PostgreSQL - retrieve all IDs in a tree for a given subnode












5















I have a non-binary tree of customer, and I need to obtain all the IDs in a tree for the given node.



The table is very simple, just an join table with a parent id and child id.
This is a representation of the tree I stored in my db.



Enter image description here



In this example if I search for node 17 I need in return 14-17. If I search for 11 I need in return 1-6-5-4-8-11-12-7-2-10-3.



The order is not important. I only need the ID to avoid circularity when adding children to a node.



I created this query.
The ancestor part works fine, I retrieve all parent nodes, but for the descendants I have some trouble. I'm only able to retrieve some part of the tree.
For example, with node 11 I retrieve 4-10-6-11-7-8, so all right part of the tree is missing.



WITH RECURSIVE
-- starting node(s)
starting (parent, child) AS
(
SELECT t.parent, t.child
FROM public.customerincustomer AS t
WHERE t.child = :node or t.parent = :node
)
,
ancestors (parent, child) AS
(
SELECT t.parent, t.child
FROM public.customerincustomer AS t
WHERE t.parent IN (SELECT parent FROM starting)
UNION ALL
SELECT t.parent, t.child
FROM public.customerincustomer AS t JOIN ancestors AS a ON t.child = a.parent
),
descendants (parent, child) AS
(
SELECT t.parent, t.child
FROM public.customerincustomer AS t
WHERE t.parent IN (SELECT parent FROM starting) or t.child in (select child from starting)
UNION ALL
SELECT t.parent, t.child
FROM public.customerincustomer AS t JOIN ancestors AS a ON t.parent = a.child
)

table ancestors
union all
table descendants


UPDATE



I see that many examples included in the tree table also the root in form (root_id, null).



In my case I don't have this record.



For example, taking the smallest tree 14->17, in my table I have only one record
parent, child



14 17










share|improve this question

























  • Depending on your workload and use case you may also want to consider ltree

    – Evan Carroll
    Dec 13 '18 at 15:05


















5















I have a non-binary tree of customer, and I need to obtain all the IDs in a tree for the given node.



The table is very simple, just an join table with a parent id and child id.
This is a representation of the tree I stored in my db.



Enter image description here



In this example if I search for node 17 I need in return 14-17. If I search for 11 I need in return 1-6-5-4-8-11-12-7-2-10-3.



The order is not important. I only need the ID to avoid circularity when adding children to a node.



I created this query.
The ancestor part works fine, I retrieve all parent nodes, but for the descendants I have some trouble. I'm only able to retrieve some part of the tree.
For example, with node 11 I retrieve 4-10-6-11-7-8, so all right part of the tree is missing.



WITH RECURSIVE
-- starting node(s)
starting (parent, child) AS
(
SELECT t.parent, t.child
FROM public.customerincustomer AS t
WHERE t.child = :node or t.parent = :node
)
,
ancestors (parent, child) AS
(
SELECT t.parent, t.child
FROM public.customerincustomer AS t
WHERE t.parent IN (SELECT parent FROM starting)
UNION ALL
SELECT t.parent, t.child
FROM public.customerincustomer AS t JOIN ancestors AS a ON t.child = a.parent
),
descendants (parent, child) AS
(
SELECT t.parent, t.child
FROM public.customerincustomer AS t
WHERE t.parent IN (SELECT parent FROM starting) or t.child in (select child from starting)
UNION ALL
SELECT t.parent, t.child
FROM public.customerincustomer AS t JOIN ancestors AS a ON t.parent = a.child
)

table ancestors
union all
table descendants


UPDATE



I see that many examples included in the tree table also the root in form (root_id, null).



In my case I don't have this record.



For example, taking the smallest tree 14->17, in my table I have only one record
parent, child



14 17










share|improve this question

























  • Depending on your workload and use case you may also want to consider ltree

    – Evan Carroll
    Dec 13 '18 at 15:05
















5












5








5


0






I have a non-binary tree of customer, and I need to obtain all the IDs in a tree for the given node.



The table is very simple, just an join table with a parent id and child id.
This is a representation of the tree I stored in my db.



