Sort of hierarchy elements

It is possible to have a descending order applied to just one element of a hierarchy?
How?
e.g.. Suppose my hierarchy is
year (2000,2001)
month (1,2,3)
Now, if I expand year and month nodes I have this:
2000
--1
--2
--3
2001
--1
--2
--3
I would rather have this:
2000
--3
--2
--1
2001
--3
--2
--1
Thanks
Luc

914898 wrote:
It is possible to have a descending order applied to just one element of a hierarchy?
How?
e.g.. Suppose my hierarchy is
year (2000,2001)
month (1,2,3)
Now, if I expand year and month nodes I have this:
2000
--1
--2
--3
2001
--1
--2
--3
I would rather have this:
2000
--3
--2
--1
2001
--3
--2
--1
Thanks
LucIf by hierarchy you mean you are producing this output with a CONNECT BY statement then yes, you can use an order by SIBLINGS to achieve this.
http://docs.oracle.com/cd/E11882_01/server.112/e26088/queries003.htm#SQLRF52335
Cheers,

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graph> init
graph> mst
[(4 5) = 51
, (12 14) = 51
, (7 9) = 70
, (10 11) = 60
, (8 9) = 75
, (7 8) = 100
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, (9 10) = 95
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, (12 13) = 100
, (3 7) = 300
, (13 15) = 200
, (13 16) = 200
, (14 15) = 170
, (10 4) = 400
, (14 16) = 251
graph>
class PQEntry implements Comparable {
     int node1 , node2 , distance;
     public PQEntry(int node1 , int node2 , int distance) {
          this.node1 = node1;
          this.node2 = node2;
          this.distance = distance;
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          return distance;
     public String toString() {
          String output = "(" + node1 + " " + node2 + ")" + " = " + distance + "\n" ;
          return output;
     public int compareTo(Object o) {
          if(o instanceof PQEntry){
               if (getPriority() > ((PQEntry) o).getPriority()) {
                    return 1;
               } else if(getPriority() == ((PQEntry) o).getPriority()) {
                    return 0;
               } else {
                    return -1;
          } else {
               throw new IllegalArgumentException("o must be an instance of PQEntry");
import java.io.*;
import java.util.*;
class Main {
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     myGraph.addEdge(3, 7, 300);
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     myGraph.addEdge(5, 6, 151);
     myGraph.addEdge(7, 8, 100);
     myGraph.addEdge(7, 9, 70);
     myGraph.addEdge(8, 9, 75);
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                    System.exit(0);
               } else if(input[0].equalsIgnoreCase("mst") && input.length == 1) {
                    int[][] tempMatrix = myGraph.getMatrix();
                    PriorityQueue<PQEntry> edges = new PriorityQueue<PQEntry>(51);
                    List<PQEntry> sortedEdges = new LinkedList<PQEntry>();
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                         for(int j = 1; j < 51; j++){
                              if (tempMatrix[i][j] > 0) {
                                        edges.add(new PQEntry(i, j, tempMatrix[i][j]));
                         sortedEdges.addAll(edges);
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                    System.out.println(myGraph.toString());
               } else if(input[0].equalsIgnoreCase("init") && input.length == 1) {
                    initGraph();
               } else {
                    System.out.println("unknown Command, enter help for lsit of valid commands");
          } while(true);
import java.util.*;
class GraphImp implements Graph {
     //contains a two-dimensional array to hold the start vertex, end vertex
     //and the distance between the two verteces.
     private int [][] matrix;
          //constructor, sets the bounds of the two-dimensional array
          public GraphImp(int size) {
               matrix = new int[size][size];
          //returns the two-dimensional array to the method calling it
          public int[][] getMatrix() {
               return matrix;
          //this method first checks whether the two nodes are within the
          //acceptable range and gives an erro if not. It then checks whether an edge exists between the nodes
          //if an edge already exists it gives the user an error message, otherwise adds the edge
          //to the two-dimensional array at the appropriate position.
          public void addEdge(int node1 , int node2 , int distance) {
               if(node1 > 0 || node1 < 51 || node2 > 0 || node2 < 51) {
                    if(matrix[node1][node2] == 0 && matrix[node2][node1] == 0) {
                         matrix[node1][node2] = distance;
                    } else {
                         System.out.println("Error! An edge already exists between the two nodes");
               } else {
                    System.out.println("Error! Atleast one of the nodes are out of range, please enter within the range 1 - 51");
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               if(node1 > 0 || node2 > 0 || node1 < 51 || node2 < 51) {
                    if(matrix[node1][node2] != 0 || matrix[node2][node1] != 0) {
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                         matrix[node2][node1] = 0;
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                    for(int i = 0; i < matrix[0].length; i++){
                         if (matrix[j] > 0){
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msjamalan wrote:
Hi, here is my semi-complete program, I have removed the methods that were irrelevant to make things simple. The program is supposed to represent Graphs, and minimal spanning trees. I first have to put edges in a priority queue of type : PQEntry (Below), which are to be sorted by their 'distance'. I have not included the interfaces as they weren't necessary here (all methods are already implemented). There is no compile error, but for example when I run the program, and enter 'mst' (it shoud print the contents of the Priority Queue in the order of their distances), it puts the first few elements in order but fails to do so on the remaining elements. below is a sample run.AFAIK, a sorted priority queue sorts the output returned by the queue's poll method. So if you iterate through the entire queue via the poll() method, what is spit out should be sorted. If you just print the queue itself via println, you wan't see this sorting.
Also, when posting code here, please use code tags so that your code will retain its formatting and thus will be readable -- after all, your goal is to get as many people to read your post and understand your code as possible, right?
To do this, highlight your pasted code (please be sure that it is already formatted when you paste it into the forum; the code tags don't magically format unformatted code) and then press the code button, and your code will have tags.
Another way to do this is to manually place the tags into your code by placing the tag [code] above your pasted code and the tag [code] below your pasted code like so:
[code]
// your code goes here
// notice how the top and bottom tags are different
[/code]Much luck!

