java - Match NOT in Tinkerpop traversal -


i'd match b1, no node b2 exists depicted below. enter image description here kind of 2-hop asymmetry. note non-existent b2 must link same c:c b1, other (a2) --> (b3) --> (a1) can ignored long b3 has no edge c:c.

i tried following...

graph.traversal().v()     .match(         as("c").haslabel("c"),          as("a").haslabel("a")         .out().haslabel("b").as("b1")         .where(out().as("c"))         .out().haslabel("a").as("a2"),          not(         as("a2")         .out().haslabel("b").as("b2")         .where(out().as("c")         .out().as("a1")     ) )

...but second match step throws exception:

java.lang.illegalstateexception: provided match pattern unsolvable: [[matchstartstep(a1), hasstep([~label.eq(a)]), vertexstep(out,vertex), hasstep([~label.eq(b)]), wheretraversalstep([wherestartstep, vertexstep(out,vertex), whereendstep(c)])@[b1], vertexstep(out,vertex), hasstep([~label.eq(a)]), matchendstep(a2)], [matchstartstep(a2), hasstep([~label.eq(b)]), wheretraversalstep([notstep([wherestartstep, vertexstep(out,edge)])]), wheretraversalstep([wherestartstep, vertexstep(out,vertex)@[c], vertexstep(out,vertex), whereendstep(a1)])@[b2], matchendstep, matchendstep]]

i can't match cypher, unfortunately.

suddenly, found solution avoids scope problems:

graph.traversal().v()     as("a").haslabel("a")     .out().haslabel("b").as("b1")     .where(out().haslabel("c"))     .out().haslabel("a").as("a2")     .where(         not(             out().haslabel("b")                 .where(out().haslabel("c"))                 .out().as("a")         )     )

Comments

Post a Comment

Popular posts from this blog

python - Operations inside variables -

this question has answer here: evaluating mathematical expression in string 12 answers i want make python script finds maximum possible output using 2 operations. problem says there syntax error @ 2nd last line, in b in there. how can fix it? x = 0 = int(input("1st number:")) c = int(input("2nd number:" )) e = int(input("3rd number:" )) in range(4): if == 0: b = "+" elif == 1: b = "-" elif == 2: b = "/" else: b = "*" j in range(4): if j == 0: d = "+" elif j == 1: d = "-" elif j == 2: d = "/" else: d = "*" k = b c d e print(k) if you're after variables e on single line, couldn...
Read more

Generic Map Parameter java -

i having lot of methods of type: public static list<eduusers> getdetails(class c, map<string, integer> params) { list<eduusers> details = null; session session = hibernateutil.getsessionfactory().opensession(); transaction tx = null; try { tx = session.begintransaction(); criteria cr = session.createcriteria(c); (map.entry<string, integer> entry : params.entryset()) { cr.add(restrictions.eq(entry.getkey(), entry.getvalue())); } details = cr.list(); tx.commit(); } catch (exception asd) { log.debug(asd.getmessage()); if (tx != null) { tx.commit(); } } { session.close(); } return details; } i trying have generic method them , have written far: public static <t> list<t> getdetails(class<t> c, class<t> b) { l...
Read more

arrays - What causes a java.lang.ArrayIndexOutOfBoundsException and how do I prevent it? -

what arrayindexoutofboundsexception mean , how rid of it? here code sample triggers exception: string[] name = {"tom", "dick", "harry"}; for(int = 0; i<=name.length; i++) { system.out.print(name[i] +'\n'); } your first port of call should documentation explains reasonably clearly: thrown indicate array has been accessed illegal index. index either negative or greater or equal size of array. so example: int[] array = new int[5]; int boom = array[10]; // throws exception as how avoid it... um, don't that. careful array indexes. one problem people run thinking arrays 1-indexed, e.g. int[] array = new int[5]; // ... populate array here ... (int index = 1; index <= array.length; index++) { system.out.println(array[index]); } that miss out first element (index 0) , throw exception when index 5. valid indexes here 0-4 inclusive. correct, idiomatic for statement here be: for (int index = 0; index <...
Read more