Enter image description here



In this example if I search for node 17 I need in return 14-17. If I search for 11 I need in return 1-6-5-4-8-11-12-7-2-10-3.



The order is not important. I only need the ID to avoid circularity when adding children to a node.



I created this query.
The ancestor part works fine, I retrieve all parent nodes, but for the descendants I have some trouble. I'm only able to retrieve some part of the tree.
For example, with node 11 I retrieve 4-10-6-11-7-8, so all right part of the tree is missing.



WITH RECURSIVE
-- starting node(s)
starting (parent, child) AS
(
SELECT t.parent, t.child
FROM public.customerincustomer AS t
WHERE t.child = :node or t.parent = :node
)
,
ancestors (parent, child) AS
(
SELECT t.parent, t.child
FROM public.customerincustomer AS t
WHERE t.parent IN (SELECT parent FROM starting)
UNION ALL
SELECT t.parent, t.child
FROM public.customerincustomer AS t JOIN ancestors AS a ON t.child = a.parent
),
descendants (parent, child) AS
(
SELECT t.parent, t.child
FROM public.customerincustomer AS t
WHERE t.parent IN (SELECT parent FROM starting) or t.child in (select child from starting)
UNION ALL
SELECT t.parent, t.child
FROM public.customerincustomer AS t JOIN ancestors AS a ON t.parent = a.child
)

table ancestors
union all
table descendants


UPDATE



I see that many examples included in the tree table also the root in form (root_id, null).



In my case I don't have this record.



For example, taking the smallest tree 14->17, in my table I have only one record
parent, child



14 17










share|improve this question
















I have a non-binary tree of customer, and I need to obtain all the IDs in a tree for the given node.



The table is very simple, just an join table with a parent id and child id.
This is a representation of the tree I stored in my db.



Enter image description here



In this example if I search for node 17 I need in return 14-17. If I search for 11 I need in return 1-6-5-4-8-11-12-7-2-10-3.



The order is not important. I only need the ID to avoid circularity when adding children to a node.



I created this query.
The ancestor part works fine, I retrieve all parent nodes, but for the descendants I have some trouble. I'm only able to retrieve some part of the tree.
For example, with node 11 I retrieve 4-10-6-11-7-8, so all right part of the tree is missing.



WITH RECURSIVE
-- starting node(s)
starting (parent, child) AS
(
SELECT t.parent, t.child
FROM public.customerincustomer AS t
WHERE t.child = :node or t.parent = :node
)
,
ancestors (parent, child) AS
(
SELECT t.parent, t.child
FROM public.customerincustomer AS t
WHERE t.parent IN (SELECT parent FROM starting)
UNION ALL
SELECT t.parent, t.child
FROM public.customerincustomer AS t JOIN ancestors AS a ON t.child = a.parent
),
descendants (parent, child) AS
(
SELECT t.parent, t.child
FROM public.customerincustomer AS t
WHERE t.parent IN (SELECT parent FROM starting) or t.child in (select child from starting)
UNION ALL
SELECT t.parent, t.child
FROM public.customerincustomer AS t JOIN ancestors AS a ON t.parent = a.child
)

table ancestors
union all
table descendants


UPDATE



I see that many examples included in the tree table also the root in form (root_id, null).



In my case I don't have this record.



For example, taking the smallest tree 14->17, in my table I have only one record
parent, child



14 17







postgresql recursive tree






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Dec 30 '18 at 1:53









Peter Mortensen

23319




23319










asked Dec 13 '18 at 11:12









Luca NittiLuca Nitti

284




284













  • Depending on your workload and use case you may also want to consider ltree

    – Evan Carroll
    Dec 13 '18 at 15:05





















  • Depending on your workload and use case you may also want to consider ltree

    – Evan Carroll
    Dec 13 '18 at 15:05



















Depending on your workload and use case you may also want to consider ltree

– Evan Carroll
Dec 13 '18 at 15:05







Depending on your workload and use case you may also want to consider ltree

– Evan Carroll
Dec 13 '18 at 15:05












2 Answers
2






active

oldest

votes


















5














A very primitive implementation:



It basically divides the problem into two subproblems:




  • First find all the ancestors of the node in question (including the node itself). If the node has no parents, then this would be just itself.