Sorting a hierarchy ...

Hi Everyone,
I have a table which is structured as a hierarchy thus ...
id parent
top -/-
GB top
London GB
Bristol GB
Manchester GB
DE top
Berlin DE
Munich DE
FR top
Paris FR
Lyon FR
... etc.
... which I need to output in the same format.
However the order of the items is completely random (in the table) so Berlin does not appear directly below DE (for example) ...
ES top
Moscow RU
Madrid ES
Edinbourgh GB
... etc.
So what I need is a sorting syntax which will retreive the table's contents as I have presented above (in the first extract) with the hierarchy displayed in its natural groups (i.e. DE followed by Berlin, Munich, etc.).
I have played around the the "PARTITION" command but cannot seem to get it working so am hoping that someone out there can point me towards the correct syntax?
Many thanks,
Alan Searle

Hi, Alan,
Use CONNECT BY:
SELECT     id
,     parent
FROM     table_x
START WITH     parent     IS NULL
CONNECT BY     parent     = PRIOR id
ORDER SIBLINGS BY     id          -- Optional
;ORDER SIBLINGS would guarantee that DE and FR (with all their cities) came before GB, and that, within each country, the cities appeared in alphabetic order, like this:
id          parent
top
DE           top
Berlin           DE
Munich           DE
FR           top
Lyon           FR
Paris           FR
GB           top
Bristol      GB
London           GB
Manchester      GB
...For an introduction to CONNECT BY queries, see
http://philip.greenspun.com/sql/trees
For several examples of CONNECT BY queries, see
http://www.java2s.com/Tutorial/Oracle/0040__Query-Select/HierarchicalQueries.htm
Naturally, CONNECT BY is documented in the SQL language manual:
http://docs.oracle.com/cd/B28359_01/server.111/b28286/queries003.htm#sthref3259
Edited by: Frank Kulash on Jan 2, 2012 5:59 AM

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Hi
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-D
Message Edited by Darren on 03-16-200610:00 AM
Darren Nattinger, CLA
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Attachments:
Sorted_cluster.jpg ‏30 KB

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