最近，我閱讀了Michal Bachman關于 Neo4j中雙向關系的文章 ，他建議對于某些關系類型，我們對關系的方向不那么感興趣，因此可以在查詢時忽略它。 他使用以下示例顯示了Neo Technology和GraphAware之間的合作關系：

兩家公司都是彼此的合作伙伴，但是由于我們可以盡快找到傳入和傳出的關系，因此我們最好在兩家公司/節(jié)點之間只有一種關系。

當我們想使圖中的隱式關系明確時，經(jīng)常會出現(xiàn)這種模式。 例如，我們可能有以下圖形描述了他們所從事的人員和項目：

我們可以使用以下密碼語法在Neo4j 2.0中創(chuàng)建該圖：

CREATE (mark:Person {name: "Mark"}) CREATE (dave:Person {name: "Dave"}) CREATE (john:Person {name: "John"})CREATE (projectA:Project {name: "Project A"}) CREATE (projectB:Project {name: "Project B"}) CREATE (projectC:Project {name: "Project C"})CREATE (mark)-[:WORKED_ON]->(projectA) CREATE (mark)-[:WORKED_ON]->(projectB) CREATE (dave)-[:WORKED_ON]->(projectA) CREATE (dave)-[:WORKED_ON]->(projectC) CREATE (john)-[:WORKED_ON]->(projectC) CREATE (john)-[:WORKED_ON]->(projectB)

如果我們想弄清楚哪些人彼此認識，我們可以編寫以下查詢：

MATCH (person1:Person)-[:WORKED_ON]-()<-[:WORKED_ON]-(person2) RETURN person1, person2==> +-------------------------------------------------------+ ==> | person1 | person2 | ==> +-------------------------------------------------------+ ==> | Node[500363]{name:"Mark"} | Node[500364]{name:"Dave"} | ==> | Node[500363]{name:"Mark"} | Node[500365]{name:"John"} | ==> | Node[500364]{name:"Dave"} | Node[500363]{name:"Mark"} | ==> | Node[500364]{name:"Dave"} | Node[500365]{name:"John"} | ==> | Node[500365]{name:"John"} | Node[500364]{name:"Dave"} | ==> | Node[500365]{name:"John"} | Node[500363]{name:"Mark"} | ==> +-------------------------------------------------------+ ==> 6 rows

我們可能要在每對人之間創(chuàng)建一個KNOWS關系：

MATCH (person1:Person)-[:WORKED_ON]-()<-[:WORKED_ON]-(person2) CREATE UNIQUE (person1)-[:KNOWS]->(person2) RETURN person1, person2

現(xiàn)在，如果我們運行一個查詢（忽略關系方向）以找出哪些人彼此認識，我們將得到很多重復的結果：

MATCH path=(person1:Person)-[:KNOWS]-(person2) RETURN person1, person2, path==> +--------------------------------------------------------------------------------------------------------------------------------+ ==> | person1 | person2 | path | ==> +--------------------------------------------------------------------------------------------------------------------------------+ ==> | Node[500363]{name:"Mark"} | Node[500364]{name:"Dave"} | [Node[500363]{name:"Mark"},:KNOWS[528536]{},Node[500364]{name:"Dave"}] | ==> | Node[500363]{name:"Mark"} | Node[500365]{name:"John"} | [Node[500363]{name:"Mark"},:KNOWS[528537]{},Node[500365]{name:"John"}] | ==> | Node[500363]{name:"Mark"} | Node[500364]{name:"Dave"} | [Node[500363]{name:"Mark"},:KNOWS[528538]{},Node[500364]{name:"Dave"}] | ==> | Node[500363]{name:"Mark"} | Node[500365]{name:"John"} | [Node[500363]{name:"Mark"},:KNOWS[528541]{},Node[500365]{name:"John"}] | ==> | Node[500364]{name:"Dave"} | Node[500363]{name:"Mark"} | [Node[500364]{name:"Dave"},:KNOWS[528538]{},Node[500363]{name:"Mark"}] | ==> | Node[500364]{name:"Dave"} | Node[500365]{name:"John"} | [Node[500364]{name:"Dave"},:KNOWS[528539]{},Node[500365]{name:"John"}] | ==> | Node[500364]{name:"Dave"} | Node[500363]{name:"Mark"} | [Node[500364]{name:"Dave"},:KNOWS[528536]{},Node[500363]{name:"Mark"}] | ==> | Node[500364]{name:"Dave"} | Node[500365]{name:"John"} | [Node[500364]{name:"Dave"},:KNOWS[528540]{},Node[500365]{name:"John"}] | ==> | Node[500365]{name:"John"} | Node[500364]{name:"Dave"} | [Node[500365]{name:"John"},:KNOWS[528540]{},Node[500364]{name:"Dave"}] | ==> | Node[500365]{name:"John"} | Node[500363]{name:"Mark"} | [Node[500365]{name:"John"},:KNOWS[528541]{},Node[500363]{name:"Mark"}] | ==> | Node[500365]{name:"John"} | Node[500363]{name:"Mark"} | [Node[500365]{name:"John"},:KNOWS[528537]{},Node[500363]{name:"Mark"}] | ==> | Node[500365]{name:"John"} | Node[500364]{name:"Dave"} | [Node[500365]{name:"John"},:KNOWS[528539]{},Node[500364]{name:"Dave"}] | ==> +--------------------------------------------------------------------------------------------------------------------------------+ ==> 12 rows