  • Then find the descendants of all those ancestors (including themselves). We may have several nodes in the ancestors result set, we may get duplicates here, so we use UNION (and not UNION ALL) to remove them.

  • Note that the query will work even if the input node is a root with has no children.

  • It will also work if the data set is not a forest of trees but an arbitrary directed graph (where nodes can have more than one parent).


The query:



WITH RECURSIVE 
ancestors (parent) AS
(
SELECT :node -- start with the given node
UNION ALL
SELECT t.parent -- and find all its ancestors
FROM public.customerincustomer AS t JOIN ancestors AS a ON t.child = a.parent
),
descendants (customer) AS
(
SELECT parent AS customer -- now start with all the ancestors
FROM ancestors
UNION
SELECT t.child -- and find all their descendants
FROM public.customerincustomer AS t JOIN descendants AS d ON t.parent = d.customer
)
SELECT customer
FROM descendants ;





share|improve this answer

































    4














    This function returns the parent level of node_id:



    There is a 'level' row due there isn't a row (id, null) for parent row.



    CREATE FUNCTION get_parent(node_id int)
    RETURNS integer AS
    $$
    WITH RECURSIVE get_parent AS
    (
    SELECT
    t1.id,
    t1.parent_id,
    t1.name,
    0 AS level
    FROM
    tree t1
    WHERE
    t1.id = node_id
    UNION ALL
    SELECT
    t2.id,
    t2.parent_id,
    t2.name,
    level+1
    FROM
    tree t2
    INNER JOIN
    get_parent ON get_parent.parent_id = t2.id
    )
    SELECT
    id
    FROM
    get_parent
    ORDER BY
    level DESC
    LIMIT 1 ;
    $$
    LANGUAGE SQL;



    select get_parent(7);




    | get_parent |
    | ---------: |
    | 6 |


    Now, next query returns the whole tree structure based on a parent node.





    WITH RECURSIVE childs AS
    (
    SELECT
    t1.id,
    t1.parent_id,
    t1.name
    FROM
    tree t1
    WHERE
    t1.id = get_parent(7)
    UNION ALL
    SELECT
    t2.id,
    t2.parent_id,
    t2.name
    FROM
    tree t2
    INNER JOIN
    childs ON childs.id = t2.parent_id
    )
    SELECT
    id,
    parent_id,
    name
    FROM
    childs;



    id | parent_id | name
    -: | --------: | :------
    6 | 1 | Node 6
    4 | 6 | Node 4
    8 | 6 | Node 8
    11 | 6 | Node 11
    7 | 11 | Node 7
    10 | 7 | Node 10


    db<>fiddle here






    share|improve this answer


























    • as i said, i don't have problem with ancestors. I need the whole tree.

      – Luca Nitti
      Dec 13 '18 at 12:04













    Your Answer








    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "182"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: false,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: null,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fdba.stackexchange.com%2fquestions%2f224871%2fpostgresql-retrieve-all-ids-in-a-tree-for-a-given-subnode%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    5














    A very primitive implementation:



    It basically divides the problem into two subproblems:




    • First find all the ancestors of the node in question (including the node itself). If the node has no parents, then this would be just itself.

    • Then find the descendants of all those ancestors (including themselves). We may have several nodes in the ancestors result set, we may get duplicates here, so we use UNION (and not UNION ALL) to remove them.

    • Note that the query will work even if the input node is a root with has no children.

    • It will also work if the data set is not a forest of trees but an arbitrary directed graph (where nodes can have more than one parent).


    The query:



    WITH RECURSIVE 
    ancestors (parent) AS
    (
    SELECT :node -- start with the given node
    UNION ALL
    SELECT t.parent -- and find all its ancestors
    FROM public.customerincustomer AS t JOIN ancestors AS a ON t.child = a.parent
    ),
    descendants (customer) AS
    (
    SELECT parent AS customer -- now start with all the ancestors
    FROM ancestors
    UNION
    SELECT t.child -- and find all their descendants
    FROM public.customerincustomer AS t JOIN descendants AS d ON t.parent = d.customer
    )
    SELECT customer
    FROM descendants ;





    share|improve this answer






























      5














      A very primitive implementation:



      It basically divides the problem into two subproblems:




      • First find all the ancestors of the node in question (including the node itself). If the node has no parents, then this would be just itself.

      • Then find the descendants of all those ancestors (including themselves). We may have several nodes in the ancestors result set, we may get duplicates here, so we use UNION (and not UNION ALL) to remove them.

      • Note that the query will work even if the input node is a root with has no children.

      • It will also work if the data set is not a forest of trees but an arbitrary directed graph (where nodes can have more than one parent).


      The query:



      WITH RECURSIVE 
      ancestors (parent) AS
      (
      SELECT :node -- start with the given node
      UNION ALL
      SELECT t.parent -- and find all its ancestors
      FROM public.customerincustomer AS t JOIN ancestors AS a ON t.child = a.parent
      ),
      descendants (customer) AS
      (
      SELECT parent AS customer -- now start with all the ancestors
      FROM ancestors
      UNION
      SELECT t.child -- and find all their descendants
      FROM public.customerincustomer AS t JOIN descendants AS d ON t.parent = d.customer
      )
      SELECT customer
      FROM descendants ;





      share|improve this answer




























        5












        5








        5







        A very primitive implementation:



        It basically divides the problem into two subproblems:




        • First find all the ancestors of the node in question (including the node itself). If the node has no parents, then this would be just itself.

        • Then find the descendants of all those ancestors (including themselves). We may have several nodes in the ancestors result set, we may get duplicates here, so we use UNION (and not UNION ALL) to remove them.

        • Note that the query will work even if the input node is a root with has no children.

        • It will also work if the data set is not a forest of trees but an arbitrary directed graph (where nodes can have more than one parent).


        The query:



        WITH RECURSIVE 
        ancestors (parent) AS
        (
        SELECT :node -- start with the given node
        UNION ALL
        SELECT t.parent -- and find all its ancestors
        FROM public.customerincustomer AS t JOIN ancestors AS a ON t.child = a.parent
        ),
        descendants (customer) AS
        (
        SELECT parent AS customer -- now start with all the ancestors
        FROM ancestors
        UNION
        SELECT t.child -- and find all their descendants
        FROM public.customerincustomer AS t JOIN descendants AS d ON t.parent = d.customer
        )
        SELECT customer
        FROM descendants ;





        share|improve this answer















        A very primitive implementation:



        It basically divides the problem into two subproblems:




        • First find all the ancestors of the node in question (including the node itself). If the node has no parents, then this would be just itself.

        • Then find the descendants of all those ancestors (including themselves). We may have several nodes in the ancestors result set, we may get duplicates here, so we use UNION (and not UNION ALL) to remove them.

        • Note that the query will work even if the input node is a root with has no children.

        • It will also work if the data set is not a forest of trees but an arbitrary directed graph (where nodes can have more than one parent).


        The query:



        WITH RECURSIVE 
        ancestors (parent) AS
        (
        SELECT :node -- start with the given node
        UNION ALL
        SELECT t.parent -- and find all its ancestors
        FROM public.customerincustomer AS t JOIN ancestors AS a ON t.child = a.parent
        ),
        descendants (customer) AS
        (
        SELECT parent AS customer -- now start with all the ancestors
        FROM ancestors
        UNION
        SELECT t.child -- and find all their descendants
        FROM public.customerincustomer AS t JOIN descendants AS d ON t.parent = d.customer
        )
        SELECT customer
        FROM descendants ;






        share|improve this answer














        share|improve this answer



        share|improve this answer








        edited Dec 13 '18 at 15:00

























        answered Dec 13 '18 at 14:11









        ypercubeᵀᴹypercubeᵀᴹ

        77.1k11134215




        77.1k11134215

























            4














            This function returns the parent level of node_id:



            There is a 'level' row due there isn't a row (id, null) for parent row.



