是否可以排除 repeat().until() 上的顶点？
Is it possible to exclude vertex on repeat().until()?
我正在尝试提出一个 gremlin 查询，我可以在其中推断出谁认识 Jack，但不包括直接来自 Employees 的信息。
谁可以在 Jack 上寻求参考？
不能从其他员工或受试者的家庭亲属那里获得推荐信。
查询应该return:
吉姆认识玛丽认识杰克
Jim 知道 Mary 受雇于 IBM 受雇于 John 受雇于 Coca Cola 受雇于 Jack
Jerry Knows John Employed Coca Cola employedd Jack
但不应该包括：
Jerry Knows Jacks 的兄弟家族 Jack
Jerry 雇用可口可乐雇用 Jack
珍妮Degree_from普林斯顿Degree_From杰克
见下图：
我提出了以下查询，但无法获得从路径中排除 Jerry 或 jane 的部分。
g.V().has('isSSEmployee',true).repeat(bothE('knows','employeed').otherV().simplePath()).until(has('name','Jack')).has('isSSEmployee',false).path().by('name').by(label)
加载数据的脚本
//only for tinkerpop
graph=TinkerGraph.open()
g=graph.traversal()
//from here on common between neptune and tinkerpop
//Setup started
//create vertexes
//create a person John
g.addV('person').property(id,'p1')
g.V('p1').property('questId','123456')
g.V('p1').property('name','John')
//create a person Jack
g.addV('person').property(id,'p2')
g.V('p2').property('questId','123457')
g.V('p2').property('name','Jack')
//create a person Mary
g.addV('person').property(id,'p3')
g.V('p3').property('questId','123458')
g.V('p3').property('name','Mary')
//create a person Jim and mark him as ssemployee
g.addV('person').property(id,'p4')
g.V('p4').property('name','Jim')
g.V('p4').property('isSSEmployee',true)
g.V('p4').property('questId','1234569')
//create a person Jerry and mark him as ssemployee
g.addV('person').property(id,'p5')
g.V('p5').property('name','Jerry')
g.V('p5').property('questId','12345700')
g.V('p5').property('isSSEmployee',true)
//create a person Jack's Brother
g.addV('person').property(id,'p6')
g.V('p6').property('name',"Jack's Brother")
g.V('p6').property('questId','1234')
//create Jeane our employee
g.addV('person').property(id,'p7')
g.V('p7').property('name', 'Jeane')
g.V('p7').property('questId', '1234580')
g.V('p7').property('isSSEmployee', true)
//create a company Coca Cola
g.addV('company').property(id,'c1')
g.V('c1').property('name','Coca Cola')
g.V('c1').property('questId','123456')
//create a company IBM
g.addV('company').property(id,'c2')
g.V('c2').property('name','IBM')
g.V('c2').property('questId','123457')
//create an university Princeton
g.addV('univeristy').property(id,'u1')
g.V('u1').property('name','Princeton')
//create edges
//Coca Cola employes John
g.addE('employeed').from(g.V('c1')).to(g.V('p1')).property(id,'c1p1')
g.E('c1p1').property('fromDate','2009-12-13').next()
g.E('c1p1').property('toDate','2019-12-13').next()
//Coca Cola employes Jack
g.addE('employeed').from(g.V('c1')).to(g.V('p2')).property(id,'c1p2')
g.E('c1p2').property('fromDate','2006-12-13').next()
g.E('c1p2').property('toDate','2009-12-12').next()
//Coca Cola employed Jerry our employee
g.addE('employeed').from(g.V('c1')).to(g.V('p5')).property(id,'c1p5')
g.E('c1p5').property('fromDate','2007-12-13').next()
g.E('c1p5').property('toDate','2008-12-13').next()
//IBM employees John
g.addE('employeed').from(g.V('c2')).to(g.V('p1')).property(id,'c2p1')
g.E('c2p1').property('fromDate','2006-12-13').next()
