下面的创建节点实例,请教一下CALL apoc. cypher .doIt如何创建多个标签?现在的方式是只能指定一个标签!


    
     UNWIND
      
[{name:"sdasdsad234fdgsasdfas33",properties:{born:1978},labels:"ables213"},{name:"ssds12ad23489gsasdfas33",properties:{born:1978,sex:"男性",age:22},labels:"label31"}] AS row 
CALL apoc.
    
     cypher
    .doIt('MERGE (n:`' + row.labels + '` {name: {name}}) SET n += {properties}', {properties: row.properties, name: row.name}) YIELD value 
RETURN 1

解决方案一:

写在set里面,set n += {properties},n:human

解决方案讨论:

@zhoujieren64 用参数的形式传递怎么写呢? 例如这个会报错:


    
     UNWIND
      
[{name:"sdasdsad234fdgsasdfas33",properties:{born:1978},labels:"ables213",labelTest:"test"},{name:"ssds12ad23489gsasdfas33",properties:{born:1978,sex:"男性",age:22},labels:"label31",labelTest:"test"}] AS row 
CALL apoc.
    
     cypher
    .doIt('MERGE (n:`' + row.labels + '` {name: {name}}) SET n += {properties},n:{labelTest}', {properties: row.properties, name: row.name,labelTest:row.labelTest}) YIELD value 
RETURN 1
原文地址:http://neo4j.com.cn/topic/5c4e7ea5cd4dafa110f1c6f1

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