            CREATE FUNCTION get_parent(node_id int)
            RETURNS integer AS
            $$
            WITH RECURSIVE get_parent AS
            (
            SELECT
            t1.id,
            t1.parent_id,
            t1.name,
            0 AS level
            FROM
            tree t1
            WHERE
            t1.id = node_id
            UNION ALL
            SELECT
            t2.id,
            t2.parent_id,
            t2.name,
            level+1
            FROM
            tree t2
            INNER JOIN
            get_parent ON get_parent.parent_id = t2.id
            )
            SELECT
            id
            FROM
            get_parent
            ORDER BY
            level DESC
            LIMIT 1 ;
            $$
            LANGUAGE SQL;



            select get_parent(7);




            | get_parent |
            | ---------: |
            | 6 |


            Now, next query returns the whole tree structure based on a parent node.





            WITH RECURSIVE childs AS
            (
            SELECT
            t1.id,
            t1.parent_id,
            t1.name
            FROM
            tree t1
            WHERE
            t1.id = get_parent(7)
            UNION ALL
            SELECT
            t2.id,
            t2.parent_id,
            t2.name
            FROM
            tree t2
            INNER JOIN
            childs ON childs.id = t2.parent_id
            )
            SELECT
            id,
            parent_id,
            name
            FROM
            childs;



            id | parent_id | name
            -: | --------: | :------
            6 | 1 | Node 6
            4 | 6 | Node 4
            8 | 6 | Node 8
            11 | 6 | Node 11
            7 | 11 | Node 7
            10 | 7 | Node 10


            db<>fiddle here






            share|improve this answer


























            • as i said, i don't have problem with ancestors. I need the whole tree.

              – Luca Nitti
              Dec 13 '18 at 12:04


















            4














            This function returns the parent level of node_id:



            There is a 'level' row due there isn't a row (id, null) for parent row.



            CREATE FUNCTION get_parent(node_id int)
            RETURNS integer AS
            $$
            WITH RECURSIVE get_parent AS
            (
            SELECT
            t1.id,
            t1.parent_id,
            t1.name,
            0 AS level
            FROM
            tree t1
            WHERE
            t1.id = node_id
            UNION ALL
            SELECT
            t2.id,
            t2.parent_id,
            t2.name,
            level+1
            FROM
            tree t2
            INNER JOIN
            get_parent ON get_parent.parent_id = t2.id
            )
            SELECT
            id
            FROM
            get_parent
            ORDER BY
            level DESC
            LIMIT 1 ;
            $$
            LANGUAGE SQL;



            select get_parent(7);




            | get_parent |
            | ---------: |
            | 6 |


            Now, next query returns the whole tree structure based on a parent node.





            WITH RECURSIVE childs AS
            (
            SELECT
            t1.id,
            t1.parent_id,
            t1.name
            FROM
            tree t1
            WHERE
            t1.id = get_parent(7)
            UNION ALL
            SELECT
            t2.id,
            t2.parent_id,
            t2.name
            FROM
            tree t2
            INNER JOIN
            childs ON childs.id = t2.parent_id
            )
            SELECT
            id,
            parent_id,
            name
            FROM
            childs;



            id | parent_id | name
            -: | --------: | :------
            6 | 1 | Node 6
            4 | 6 | Node 4
            8 | 6 | Node 8
            11 | 6 | Node 11
            7 | 11 | Node 7
            10 | 7 | Node 10


            db<>fiddle here






            share|improve this answer


























            • as i said, i don't have problem with ancestors. I need the whole tree.

              – Luca Nitti
              Dec 13 '18 at 12:04
















            4












            4








            4







            This function returns the parent level of node_id:



            There is a 'level' row due there isn't a row (id, null) for parent row.



