Edge Patterns Have a Direction in Graph Query Language

SQL Show and Tell
1

In the Introduction to Graph Query Language post we were looking at the circular relationship between a Customer and Tender (mode of payment) as used in Transactions

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To recap:

  • Customer may use a Tender (mode of payment e.g. Credit Card, Gift Card etc);
  • A Tender (mode of payment) can be used by Customer

Here the the relationship is bi-directional with different context. Note that Customer may use several different Tenders in different transactions e.g. The Customer may use a Gift Card for Transaction 1 and Credit Card for Transaction 2. Also Note that a Single Tender may be used by several Customer for e.g. a Husband and Wife using the same Gift Card to pay for two different Transactions.

We had used following to retrieve the list of “Shared” Tenders

SELECT distinct *
FROM GRAPH_TABLE( PAYMENT_TENDER_GRAPH
MATCH (customer1 is customer) - [e1 is customer_paid_using] -> 
    (shared_tender is tender) - [e2 is tender_used_by] -> 
    (customer2 is customer)
where customer1.customer_id != customer2.customer_id
COLUMNS ( customer1.first_name as customer_1_name
          , shared_tender.card_number as shared_tender
          , customer2.first_name as customer_2_name)
);

Note that in the above MATCH clause the pattern was defined as (node) – [edge] → (node) → [edge] → (node). This is going in one direction with using both customer_paid_using and tender_used_by edges.

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Graph theory says that the same pattern also be redefined as (node) – [edge] -> (node) <- [edge] - (node) using just the customer_paid_using edge. This is why it is Edge Pattern Definitions have a direction in GQL.

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An edge pattern has a direction, as edges in graphs do. Thus, (a) <-[]- (b) specifies a case where b has an edge pointing at a, whereas (a) -[]-> (b) looks for an edge in the opposite direction.

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SELECT distinct *
FROM GRAPH_TABLE( PAYMENT_TENDER_GRAPH
MATCH (customer1 is customer) - [e1 is customer_paid_using] -> 
  (shared_tender is tender) <- [e2 is customer_paid_using] 
  - (customer2 is customer)
where customer1.customer_id != customer2.customer_id
COLUMNS (customer1.first_name as customer_1_name
         , shared_tender.card_number as shared_tender
         , customer2.first_name as customer_2_name)
);

This will produce the same result:
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