修改密码查询
修改密码查询
提问于 2017-08-02 05:28:43
我有如下所示的neo4j数据：
这里我有COUNTEY_PROVINCE关系，country和province，PROVINCE_CITY关系，province和city，COUNTRY_CITY关系，country和city。当用户用country name调用api时，我想用city返回所有province。为此，我运行了以下查询：
MATCH path=(cn:Country { name: "Bangladesh" })-[:COUNTRY_PROVINCE]->(pv:Province)-[:PROVINCE_CITY]->(ct:City)
RETURN { x: nodes(path) }我得到了以下结果：(国家、省、市)：
{ x: nodes(path) }
{ "x": [ { "name": "Bangladesh"}, { "name": "Dhaka" }, { "name": "Dhaka" } ] }
{ "x": [ { "name": "Bangladesh"}, { "name": "Dhaka" }, { "name": "Narayanganj" } ] }
{ "x": [ { "name": "Bangladesh"}, { "name": "Sylhet" }, { "name": "Sylhet" } ] }
{ "x": [ { "name": "Bangladesh"}, { "name": "Khulna" }, { "name": "Khulna" } ] }
{ "x": [ { "name": "Bangladesh"}, { "name": "Khulna" }, { "name": "Jessore" } ] }
{ "x": [ { "name": "Bangladesh"}, { "name": "Chittagong" }, { "name": "Chittagong" } ] }
{ "x": [ { "name": "Bangladesh"}, { "name": "Chittagong" }, { "name": "Comilla" } ] }现在我的问题是，我怎样才能得到与省和联系城市的国家名单，例如：
[
{
country: {
name: "Bangladesh",
province: [
{
name: "Dhaka",
city: [
{ name: "Dhaka" },
{ name: "Narayanganj" }
]
},
{
name: "Sylhet",
city: [
{ name: "Sylhet" }
]
},
{
name: "Chittagong",
city: [
{ name: "Chittagong" },
{ name: "Comilla" }
]
},
{
name: "Khulna",
city: [
{ name: "Khulna" },
{ name: "Jessore" }
]
}
]
}
}
]回答 2
Stack Overflow用户
回答已采纳
发布于 2017-08-02 11:35:09
我在这里模拟了你的情景。
加载初始数据集(类似于问题中描述的数据集)：
CREATE (c:Country {name:"Country A"})
CREATE (p1:Province {name:"Province A"})
CREATE (p2:Province {name:"Province B"})
CREATE (c1:City {name:"City A"})
CREATE (c2:City {name:"City B"})
CREATE (c3:City {name:"City C"})
CREATE (c4:City {name:"City D"})
CREATE (c)-[:COUNTRY_PROVINCE]->(p1)
CREATE (p1)-[:PROVINCE_CITY]->(c1)
CREATE (p1)-[:PROVINCE_CITY]->(c2)
CREATE (c)-[:COUNTRY_PROVINCE]->(p2)
CREATE (p2)-[:PROVINCE_CITY]->(c3)
CREATE (p2)-[:PROVINCE_CITY]->(c4)查询：
MATCH (cn:Country { name: "Country A" })-[:COUNTRY_PROVINCE]->(pv:Province)-[:PROVINCE_CITY]->(ct:City)
WITH cn, pv, collect({name : ct.name}) as cities
RETURN {coutry : {name : cn.name, province : collect( distinct { name:pv.name, city :cities }) } }结果：
{
"coutry":{
"name":"Country A",
"province":[
{
"name":"Province A",
"city":[
{
"name":"City B"
},
{
"name":"City A"
}
]
},
{
"name":"Province B",
"city":[
{
"name":"City D"
},
{
"name":"City C"
}
]
}
]
}
}Stack Overflow用户
发布于 2017-08-02 05:36:51
好吧，如果你要返回路径的节点，很明显这就是你要得到的。你有几个不同的选择：
- 例如，查看一下apoc.convert.toTree，托杰森可能会使您非常接近您想要的内容。
- 自己的事情自己做。从任何客户端应用程序和进程(+格式)自己发送查询结果(用Java、Python、C#、.)
- ..。
希望这能有所帮助。
你好，汤姆
页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/45452169
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