SQL Order random rows based on 2 columns - sql

Question

How to sort this table in Oracle9:
START | END | VALUE
A | F | 1
D | H | 9
F | C | 8
C | D | 12
To make it look like this?:
START | END | VALUE
A | F | 1
F | C | 12
C | D | 8
D | H | 9
Goal is to start every next row with the end from the previous row.

Answer 1

This cannot be done with the order by clause alone, as it would have to find the record without a predecessor first, then find the next record comparing end and start column of the two records etc. This is an iterative process for which you need a recursive query.
That recursive query would find the first record, then the next and so on, giving them sequence numbers. Then you'd use the result and order by those generated numbers.
Here is how to do it in standard SQL. This is supported from Oracle 11g onwards only, however. In Oracle 9 you'll have to use CONNECT BY with which I am not familiar. Hopefully you or someone else can convert the query for you:
with chain(startkey, endkey, value, pos) as
(
select startkey, endkey, value, 1 as pos
from mytable
where not exists (select * from mytable prev where prev.endkey = mytable.startkey)
union all
select mytable.startkey, mytable.endkey, mytable.value, chain.pos + 1 as pos
from chain
join mytable on mytable.startkey = chain.endkey
)
select startkey, endkey, value
from chain
order by pos;
UPDATE: As you say the data is cyclic, you'd have to change above query so as to start with an arbitrarily chosen row and stop when through:
with chain(startkey, endkey, value, pos) as
(
select startkey, endkey, value, 1 as pos
from mytable
where rownum = 1
union all
select mytable.startkey, mytable.endkey, mytable.value, chain.pos + 1 as pos
from chain
join mytable on mytable.startkey = chain.endkey
)
cycle startkey set cycle to 1 default 0
select startkey, endkey, value
from chain
where cycle = 0
order by pos;