How to stop traversing after edge class?
The graph for the query TRAVERSE bothE(), bothV() FROM ( SELECT FROM Title WHERE title_id IN [12] ) looks like this:
If I wrote (for blue graph):
TRAVERSE bothE("E14", "E5", "E6"), bothV() FROM ( SELECT FROM Title WHERE title_id IN [12] )
it don't gave me V 970.
How to construct a query that will return the part I need? I would like the whole graph to be returned if there exists an orange route.
I tested: TRAVERSE inE(..), outE(...), bothE() and WHILE $parent.$current.#class != "E13" bit it don't works.
I've seen https://stackoverflow.com/a/43776591/1194525, but if I understand correctly, MATCH can be used at a well known well.
I have an idea how to do it in 2 queries, but could use improvement, eg .: Doing this in 1 query.
step 1: Traverse graph by all "right" edges.
TRAVERSE bothE("Prequel", "Sequel"), bothV() FROM (
SELECT FROM Title WHERE title_id IN [12] )
step 2: Get all Vertex rid from response (in client)
step 3: Query by "stop" edges for all Vertex rid from previous query.
TRAVERSE bothE("Character"), bothV() FROM (
SELECT FROM Title WHERE title_id IN [12, 431, 432, 433, ...] )
MAXDEPTH 2
step 4: Merge responses (Vertex + Edges) in client.
Related
I have an array of strings, some of which may be repeated. I am trying to build a query which returns a single json object where the keys are the distinct values in the array, and the values are the count of times each value appears in the array.
I have built the following query;
WITH items (item) as (SELECT UNNEST(ARRAY['a','b','c','a','a','a','c']))
SELECT json_object_agg(distinct_values, counts) item_counts
FROM (
SELECT
sub2.distinct_values,
count(items.item) counts
FROM (
SELECT DISTINCT items.item AS distinct_values
FROM items
) sub2
JOIN items ON items.item = sub2.distinct_values
GROUP BY sub2.distinct_values, items.item
) sub1
DbFiddle
Which provides the result I'm looking for: { "a" : 4, "b" : 1, "c" : 2 }
However, it feels like there's probably a better / more elegant / less verbose way of achieving the same thing, so I wondered if any one could point me in the right direction.
For context, I would like to use this as part of a bigger more complex query, but I didn't want to complicate the question with irrelevant details. The array of strings is what one column of the query currently returns, and I would like to convert it into this JSON blob. If it's easier and quicker to do it in code then I can, but I wanted to see if there was an easy way to do it in postgres first.
I think a CTE and json_object_agg() is a little bit of a shortcut to get you there?
WITH counter AS (
SELECT UNNEST(ARRAY['a','b','c','a','a','a','c']) AS item, COUNT(*) AS item_count
GROUP BY 1
ORDER BY 1
)
SELECT json_object_agg(item, item_count) FROM counter
Output:
{"a":4,"b":1,"c":2}
I have an employee , manager hierarchy which could end up being circular.
Ex:
28397468N>88518119N>87606705N>28397468N
Create Table emp_manager ( Emp_id varchar(30), Manager_id varchar(30));
Insert into emp_manager values ('28397468N','88518119N');
Insert into emp_manager values ('88518119N','87606705N');
Insert into emp_manager values ('87606705N','28397468N');
My requirement is:
When my proc is called and there are circular hierarchies in the emp_manager table, we should return an error listing the employees in the hierarchy.
The below link contains some useful info:
https://mccalljt.io/blog/2017/01/postgres-circular-references/
I have modified it as below:
select * from (
WITH RECURSIVE circular_managers(Emp_id, Manager_id, depth, path, cycle) AS (
SELECT u.Emp_id, u.Manager_id, 1,
ARRAY[u.Emp_id],
false
FROM emp_manager u
UNION ALL
SELECT u.Emp_id, u.Manager_id, cm.depth + 1,
(path || u.Emp_id)::character varying(32)[],
u.Emp_id = ANY(path)
FROM emp_manager u, circular_managers cm
WHERE u.Emp_id = cm.Manager_id AND NOT cycle
)
select
distinct (path) d
FROM circular_managers
WHERE cycle
AND path[1] = path[array_upper(path, 1)]) cm
BUT, the problem is, it is returning all combinations of the hierarchy:
{28397468N,88518119N,87606705N,28397468N}
{87606705N,28397468N,88518119N,87606705N}
{88518119N,87606705N,28397468N,88518119N}
I need a simple answer like this:
28397468N>88518119N>87606705N>28397468N
even this will do:
28397468N>88518119N>87606705N
Please help!
So all references:
{28397468N,88518119N,87606705N,28397468N}
{87606705N,28397468N,88518119N,87606705N}
{88518119N,87606705N,28397468N,88518119N}
are correct but just start from different element.
I need a simple answer like this: 28397468N>88518119N>87606705N>28397468N
So what's needed is a filter for the same circle refs.
Let's do that in a way:
sort distinct items in arrays
aggregate them back - so for all references it will be '{28397468N,87606705N,88518119N}'
use produced value for DISTINCT FIRST_VALUE
WITH D (circle_ref ) AS (
VALUES
('{28397468N,88518119N,87606705N,28397468N}'::text[]),
('{87606705N,28397468N,88518119N,87606705N}'::text[]),
('{88518119N,87606705N,28397468N,88518119N}'::text[])
), ordered AS (
SELECT
D.circle_ref,
(SELECT ARRAY_AGG(DISTINCT el ORDER BY el) FROM UNNEST(D.circle_ref) AS el ) AS ordered_circle
FROM
D
)
SELECT DISTINCT
FIRST_VALUE (circle_ref) OVER (PARTITION BY ordered_circle ORDER BY circle_ref) AS circle_ref
FROM
ordered;
circle_ref
{28397468N,88518119N,87606705N,28397468N}
DB Fiddle: https://www.db-fiddle.com/f/6ytb2v11s8T95PPLoTZZed/0
To prevent circular references, you can use a closure table and a trigger - as explained in https://stackoverflow.com/a/38701519/5962802
The closure table will also allow you to easily get all subordinates for a given supervisor (no matter how deep in the hierarchy) - or all direct bosses of a given employee (up to the root).
