您可以使用纯Cypher方法，如下所示：

FOREACH (i IN range(0, size($module_path) - 2) | 
           MERGE (l:Module {name: $module_path[i]})
           MERGE (r:Module {name: $module_path[i+1]})          
           MERGE (l)-[:contains]->(r) );

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你可以用问题中的数据来测试这是如何工作的：

UNWIND [['a','b','c','d'],['a','b','c','e'],['a'],['a','b','c'],
        ['a','b']] AS modules
FOREACH (i IN range(0, size(modules) - 2) | 
            MERGE (l:Module {name: modules[i]})
            MERGE (r:Module {name: modules[i+1]})          
            MERGE (l)-[:contains]->(r) )

型
结果如下：

它可以通过以下查询实现（添加了注解）

// simulate your input
WITH "a.b.c.d" AS pkg 
// split on the dot, producing an array 
WITH split(pkg, ".") AS parts ['a','b','c','d']
// iterating of the array
UNWIND parts AS part 
// MERGE (match or create) the module with the name as id ( up to you which property to use )
MERGE (m:Module {id: part}) 
// break and introduce a variable being an array of the nodes created or matched
WITH collect(m) AS modules 
// use APOC to link those nodes together with the NEXT relationship
CALL apoc.nodes.link(modules,'NEXT')

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它产生这样的图形

的数据
稍后，如果您尝试对已经存在的模块路径执行相同的操作，它将不会执行任何操作，例如，在执行上述操作之后，以下查询将不会创建任何内容

WITH "a.b.c" AS pkg
WITH split(pkg, ".") AS parts
UNWIND parts AS part
MERGE (m:Module {id: part})
WITH collect(m) AS modules
CALL apoc.nodes.link(modules,'NEXT')

型

的