Exploring OrientDB, I am stuck with the following query:
I've a set of classes Package, Library, Application, each with a name as attribute.
I've also a class Person with name and email as attributes.
Relations are defined as classes 'require' or 'writtenBy' from E.
Applications vertices may have 'require' edges to one or more Librarys.
Librarys vertices may have 'require' edges to one or more Packages.
Packages vertices also may have 'require' edges to one or more Packages !
Applications , Librarys, Packages have 'writtenBy' edges to 1 Person; one Person may write several Packages/Library/Application
Given a specific Application/Library or Package rid, I would like to select, as an array, all Persons (name and email) writing its constituents (identified by the 'require' edges) .
I've tried different strategies in OrientDB Studio but can't find a solution. Thanks for any help !
Found it.
Traverse is the way to go:
the solution is something like:
select expand(writtenBy) from (traverse out('require') from #rid )
Related
We have a set of tables shown below we use for our other tables to reference for location data. Some examples are:
Find all companies within X miles of X City
Create a company profile's location as X City
We solve the problem of multiple cities with similar names by matching with State as well, but now we ran into a different set of problems. We use Google's Place Autocomplete for both Geocoding and matching up a users query with our Cities. This works fairly well until Google's format deviates from ours.
Example:
St. Louis !== Saint Louis and
Ameca del Torro !== Ameca Torro
Is there a way to fuzzy match cities in our queries?
Our query to match cities now looks like:
SELECT c.id
FROM city c
INNER JOIN state s
ON s.id = c.state_id
WHERE c.name = 'Los Angeles' AND s.short_name = 'CA'
I've also considered the denormalizing city and simply storing coordinates to still accomplish the radius search. We have around 2 million rows in our company table now so a radius search would be performed on that rather than by city table with a JOIN on company. This would also mean we wouldn't be able to create custom regions (simply anyway) for cities, and add other attributes to cities in the future.
I found this answer but it is basically affirming our way of normalizing input is a good method, but not how we match to our local Table (unless Google offers a City Name export I don't know about).
The short answer is that you can use Postgres's full text search functionality, with a customized search configuration.
Since your dealing with place names, your probably want to avoid stemming, so you can use the simple configuration as a starting point. You can also add stop-words that make sense for place names (with the examples above, you can probably consider "St.", "Saint", and "del" as stop-words).
A pretty basic outline of setting up your customized is below:
Create a stopwords file and put it in your $SHAREDIR/tsearch_data Postgres directory. See https://www.postgresql.org/docs/9.1/static/textsearch-dictionaries.html#TEXTSEARCH-STOPWORDS.
Create a dictionary that uses this stopwords list (you can probably use the pg_catalog.simple as your template dictionary). See https://www.postgresql.org/docs/9.1/static/textsearch-dictionaries.html#TEXTSEARCH-SIMPLE-DICTIONARY.
Create a search configuration for place names. See https://www.postgresql.org/docs/9.1/static/textsearch-configuration.html.
Alter your search configuration to use the dictionary you created in Step 2 (cf. the link above).
Another consideration is how to consider internationalization. It seems that the issue for your second example (Ameca del Torro vs. Ameca Torro) might be a Spanish vs. English representation of the name. If that's the case, you could also consider storing both a "localized" and "universal" (e.g. English) version of the city name.
At the end, your query (using full-text search) might look like this (where the 'places' is the name of your search configuration):
SELECT cities."id"
FROM cities
INNER JOIN "state" ON "state".id = cities.state_id
WHERE
"state".short_name = 'CA'
AND TO_TSVECTOR('places', cities.name) ## TO_TSQUERY('places', 'Los & Angeles')
So I'm new to OrientDB, and while I'm pretty good at SQL the syntax to get what I want in OrientDB is escaping me.
I know I can do something like select *, in().size() as size from Users order by size desc to find the most connected node of a certain class (Users in this case), but how do I find the most connected children a couple levels down?