            CREATE FUNCTION get_parent(node_id int)
            RETURNS integer AS
            $$
            WITH RECURSIVE get_parent AS
            (
            SELECT
            t1.id,
            t1.parent_id,
            t1.name,
            0 AS level
            FROM
            tree t1
            WHERE
            t1.id = node_id
            UNION ALL
            SELECT
            t2.id,
            t2.parent_id,
            t2.name,
            level+1
            FROM
            tree t2
            INNER JOIN
            get_parent ON get_parent.parent_id = t2.id
            )
            SELECT
            id
            FROM
            get_parent
            ORDER BY
            level DESC
            LIMIT 1 ;
            $$
            LANGUAGE SQL;



            select get_parent(7);




            | get_parent |
            | ---------: |
            | 6 |


            Now, next query returns the whole tree structure based on a parent node.





            WITH RECURSIVE childs AS
            (
            SELECT
            t1.id,
            t1.parent_id,
            t1.name
            FROM
            tree t1
            WHERE
            t1.id = get_parent(7)
            UNION ALL
            SELECT
            t2.id,
            t2.parent_id,
            t2.name
            FROM
            tree t2
            INNER JOIN
            childs ON childs.id = t2.parent_id
            )
            SELECT
            id,
            parent_id,
            name
            FROM
            childs;



            id | parent_id | name
            -: | --------: | :------
            6 | 1 | Node 6
            4 | 6 | Node 4
            8 | 6 | Node 8
            11 | 6 | Node 11
            7 | 11 | Node 7
            10 | 7 | Node 10


            db<>fiddle here






            share|improve this answer















            This function returns the parent level of node_id:



            There is a 'level' row due there isn't a row (id, null) for parent row.



            CREATE FUNCTION get_parent(node_id int)
            RETURNS integer AS
            $$
            WITH RECURSIVE get_parent AS
            (
            SELECT
            t1.id,
            t1.parent_id,
            t1.name,
            0 AS level
            FROM
            tree t1
            WHERE
            t1.id = node_id
            UNION ALL
            SELECT
            t2.id,
            t2.parent_id,
            t2.name,
            level+1
            FROM
            tree t2
            INNER JOIN
            get_parent ON get_parent.parent_id = t2.id
            )
            SELECT
            id
            FROM
            get_parent
            ORDER BY
            level DESC
            LIMIT 1 ;
            $$
            LANGUAGE SQL;



            select get_parent(7);




            | get_parent |
            | ---------: |
            | 6 |


            Now, next query returns the whole tree structure based on a parent node.





            WITH RECURSIVE childs AS
            (
            SELECT
            t1.id,
            t1.parent_id,
            t1.name
            FROM
            tree t1
            WHERE
            t1.id = get_parent(7)
            UNION ALL
            SELECT
            t2.id,
            t2.parent_id,
            t2.name
            FROM
            tree t2
            INNER JOIN
            childs ON childs.id = t2.parent_id
            )
            SELECT
            id,
            parent_id,
            name
            FROM
            childs;



            id | parent_id | name
            -: | --------: | :------
            6 | 1 | Node 6
            4 | 6 | Node 4
            8 | 6 | Node 8
            11 | 6 | Node 11
            7 | 11 | Node 7
            10 | 7 | Node 10


            db<>fiddle here







            share|improve this answer














            share|improve this answer



            share|improve this answer








            edited Dec 13 '18 at 16:14

























            answered Dec 13 '18 at 11:39









            McNetsMcNets

            16k42161




            16k42161













            • as i said, i don't have problem with ancestors. I need the whole tree.

              – Luca Nitti
              Dec 13 '18 at 12:04





















            • as i said, i don't have problem with ancestors. I need the whole tree.

              – Luca Nitti
              Dec 13 '18 at 12:04



















            as i said, i don't have problem with ancestors. I need the whole tree.

            – Luca Nitti
            Dec 13 '18 at 12:04







            as i said, i don't have problem with ancestors. I need the whole tree.

            – Luca Nitti
            Dec 13 '18 at 12:04




















            draft saved

            draft discarded




















































            Thanks for contributing an answer to Database Administrators Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fdba.stackexchange.com%2fquestions%2f224871%2fpostgresql-retrieve-all-ids-in-a-tree-for-a-given-subnode%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            Popular posts from this blog

            Bundesstraße 106

            Verónica Boquete

            Ida-Boy-Ed-Garten