g.E('c2p1').property('toDate','2009-12-13').next()
//IBM employes Mary
g.addE('employeed').from(g.V('c2')).to(g.V('p3')).property(id,'c2p3')
g.E('c2p3').property('fromDate','2006-10-11').next()
g.E('c2p3').property('toDate','2009-10-11').next()
//Jim our employee knows Mary
g.addE('knows').from(g.V('p4')).to(g.V('p3')).property(id,'p4p3')
//Jerry our employee knows John
g.addE('knows').from(g.V('p5')).to(g.V('p1')).property(id,'p5p1').next()
//Mary knows Jack
g.addE('knows').from(g.V('p3')).to(g.V('p2')).property(id,'p3p2')
//Jerry our employee knows Jack's Brother
g.addE('knows').from(g.V('p5')).to(g.V('p6')).property(id,'p5p6')
//Jack's Brother is Jack's Brother (family relation)
g.addE('family').from(g.V('p6')).to(g.V('p2')).property(id,'p6p2')
//Jeane our employee got a degree from Princeton
g.addE('degree_from').from(g.V('p7')).to(g.V('u1')).property(id,'p7u1')
g.E('p7u1').property('class_of',1989)
//Jack got a degree from Princeton
g.addE('degree_from').from(g.V('p2')).to(g.V('u1')).property(id,'p2u1')
g.E('p2u1').property('class_of',1989)
g.V().has('isSSEmployee',true).
repeat(bothE().otherV().simplePath()).
until(has('name','Jack')).
path().
by('name').
by(label)
如果我没有正确理解您的规则，那么这应该是您要查找的查询：
gremlin> g.V().has('person','name','Jack').
bothE('knows','employeed').otherV().
sideEffect(hasLabel('company').aggregate('e')).barrier().
repeat(bothE('knows','employeed').otherV().simplePath()).
until(has('isSSEmployee',true).and().
not(__.in('employeed').where(within('e')))).
path().
by('name').
by(label).
map {it.get().objects().reverse()}
==>[Jim,knows,Mary,knows,Jack]
==>[Jim,knows,Mary,employeed,IBM,employeed,John,employeed,Coca Cola,employeed,Jack]
==>[Jim,knows,Mary,employeed,IBM,employeed,John,knows,Jerry,employeed,Coca Cola,employeed,Jack]
但是，这与您的预期结果不符，所以我认为我们仍然需要一些说明（一旦弄清楚，我会更新答案）。
UPDATE（路径必须包含未受雇于 Jack 的任何员工的人员）：
gremlin> g.V().has('person','name','Jack').
......1> sideEffect(__.in('employeed').aggregate('c')).barrier().
......2> repeat(bothE('knows','employeed').as('e').otherV().simplePath().as('p')).
......3> emit(__.and(has('isSSEmployee',true),
......4> select(all,'e').and(unfold().hasLabel('knows'),
......5> count(local).is(gt(1))))).
......6> filter(select(all, 'p').unfold().
......7> not(__.in('employeed').where(within('c')))).
......8> path().
......9> by('name').
.....10> by(label).
.....11> map {it.get().objects().reverse()}
==>[Jim,knows,Mary,knows,Jack]
==>[Jerry,knows,John,employeed,Coca Cola,employeed,Jack]
==>[Jerry,knows,John,employeed,IBM,employeed,Mary,knows,Jack]
==>[Jerry,employeed,Coca Cola,employeed,John,employeed,IBM,employeed,Mary,knows,Jack]
==>[Jim,knows,Mary,employeed,IBM,employeed,John,employeed,Coca Cola,employeed,Jack]
==>[Jim,knows,Mary,employeed,IBM,employeed,John,knows,Jerry,employeed,Coca Cola,employeed,Jack]
这是我得出的正确答案
g.V().has('person','name','Jack').
repeat(bothE().has(label,neq('family')).otherV().hasNot('isSSEmployee').simplePath()).
until(has(label,'person')).
bothE().otherV().has('isSSEmployee',true).path().
