Question

这与此问题有关，但不完全相同：Performing a MATCH on all nodes on a path

与上一个问题一样，我也有类似的设置：

(a)-->(b)-->(c)-->(d)
 ^     ^     ^     ^
 |r    |r    |r    |r
(aa)  (bb)  (cc)  (dd)

示例创建（请参阅控制台http://console.neo4j.org/r/vx99q）：

CREATE 
    (a {name:'a'}),
    (b {name:'b'}),
    (c {name:'c'}),
    (d {name:'d'}),
    (aa {name:'aa'}),
    (bb {name:'bb'}),
    (cc {name:'cc'}),
    (dd {name:'dd'}),
    a-[:rel1]->b,
    b-[:rel1]->c,
    c-[:rel1]->d,
    a-[:r]->aa,
    b-[:r]->bb,
    c-[:r]->cc,
    d-[:r]->dd;

假设我有a为7，d为4.我会得到这样的路径：

START a=node(7), d=node(4) 
MATCH path = a-[:rel1*]->d 
RETURN path

我想要的是aa，bb，cc和dd。

我很难过。

Answer 1

我花了很多时间围绕着这个，但我想通了。

START n=node(*), a=node(1), z=node(4) 
MATCH path = a-[:rel1*]->z 
WITH n, path 
MATCH n-[:r]->o 
WHERE n IN path 
RETURN o

虽然看起来确实很慢，但我认为可能有更好的方法。

输出：

o
(5 {name:"aa"})
(6 {name:"bb"})
(7 {name:"cc"})
(8 {name:"dd"})

编辑： yhw42提到如果像这样修改，查询可以运行得更快：

START a=node(1), z=node(4) 
MATCH path = a-[:rel1*]->z 
WITH path 
MATCH n-[:r]->o 
WHERE n IN path 
RETURN o

在样本中，确实如此。我的方式是：

Query Results

+--------------------+
| o                  |
+--------------------+
| Node[5]{name:"aa"} |
| Node[6]{name:"bb"} |
| Node[7]{name:"cc"} |
| Node[8]{name:"dd"} |
+--------------------+
4 rows
344 ms

Execution Plan

ColumnFilter(symKeys=["n", "path", "o", "  UNNAMED88"], returnItemNames=["o"], _rows=4, _db_hits=0)
PatternMatch(g="(n)-['  UNNAMED88']-(o)", _rows=4, _db_hits=0)
  Filter(pred="any(-_-INNER-_- in path where n == -_-INNER-_-)", _rows=4, _db_hits=0)
    ColumnFilter(symKeys=["path", "n", "a", "  UNNAMED53", "z"], returnItemNames=["n", "path"], _rows=9, _db_hits=0)
      ExtractPath(name="path", patterns=["  UNNAMED53=a-[:rel1*]->z"], _rows=9, _db_hits=0)
        PatternMatch(g="(a)-['  UNNAMED53']-(z)", _rows=9, _db_hits=0)
          NodeById(name="Literal(List(4))", identifier="z", _rows=9, _db_hits=9)
            NodeById(name="Literal(List(1))", identifier="a", _rows=9, _db_hits=9)
              AllNodes(identifier="n", _rows=9, _db_hits=9)

他的方式是：

Query Results

+--------------------+
| o                  |
+--------------------+
| Node[5]{name:"aa"} |
| Node[6]{name:"bb"} |
| Node[7]{name:"cc"} |
| Node[8]{name:"dd"} |
+--------------------+
4 rows
174 ms

Execution Plan

ColumnFilter(symKeys=["path", "n", "o", "  UNNAMED74"], returnItemNames=["o"], _rows=4, _db_hits=0)
PatternMatch(g="(n)-['  UNNAMED74']-(o)", _rows=4, _db_hits=0)
  Filter(pred="any(-_-INNER-_- in path where n == -_-INNER-_-)", _rows=4, _db_hits=0)
    AllNodes(identifier="n", _rows=9, _db_hits=9)
      ColumnFilter(symKeys=["a", "z", "  UNNAMED42", "path"], returnItemNames=["path"], _rows=1, _db_hits=0)
        ExtractPath(name="path", patterns=["  UNNAMED42=a-[:rel1*]->z"], _rows=1, _db_hits=0)
          PatternMatch(g="(a)-['  UNNAMED42']-(z)", _rows=1, _db_hits=0)
            NodeById(name="Literal(List(4))", identifier="z", _rows=1, _db_hits=1)
              NodeById(name="Literal(List(1))", identifier="a", _rows=1, _db_hits=1)

Neo4J从路径中的节点获取节点

1 个答案: