Py2neo：查找關系並返回節點

Question

你如何從py2neo中的neo4j web模塊中的電影圖形示例中執行此Cypher查詢而不使用graph.cypher.execute（或者獲取graph.cypher.execute以返回一組節點而不是它具有的丑陋的返回字符串）

MATCH (tom:Person {name: "Tom Hanks"})-[:ACTED_IN]->(tomHanksMovies) RETURN tom,tomHanksMovies

我想要的是：

(n4358:Person {born:1956,name:"Tom Hanks"}, {'PLAYED_IN', 'year': 1990}, n4354:Movie {released:1998,tagline:"At odds in life... in love on-line.",title:"You've Got Mail"})

其中a [0]給出tom hanks節點，a [1]表示關系，a [2]表示電影。

---

編輯：添加“錯誤”示例輸出

>>> print('List all Tom Hanks movies...')
>>> a = graph.cypher.execute('MATCH (tom:Person {name: "Tom Hanks"})-[:ACTED_IN]->(tomHanksMovies) RETURN tomHanksMovies')
>>> print(a)

List all Tom Hanks movies...
    | tomHanksMovies                                                                                                                                                                 
----+---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
  1 | (n4354:Movie {released:1998,tagline:"At odds in life... in love on-line.",title:"You've Got Mail"})                                                                            
  2 | (n4360:Movie {released:1993,tagline:"What if someone you never met, someone you never saw, someone you never knew was the only someone for you?",title:"Sleepless in Seattle"})
  3 | (n4365:Movie {released:1990,tagline:"A story of love, lava and burning desire.",title:"Joe Versus the Volcano"})                                                               
  4 | (n4372:Movie {released:1996,tagline:"In every life there comes a time when that thing you dream becomes that thing you do",title:"That Thing You Do"})                         
  5 | (n4392:Movie {released:2012,tagline:"Everything is connected",title:"Cloud Atlas"})                                                                                            
  6 | (n4398:Movie {released:2006,tagline:"Break The Codes",title:"The Da Vinci Code"})                                                                                              
  7 | (n4417:Movie {released:1999,tagline:"Walk a mile you'll never forget.",title:"The Green Mile"})                                                                                
  8 | (n4431:Movie {released:1995,tagline:"Houston, we have a problem.",title:"Apollo 13"})                                                                                          
  9 | (n4437:Movie {released:2000,tagline:"At the edge of the world, his journey begins.",title:"Cast Away"})                                                                        
 10 | (n4446:Movie {released:2007,tagline:"A stiff drink. A little mascara. A lot of nerve. Who said they couldn't bring down the Soviet empire.",title:"Charlie Wilson's War"})     
 11 | (n4448:Movie {released:2004,tagline:"This Holiday Season… Believe",title:"The Polar Express"})                                                                                 
 12 | (n4449:Movie {released:1992,tagline:"Once in a lifetime you get a chance to do something different.",title:"A League of Their Own"})   


>>> print(a[0])

 tomHanksMovies                                                                                     
-----------------------------------------------------------------------------------------------------
 (n4354:Movie {released:1998,tagline:"At odds in life... in love on-line.",title:"You've Got Mail"})

>>> print(type(a[0]))

<class 'py2neo.cypher.core.Record'>

Answer 1

我假設你意識到execute實際上並沒有返回一個字符串？ 你在a看到a是文檔中指定的RecordList的表示：

http://py2neo.org/2.0/cypher.html#py2neo.cypher.CypherResource.execute

a[0]然后給你RecordList的第一個Record 。 然后可以通過名稱訪問該Record值，例如a[0]['tomHanksMovies'] 。

Record對象的詳細信息如下：

http://py2neo.org/2.0/cypher.html#records

Answer 2

如果您已經結婚使用py2neo，您可以做的是稍微修改您的查詢以返回您想要的所有信息。 例如，

a = graph.cypher.execute("MATCH (a:Person {name:"Tom Hanks"})-[acted:ACTED_IN]->(movies:Movie) RETURN a, acted, movies")

這應該做的是給你一個列表中的結果，就像你說你不想要的那樣。 但是，從這里你可以索引結果以獲得你想要的每個部分。 例如，a [0]會給你第一行結果，[0] [0]會給你第一行結果的人節點，[0] [0] [0]會給你第一行中第一個節點的第一個屬性，等等。從這里你可以運行一個for循環來將結果組織成一個你更感興趣的形式。

希望這可以幫助。