relevant code / SQL query you are dealing with.
Also on what basis are you reorganizing your records every week, you have just shown us how it should appear not why or the Logic behind it.
Also please keep in mind to read essays like these, they will be very helpful in teaching you how to construct your questions so that more people will actually pay attention to you.

I kind of think he's trying to make us write those.

Table-> cat_users

id cat_id user_id
81 7 12
160 7 22
193 7 26
213 7 29
223 7 30

Table-> categories

id parent_id cat
7 NULL Stive
11 0 Akshay
15 NULL Shooting
24 NULL Fighting
52 NULL Puzzle

Table-> Users
id valid name date_added
1 1 Holly NULL
3 1 Tom 2007-11-01
7 1 Stive 2007-11-02
10 1 Elite 2007-11-02
11 1 Akshay 2007-11-02


Table-> ratings

id user_id rating date_added
1 7 4 2008-07-18
2 7 3 2008-07-18
3 3 4 2008-07-18
4 5 3 2008-07-18
5 10 2 2008-07-18
6 6 4 2008-07-18
7 1922 5 2008-07-18
15 1322 3 2008-08-02
16 2055 3 2008-08-07
81 74 4 2008-08-25
160 169 5 2008-08-25


$sql1 = "select users.*, round(avg(rating)) as rating
, date_format(date_added,'%Y-%m-%d') as `date`
from users
left join ratings on (ratings.user_id = users.id)
inner join cat_users on cat_users.user_id = users.id
where
valid = 1
and users.id in (
select user_id
from cat_users
where cat_id = $cat_id
)
and users.date_added <= now()
group by users.id
order by date_added desc, rating desc, name
";
$max = GetSQL("select count(*) from ($sql1) counter");


I am not giving all database it's very large
so got output like ....

id user_id rating date_created
123 1 4 2008-12-09
211 1 3 2008-12-08
33 1 4 2008-11-30
45 1 3 2008-10-31
57 1 2 2008-10-31
666 1 4 2008-08-25
723 1 5 2008-08-25
15 1 3 2008-08-25
16 1 3 2008-08-25
81 1 4 2008-08-25
160 1 5 2008-08-25

after that i want to display...after 1 week
like...

id user_id rating date_created
160 1 5 2008-08-25 .....change
123 1 4 2008-12-09 last data take 1st position
211 1 3 2008-12-08
33 1 4 2008-11-30
45 1 3 2008-10-31
57 1 2 2008-10-31
666 1 4 2008-08-25
723 1 5 2008-08-25
15 1 3 2008-08-25
16 1 3 2008-08-25
81 1 4 2008-08-25

again after 1 week ...
id user_id rating date_created
81 1 4 2008-08-25 .....change
160 1 5 2008-08-25 last data take 1st position
123 1 4 2008-12-09
211 1 3 2008-12-08
33 1 4 2008-11-30
45 1 3 2008-10-31
57 1 2 2008-10-31
666 1 4 2008-08-25
723 1 5 2008-08-25
15 1 3 2008-08-25
16 1 3 2008-08-25


How to do that??
can i add column like postion_id so that change position of data??
Reason behind change position is old user viewed there profile by another users.

if you want to sort your output based on the data you already have, there is no need to add another column.
read it into an array of objects and sort those before printing

An "order by" clause ?

Your question has nothing to do with any programming language, what-so-ever. Find an SQL forum.

I edit my table add one column

id user_id rating date_created change_id
123 1 4 2008-12-09
211 1 3 2008-12-08
33 1 4 2008-11-30
45 1 3 2008-10-31
57 1 2 2008-10-31
666 1 4 2008-08-25

which query to write....so that i got ..

id user_id rating date_created change_id
123 1 4 2008-12-09 1
211 1 3 2008-12-08 2
33 1 4 2008-11-30 3
45 1 3 2008-10-31 4
57 1 2 2008-10-31 5
666 1 4 2008-08-25 6

add data in change_id column with increment value...
how to put data for change_id with increment value in column??

you might want to read this answer first.
obviously, masjiade did more effort into reading your code than I did

i am edit table.....add column change_id
And i want to write data...means update table.....

original table
id user_id rating date_created
123 1 4 2008-12-09
211 1 3 2008-12-08
33 1 4 2008-11-30
45 1 3 2008-10-31
57 1 2 2008-10-31
666 1 4 2008-08-25

after update
id user_id rating date_created change_id
123 1 4 2008-12-09 1
211 1 3 2008-12-08 2
33 1 4 2008-11-30 3
45 1 3 2008-10-31 4
57 1 2 2008-10-31 5
666 1 4 2008-08-25 6

plz look into this table..just add number in this column in increment form

What about it? This is not an SQL forum, and your's is an SQL question.

As I have already said
Find an SQL Forum.