by('name').
by(label)
我正在尝试提出一个 gremlin 查询，我可以在其中推断出谁认识 Jack，但不包括直接来自 Employees 的信息。
谁可以在 Jack 上寻求参考？ 不能从其他员工或受试者的家庭亲属那里获得推荐信。
查询应该return:
吉姆认识玛丽认识杰克
Jim 知道 Mary 受雇于 IBM 受雇于 John 受雇于 Coca Cola 受雇于 Jack
Jerry Knows John Employed Coca Cola employedd Jack
但不应该包括：
Jerry Knows Jacks 的兄弟家族 Jack
Jerry 雇用可口可乐雇用 Jack
珍妮Degree_from普林斯顿Degree_From杰克
见下图：
我提出了以下查询，但无法获得从路径中排除 Jerry 或 jane 的部分。
g.V().has('isSSEmployee',true).repeat(bothE('knows','employeed').otherV().simplePath()).until(has('name','Jack')).has('isSSEmployee',false).path().by('name').by(label)
加载数据的脚本
//only for tinkerpop
graph=TinkerGraph.open()
g=graph.traversal()
//from here on common between neptune and tinkerpop
//Setup started
//create vertexes
//create a person John
g.addV('person').property(id,'p1')
g.V('p1').property('questId','123456')
g.V('p1').property('name','John')
//create a person Jack
g.addV('person').property(id,'p2')
g.V('p2').property('questId','123457')
g.V('p2').property('name','Jack')
//create a person Mary
g.addV('person').property(id,'p3')
g.V('p3').property('questId','123458')
g.V('p3').property('name','Mary')
//create a person Jim and mark him as ssemployee
g.addV('person').property(id,'p4')
g.V('p4').property('name','Jim')
g.V('p4').property('isSSEmployee',true)
g.V('p4').property('questId','1234569')
//create a person Jerry and mark him as ssemployee
g.addV('person').property(id,'p5')
g.V('p5').property('name','Jerry')
g.V('p5').property('questId','12345700')
g.V('p5').property('isSSEmployee',true)
//create a person Jack's Brother
g.addV('person').property(id,'p6')
g.V('p6').property('name',"Jack's Brother")
g.V('p6').property('questId','1234')
//create Jeane our employee
g.addV('person').property(id,'p7')
g.V('p7').property('name', 'Jeane')
g.V('p7').property('questId', '1234580')
g.V('p7').property('isSSEmployee', true)
//create a company Coca Cola
g.addV('company').property(id,'c1')
g.V('c1').property('name','Coca Cola')
g.V('c1').property('questId','123456')
//create a company IBM
g.addV('company').property(id,'c2')
g.V('c2').property('name','IBM')
g.V('c2').property('questId','123457')
//create an university Princeton
g.addV('univeristy').property(id,'u1')
g.V('u1').property('name','Princeton')
//create edges
//Coca Cola employes John
g.addE('employeed').from(g.V('c1')).to(g.V('p1')).property(id,'c1p1')
g.E('c1p1').property('fromDate','2009-12-13').next()
g.E('c1p1').property('toDate','2019-12-13').next()
//Coca Cola employes Jack
g.addE('employeed').from(g.V('c1')).to(g.V('p2')).property(id,'c1p2')
g.E('c1p2').property('fromDate','2006-12-13').next()
g.E('c1p2').property('toDate','2009-12-12').next()
//Coca Cola employed Jerry our employee
g.addE('employeed').from(g.V('c1')).to(g.V('p5')).property(id,'c1p5')
g.E('c1p5').property('fromDate','2007-12-13').next()
g.E('c1p5').property('toDate','2008-12-13').next()
//IBM employees John
g.addE('employeed').from(g.V('c2')).to(g.V('p1')).property(id,'c2p1')
g.E('c2p1').property('fromDate','2006-12-13').next()
