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All the Perl that's Practical to Extract and Report

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  • Too lazy to test it by would something like

    ORDER BY FIELD( id, 5,7,1,48)

    do the job?

  • This seems obvious?

        ORDER BY id=5, id=7, id=1, id=48

    --
    rjbs
    • Is that SQL standard? The solution I've gone with is actually much simpler.

      • I have never seen a SQL where that didn't work. I am also interested to see what could be much simpler than that. :)

        --
        rjbs
        • OK, a few people on Twitter have struck out on this and Abigail guessed something similar to what you and Adrian guessed.

          The solution is:

          ORDER BY length(path)

          Turns out it's a deterministic emergent property of materialized paths. There is one caveat I realized, though. If you have multiple trees and you want to see the root nodes, you can do this:

          SELECT id FROM tree WHERE id = path;

          However, some people omit the ID from the path. So you'd have this (in particular, note the path for the 5 node):

          node |

          • Seems problematic:

            1.23.45.67

            1.2.3.4.5

            Perhaps you could count dots.

            --
            rjbs
            • No, you can guarantee that conflicting paths won't exist as you walk up the tree.

            • _"Seems problematic:"_

              Nah - the problem is selecting all the parents of a child in order. The length of the path is guaranteed to increase.

            • The advantage of explicitly counting dots (rather than relying on string length to always increase with the number of dots) is that it makes path-lengths comparable. Not directly applicable to this problem, but not uncommon.
          • Wouldn't it be enough to just ORDER BY path?
  • Depending on the size of the tree, I've found that the nested sets representation is very nice for queries.

    The query that you need is given as an example in this page:

    http://dev.mysql.com/tech-resources/articles/hierarchical-data.html [mysql.com]

    (Although this is a mysql.com site, there is nothing MySQL specific with the solution)

    The problem with nested sets is updates: they might change a large number of records - worst case, inserting a new node at the left-most place will update all the rows.

    But smaller trees, they

    --
    life is short