要在SQL中获取以前的值，请使用lag窗口函数，例如

SELECT 
  x, 
  lag(x) OVER (ORDER BY id) as prev_x, ...
FROM ...
ORDER BY id;

这样你就可以得到给定线段在三维空间中的上一个和下一个点，然后你就可以用regular geometric maths简单地计算线段的长度。
现在，您将获得每个段的长度（sqlfiddle查询），您可以使用SELECT ... FROM (SELECT ...)子查询或CTE（WITH ....）项将其作为其他查询的输入。
从节点段长度到节点到节点的长度是非常困难的，需要使用递归CTE或a window function创建一个包含空条目的表。
我最后得到了这个怪物：

SELECT
  array_agg(from_id) AS seg_ids,
  -- 'max' is used here like 'coalese' for an aggregate,
  -- since non-null is greater than null
  max(from_node) AS from_node,
  max(to_node) AS to_node,
  sum(seg_length) AS seg_length
FROM (
  -- lengths of all sub-segments with the null last segment
  -- removed and a partition counter added
  SELECT
   *,
   -- A running counter that increments when the
   -- node ID changes. Allows us to group by series
   -- of nodes in the outer query.
   sum(CASE WHEN from_node IS NULL THEN 0 ELSE 1 END) OVER (ORDER BY from_id) AS partition_id
  FROM
  (
    -- lengths of all sub-segments
    SELECT
      id AS from_id,
      lead(id, 1) OVER (ORDER BY id) AS to_id,
      -- length of sub-segment
      sqrt(
        (x - lead(x, 1) OVER (ORDER BY id)) ^ 2 +
        (y - lead(y, 1) OVER (ORDER BY id)) ^ 2 +
        (z - lead(z, 1) OVER (ORDER BY id)) ^ 2
      ) AS seg_length,
      node AS from_node,
      lead(node, 1) OVER (ORDER BY id) AS to_node
    FROM
      Table1
  ) sub
  -- filter out the last row
  WHERE to_id IS NOT NULL
) seglengths
-- Group into series of sub-segments between two nodes
GROUP BY partition_id;

分区技巧归功于How do I efficiently select the previous non-null value?。
结果：

seg_ids | to_node | from_node | seg_length 
---------+---------+---------+------------
 {1,2,3} |     100 |     200 |          3
 {4,5,6} |     200 |     300 |          3
(2 rows)

若要直接插入另一个数据表，请使用INSERT INTO ... SELECT ...。