I.e., let's say I have Organizations --> PROMOTES (edge) --> Platform --> MANAGES (edge) --> Suggestion
How do I find the most connected Suggestions at the Organization level? I.e., I know I can easily find the most connected suggestions one level out using the query I shared, but what about the most connected another level beyond that?
I'd ultimately like a result which lists each Suggestion along with how indirectly connected (number of edges) it is to Organizations.
Thank you!
Use TRAVERSE
You should use the command TRAVERSE to do that
SELECT out(PROMOTES).size() AS connectedOrg,*
FROM (
TRAVERSE out(MANAGES)
FROM Suggestion WHILE $depth < 2
)
WHERE $depth > 0
Result will be the Platforms linked to a Suggestion. Records can be duplicated.
Along with each Platform, you get the number of Organisation called connectedOrg.
About traverse :
Traverse follow the record ids present in a record and aggregates them in the results. With WHILE $depth < 2 you can limit search to only one level and with WHERE $depth > 0 you can remove the original record from the result. More info here.
Use OrientDB Functions
If you need to know how many Organization is linked to each Suggestion (through Plateform), use this syntax.
SELECT *, set(out(MANAGES).out(PROMOTES), null).size() FROM Suggestion
Note : , null allows to switch from AGGREGATE to INLINE. It prevents from having all sizes aggregated in a single record. See the doc for more info.
I have been struggling for some time now and I can't really find anyone having done the same thing before.
I'm creating a backend module in TYPO3 7.6 which belongs to a shop extension.
The shop extension with the backend module was created with the extension builder. The shop has the following three models:
Product (products which can be ordered through the shop)
Productsorder (link to the customer)
ProductsorderPosition (the ordered product, the ordered amount and size and the link to the Productsorder)
The customers are of a model type from a different extension. These customers are linked to fe_users.
Now what I wanna do in my backend module is getting an overview to all these orders listed with the customer, some information about the fe_user and of course the product. I have created a sql-query, which does exactly that:
SELECT p.productname, p.productpriceperpiece,
pop.amount, pop.size,
h.name, h.address, h.zipcode, h.city, h.email, h.phone,
f.first_name, f.last_name, f.email
FROM `tx_gipdshop_domain_model_productorderposition` AS pop
JOIN `tx_gipdshop_domain_model_product` AS p ON pop.products = p.uid
JOIN `tx_gipdshop_domain_model_productsorder` AS po ON pop.productorder = po.uid
JOIN `tx_gipleasedisturbhotels_domain_model_hotel` AS h ON po.hotel = h.uid
JOIN `fe_users` AS f ON h.feuser = f.uid
If I use this query from the product repository it gives back the right amount of data records but they're of type product and the products are all "empty" (uid = 0 etc).
I've added an additional action for this in the product controller (getOrdersAction) and in the repository containing the query I've added a method findAllOrders.
I'm still rather a beginner in TYPO3 but I can somehow understand why it returns data sets of type Product when the query is called from the ProductRepository. But what I do not know is how I can get all the information from the query above and list it in the backend module.
I've already thought about moving the query to the ProductsorderPositionRepository but I would probably be faced with a similar problem, it would only return the information from the ProductsorderPosition and everything else would be left out.
Can someone point me to the right direction?
Would I need to create another model with separate repository and controller? Isn't there an easier way?
If you need more information, just ask! ;)
First of all, you are doing a joined query with subsets of data mixed from multiple tables. There is nothing against this.
Because of this, there is no "model" which has the mixed datasets.
If you are using the default query thing in a repository, the magic behind the repository assumes that the result of the query statement reflects the defined base model for this repository.
Moving the query function to another repository does not solve the problem.
You have not provided the code snippet you are executing the sql statement, so I assume you have used the query thing in the repository to execute the statement. Something like this:
$result = $query->statement('
SELECT p.productname, p.productpriceperpiece,
pop.amount, pop.size,
h.name, h.address, h.zipcode, h.city, h.email, h.phone,
f.first_name, f.last_name, f.email
FROM `tx_gipdshop_domain_model_productorderposition` AS pop
JOIN `tx_gipdshop_domain_model_product` AS p ON pop.products = p.uid
JOIN `tx_gipdshop_domain_model_productsorder` AS po ON pop.productorder = po.uid
JOIN `tx_gipleasedisturbhotels_domain_model_hotel` AS h ON po.hotel = h.uid
JOIN `fe_users` AS f ON h.feuser = f.uid', NULL);
or have used the query building stuff.
