【发布时间】:2019-11-26 09:52:51
【问题描述】:
我想做一个请求,它不起作用:
这里是代码(在 NEO4J 沙箱中,csv 文件在谷歌表格中,可公开访问):
LOAD CSV WITH HEADERS FROM 'https://docs.google.com/spreadsheets/d/17UhQ3m2dE3wS4IsBjgv0ozcW_zH2rSBs2ZUdqeT4cQw/export?format=csv&id=17UhQ3m2dE3wS4IsBjgv0ozcW_zH2rSBs2ZUdqeT4cQw&gid=726094387' AS line
MERGE (cp:cp {Name:line.cp})
MERGE (commune:commune {Name:line.commune})
MERGE (voie:voie {Name:line.rao_libelle_voie})
MERGE (adr:adr {Name:line.adr_lib_ap_ocr_postal_0})
MERGE (name:name {Name:line.adr_l1_name})
MERGE (cp) -[:TO {dist:line.count} ]-> (commune)
MERGE (commune) -[:TO {dist:line.count} ]-> (voie)
MERGE (voie) -[:TO {dist:line.count} ]-> (adr)
MERGE (adr) -[:TO {dist:line.count} ]-> (name)
这是我的请求(我想要(rao_libelle_voie,变量:voie)中一条街道的居民姓名(adr_l1_name,变量:name);
这是 SQL 请求(来自其他来源):
SELECT `adr_l1_name`
FROM `database`
WHERE ` rao_libelle_voie ` LIKE '%RUE DE GLAIRE%'
结果有效;
在 NEO4J 沙箱中,我提出这个请求:
MATCH (n:name)
WHERE (n:voie) = 'RUE DE GLAIRE'
RETURN (n:name)
没有错误信息,但没有出现结果;
【问题讨论】:
标签: neo4j where-clause sql-like