[英]Py2neo: Find relations and return nodes
你如何从py2neo中的neo4j web模块中的电影图形示例中执行此Cypher查询而不使用graph.cypher.execute(或者获取graph.cypher.execute以返回一组节点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]给出tom hanks节点,a [1]表示关系,a [2]表示电影。
编辑:添加“错误”示例输出
>>> 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'>
我假设你意识到execute
实际上并没有返回一个字符串? 你在a
看到a
是文档中指定的RecordList
的表示:
http://py2neo.org/2.0/cypher.html#py2neo.cypher.CypherResource.execute
a[0]
然后给你RecordList
的第一个Record
。 然后可以通过名称访问该Record
值,例如a[0]['tomHanksMovies']
。
Record
对象的详细信息如下:
如果您已经结婚使用py2neo,您可以做的是稍微修改您的查询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")
这应该做的是给你一个列表中的结果,就像你说你不想要的那样。 但是,从这里你可以索引结果以获得你想要的每个部分。 例如,a [0]会给你第一行结果,[0] [0]会给你第一行结果的人节点,[0] [0] [0]会给你第一行中第一个节点的第一个属性,等等。从这里你可以运行一个for循环来将结果组织成一个你更感兴趣的形式。
希望这可以帮助。
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