Before using the rebuild_tree stored procedure you will have to remove all circular references from the hierarchy.
I am trying to set some variables on the top of my query via CTEs to make maintenance of a long query more easy to handle.
I have extracted an example of what I am trying to achieve. I am not managing to make 'tags' be perceived as a list rather than a whole string. I have tried split_part but have not managed to get what I require.
WITH tmp AS (
SELECT
'tag1, tag2, tag3' as tags
)
select
CASE WHEN 'tag1' in (select tags from tmp) THEN 1 ELSE 0 END matched_tags
Basically what I need is to have a string 'tag1' and see if it exists in the list 'tag1','tag2' or 'tag3'. This should give me 1 as there is a match
This is obviously not working because it is taking the 'tag1, tag2, tag3' as one string so there is no match.
Can anyone help me with this?
The STRPOS() function should do what you want. https://docs.aws.amazon.com/redshift/latest/dg/r_STRPOS.html
Something like this:
WITH tmp AS (
SELECT
'tag1, tag2, tag3' as tags
)
SELECT
CASE WHEN STRPOS(tags, 'tag1') > 0 THEN 1 ELSE 0 END as matched_tags
FROM tmp;
I'm trying to find the pairs of vertices that have the greatest number of common vertices between them. It is very similar to the 'number of mutual friends' example used in many graph database demos. I can determine the number of mutual vertices between a pair of known vertices using this:
SELECT Expand($query) LET
$query1 = (SELECT Expand(outE().in) FROM #1:2,
$query2 = (SELECT Expand(OutE().in) FROM #1:3,
$query = Intersect($query1,$query2);
The Count() of the above query's result is the number of common vertices.
However, I can't figure out how to aggregate that query across my entire data set. My best solution has been a brute force, where I iterate through each vertex and run the above query against all other vertices (technically, I do all the vertices 'after' that vertex).
My solution is inefficient and had to be coded in C# rather than done entirely in SQL. How can this be done using OrientDb's SQL?
You can use a SELECT with a MATCH:
SELECT FROM (
SELECT a, b, count(friend) as nFriends from (
MATCH
{class:Person, as:a} -FriendOf- {as:friend} -FriendOf-{as:b, where:($matched.a != $currentMatch)}
RETURN a, b, friend
)
) ORDER BY nFriends
Slight modification to #Luigi's answer:
SELECT a, b, Count(friend) AS nFriends FROM (
MATCH
{class:Person, as:a} -E- {as:friend} -E- {class:Person, as:b, where:($matched.a != $currentMatch)}
RETURN a, b, friend
) GROUP BY a, b ORDER BY nFriends DESC
I needed the GROUP BY or I just get one big count.
I haven't been able to find an answer for the SQL for this.
Given pairs of vertices (record ids) and edge types between them, I want to check if all pairs exists.
V1 --E1--> V2
V3 --E2--> V4
... and so on. The answer I want is true / false or something equivalent. ALL connections must be present in order to evaluate to true, so at least one edge (of correct type) must exist for each pair.
Pseudo, the question would be:
Does V1 have edge <E1EdgeType> to V2?
AND
Does V3 have edge <E2EdgeType> to V4?
AND
... and so on
Does anyone know what the orientDB SQL would be to achieve this?
UPDATE
I did already have one way of checking if one single edge exists between known vertices. It's perhaps not very pretty either, but it works:
SELECT FROM (
SELECT EXPAND(out('TestEdge')) FROM #12:0
) WHERE #rid=#12:1
This will return the destination record (#12:0) if an edge of type 'TestEdge' exists from #12:0 to #12:1. However, if I have two of those, how can I query for one single result for both queries. Something like:
SELECT <something with $c> LET
$a = (SELECT FROM (SELECT EXPAND(out('TestEdge')) FROM #12:0) WHERE #rid=#12:1)
$b = (SELECT FROM (SELECT EXPAND(out('AnotherTestEdge')) FROM #12:2) WHERE #rid=#12:3)
$c = <something that checks that both a and b yield results>
That's what I aim towards doing. Please tell me if I'm solving this the wrong way. I'm not even sure what the gain is to merge queries like this compared to just repeat queries.
Given a pair of vertices, say #11:0 and #12:0, the following query will effectively check whether there is an edge of type E from #11:0
to #12:0
select from (select #this, out(E) from #11:0 unwind out) where out = #12:0
----+------+-----+-----
# |#CLASS|this |out
----+------+-----+-----
0 |null |#11:0|#12:0
----+------+-----+-----
This is highly inelegant and I would encourage you to think about formulating an enhancement request accordingly at https://github.com/orientechnologies/orientdb/issues
One way to incorporate the boolean tests you have in mind is illustrated by the following:
select from
(select $a.size() as a, $b.size() as b
let a=(select count(*) as e from (select out(E) from #11:0 unwind out)
where out = #12:0),
b=(select count(*) as e from (select out(E) from #11:1 unwind out)
where out = #12:2))
where a > 0 and b > 0
Yes, inelegance again :-(
It might be useful to you the following query
SELECT eval('sum($a.size(),$b.size())==2') as existing_edges
let $a = ( SELECT from TestEdge where out = #12:0 and in = #12:1 limit 1),
$b = ( SELECT from AnotherTestEdge where out = #12:2 and in = #12:3 limit 1)
Hope it helps.