g.E('c2p1').property('toDate','2009-12-13').next()
//IBM employes Mary
g.addE('employeed').from(g.V('c2')).to(g.V('p3')).property(id,'c2p3')
g.E('c2p3').property('fromDate','2006-10-11').next()
g.E('c2p3').property('toDate','2009-10-11').next()
//Jim our employee knows Mary
g.addE('knows').from(g.V('p4')).to(g.V('p3')).property(id,'p4p3')
//Jerry our employee knows John
g.addE('knows').from(g.V('p5')).to(g.V('p1')).property(id,'p5p1').next()
//Mary knows Jack
g.addE('knows').from(g.V('p3')).to(g.V('p2')).property(id,'p3p2')
//Jerry our employee knows Jack's Brother
g.addE('knows').from(g.V('p5')).to(g.V('p6')).property(id,'p5p6')
//Jack's Brother is Jack's Brother (family relation)
g.addE('family').from(g.V('p6')).to(g.V('p2')).property(id,'p6p2')
//Jeane our employee got a degree from Princeton
g.addE('degree_from').from(g.V('p7')).to(g.V('u1')).property(id,'p7u1')
g.E('p7u1').property('class_of',1989)
//Jack got a degree from Princeton
g.addE('degree_from').from(g.V('p2')).to(g.V('u1')).property(id,'p2u1')
g.E('p2u1').property('class_of',1989)
g.V().has('isSSEmployee',true).
repeat(bothE().otherV().simplePath()).
until(has('name','Jack')).
path().
by('name').
by(label)
如果我没有正确理解您的规则，那么这应该是您要查找的查询：
gremlin> g.V().has('person','name','Jack').
bothE('knows','employeed').otherV().
sideEffect(hasLabel('company').aggregate('e')).barrier().
repeat(bothE('knows','employeed').otherV().simplePath()).
until(has('isSSEmployee',true).and().
not(__.in('employeed').where(within('e')))).
path().
by('name').
by(label).
map {it.get().objects().reverse()}
==>[Jim,knows,Mary,knows,Jack]
==>[Jim,knows,Mary,employeed,IBM,employeed,John,employeed,Coca Cola,employeed,Jack]
==>[Jim,knows,Mary,employeed,IBM,employeed,John,knows,Jerry,employeed,Coca Cola,employeed,Jack]
但是，这与您的预期结果不符，所以我认为我们仍然需要一些说明（一旦弄清楚，我会更新答案）。
UPDATE（路径必须包含未受雇于 Jack 的任何员工的人员）：
gremlin> g.V().has('person','name','Jack').
......1> sideEffect(__.in('employeed').aggregate('c')).barrier().
......2> repeat(bothE('knows','employeed').as('e').otherV().simplePath().as('p')).
......3> emit(__.and(has('isSSEmployee',true),
......4> select(all,'e').and(unfold().hasLabel('knows'),
......5> count(local).is(gt(1))))).
......6> filter(select(all, 'p').unfold().
......7> not(__.in('employeed').where(within('c')))).
......8> path().
......9> by('name').
.....10> by(label).
.....11> map {it.get().objects().reverse()}
==>[Jim,knows,Mary,knows,Jack]
==>[Jerry,knows,John,employeed,Coca Cola,employeed,Jack]
==>[Jerry,knows,John,employeed,IBM,employeed,Mary,knows,Jack]
==>[Jerry,employeed,Coca Cola,employeed,John,employeed,IBM,employeed,Mary,knows,Jack]
==>[Jim,knows,Mary,employeed,IBM,employeed,John,employeed,Coca Cola,employeed,Jack]
==>[Jim,knows,Mary,employeed,IBM,employeed,John,knows,Jerry,employeed,Coca Cola,employeed,Jack]
这是我得出的正确答案
g.V().has('person','name','Jack').
repeat(bothE().has(label,neq('family')).otherV().hasNot('isSSEmployee').simplePath()).
until(has(label,'person')).
bothE().otherV().has('isSSEmployee',true).path().
by('name').
by(label)