目前我有这个gremlin/groovy代码:
if(!g.V().has("Number","number","3").hasNext()) {
g.addV("Number").property("number","3")
}
可以在不使用多个遍历的情况下具有相同的结果?
我尝试了此操作,但它不起作用(它不添加数字或用户顶点)
g.V().choose(has("Number","number", "3"),
addV("Number").property("number", "3"),
has("Number","number", "3")
).as("number")
.addV("User").property("uuid","test uuuid")
.forEachRemaining(System.out::println);
我尝试了这里建议的内容(https://stackoverflow.com/a/33965737/986160),但它不允许我继续在DSE的单个交易中添加另一个用户:
g.V()
.has("Number","number", "3")
.tryNext()
.orElseGet(
() -> g.addV("Number")
.property("number", "3").next()
);
谢谢!
不幸的是我们还没有g.coalesce()
,但是有一个解决方法:
gremlin> g = TinkerGraph.open().traversal()
==>graphtraversalsource[tinkergraph[vertices:0 edges:0], standard]
gremlin> g.inject(1).coalesce(V().has("Number", "number", 3), addV("Number").property("number", 3))
==>v[0]
gremlin> g.inject(1).coalesce(V().has("Number", "number", 3), addV("Number").property("number", 3))
==>v[0]
gremlin> g.inject(1).coalesce(V().has("Number", "number", 3), addV("Number").property("number", 3))
==>v[0]
gremlin> g.V().valueMap()
==>[number:[3]]