First solution
The first and simpliest solution would be to retrieve the result as plain php array. Before TYPO3 7.0 you could have done this by using this:
$query->getQuerySettings()->setReturnRawQueryResult(TRUE);
With TYPO3 7.0 this deprecated method was removed from the core.
The only way is to define the query and call $query->execute(TRUE); for now.
This should return the data in pure array form.
This is the simpliest one, but as we are in the extbase context this should not be suffering enough.
Second Solution - no, just an idea that I would try next
The second solution means that you have some work to do and is for now only a suggestion, because I have not tried this by myself.
Create a model with the properties and getter/setters for the result columns of your query
Create a corresponding repository
Third solution
Not nice, but if nothing else works, fall back to the old TYPO3 v4 query methods:
$GLOBALS['TYPO3_DB']->exec_SELECTgetRows([...]))
and replace this with the QueryBuilder in/for TYPO3 v8.
This is really not nice.
I hope I could direct you to the right way, even if not giving a full solving solution.
I have a question about the graph editor. How can I keep the vertices fixed when I select some vertices? Now if I select some new vertices, the old vertices will move automatically.
If you want see the vertices from Car and Company and both classes extends V you can run this command:
select from V
this is the output:
Instead if you have some edges between this two classes you can still use the command above or, if you want to see which Company made a Car, you can use this command:
select name,out('made_by').name as Company from Car unwind Company
this is the output:
Hope it helps
Regards
For all I know, you can't keep fixed the old vertices.
OrientDB Server v2.0.10 ,
I am trying to come up with a query for the following scenario.
I have 2 hierarchies: A->B->C and D->E->F
The number of nodes in the hierarchy can change.
The node in 1st hierarchy can be connected to the other hierarchy using some relation say 'Assigned'.
What I want is the parent node of the 2nd hierarchy if there is any incoming edge to any of the node in that 2nd hierarchy from the 1st.
For example, say we have Car-Child->Engine-Child->Piston and Country-Child->State-Child->City
And a relationship Made_In which relates Car or Engine or Piston to either Country or State or City
So if there is a relation with either of Country or State or City, the Country should be returned. Example, Engine1-Made_In->Berlin, this would return Germany.
Sorry for such a toyish example. I hope it is clear.
Thanks.
You should consider reading the chapter about "traversing" - that should be the missing link to answer your question. You can find it here: http://orientdb.com/docs/last/SQL-Traverse.html
Basically, if you think of your graph as a family tree, you want to achieve 3 things:
Find all children, grand-children, grand-grand-children (and so on) from tree 1 for a given family member (=Hierarchy1)
Find those who have relations to members of another family tree (=ASSIGNED)
Show me who's on top of this tree (=Hierarchy2)
One of the possible solutions should look a little something like this:
Since you want to end up on top of hierarchy2, you have to start on the other side, i.e. hierarchy1.
Get hierarchy1 (top-to-bottom)
TRAVERSE out("CHILD") FROM Car
Choose all relations
SELECT out("MADE_IN) FROM ([1])
and from those, go bottom-to-top
TRAVERSE in("CHILD") FROM ([2])
Who's on top?
SELECT FROM ([3]) WHERE #class="Country"
Combined into one sql, it looks as ugly as this:
SELECT FROM (
TRAVERSE in("CHILD") FROM (
SELECT out("MADE_IN") FROM (
TRAVERSE out("CHILD") FROM Car
)
)
) WHERE #class="Country"
You could replace Car with any #rid in hierarchy1 to get a list of countries it or any part of it was made in.
There might be better solutions for sure. But at least this one should work, so I hope it will help.