每對人出現(xiàn)4次，如果我們以Mark和Dave為例，我們可以看到原因：

MATCH path=(person1:Person)-[:KNOWS]-(person2) WHERE person1.name = "Mark" AND person2.name = "Dave" RETURN person1, person2, path==> +--------------------------------------------------------------------------------------------------------------------------------+ ==> | person1 | person2 | path | ==> +--------------------------------------------------------------------------------------------------------------------------------+ ==> | Node[500363]{name:"Mark"} | Node[500364]{name:"Dave"} | [Node[500363]{name:"Mark"},:KNOWS[528536]{},Node[500364]{name:"Dave"}] | ==> | Node[500363]{name:"Mark"} | Node[500364]{name:"Dave"} | [Node[500363]{name:"Mark"},:KNOWS[528538]{},Node[500364]{name:"Dave"}] | ==> +--------------------------------------------------------------------------------------------------------------------------------+ ==> 2 rows

如果我們看一下路徑欄下有兩個不同的KNOWS關系（與IDS 528536和528538），馬克和戴夫，一個去之間從馬克戴夫和其他來自Dave馬克。

正如Michal在他的帖子中指出的那樣，在這種情況下，不需要兩個關系。 我們只需要一種關系，可以通過創(chuàng)建KNOWS關系時不指定方向來實現(xiàn)：

MATCH (person1:Person)-[:WORKED_ON]-()<-[:WORKED_ON]-(person2) CREATE UNIQUE (person1)-[:KNOWS]-(person2) RETURN person1, person2

現(xiàn)在，如果我們重新運行查詢以檢查Mark和Dave之間的關系，則只有一個：

MATCH path=(person1:Person)-[:KNOWS]-(person2) WHERE person1.name = "Mark" AND person2.name = "Dave" RETURN person1, person2, path==> +--------------------------------------------------------------------------------------------------------------------------------+ ==> | person1 | person2 | path | ==> +--------------------------------------------------------------------------------------------------------------------------------+ ==> | Node[500375]{name:"Mark"} | Node[500376]{name:"Dave"} | [Node[500375]{name:"Mark"},:KNOWS[528560]{},Node[500376]{name:"Dave"}] | ==> +--------------------------------------------------------------------------------------------------------------------------------+ ==> 1 row

在這種情況下，關系從Mark到Dave，可以通過執(zhí)行一些考慮方向的查詢來看到：

MATCH path=(person1:Person)-[:KNOWS]->(person2) WHERE person1.name = "Mark" AND person2.name = "Dave" RETURN person1, person2, path==> +--------------------------------------------------------------------------------------------------------------------------------+ ==> | person1 | person2 | path | ==> +--------------------------------------------------------------------------------------------------------------------------------+ ==> | Node[500375]{name:"Mark"} | Node[500376]{name:"Dave"} | [Node[500375]{name:"Mark"},:KNOWS[528560]{},Node[500376]{name:"Dave"}] | ==> +--------------------------------------------------------------------------------------------------------------------------------+ ==> 1 rowMATCH path=(person1:Person)<-[:KNOWS]-(person2) WHERE person1.name = "Mark" AND person2.name = "Dave" RETURN person1, person2, path==> +--------------------------+ ==> | person1 | person2 | path | ==> +--------------------------+ ==> +--------------------------+ ==> 0 row

參考： Neo4j：通過Mark Needham博客博客的JCG合作伙伴 Mark Needham 使隱式關系成為顯式和雙向關系 。

翻譯自: https://www.javacodegeeks.com/2013/10/neo4j-making-implicit-relationships-explicit-bidirectional-relationships.html

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