访问连接的顶点在OrientDB JavaScript函数(Accessing Connected

我试图在OrientDB工作室使用的生成在功能组工作站不在一个人使用。让这些顶点查询工作正常，但我试图避免导线，因为它是很慢 - 太慢了在生产中使用。而不是通过每个自由站迭代，并与所有它的邻居们一起分组它，保持每个分组“命名”在设定最小@rid。

var groups = {};            //The list of groups of workpoints. The key is the lowest RID in the group
var mappedDesks = {};       //Every desk's RID is in this object with it's matching value being the group name they're in    

//Get all Workpoints that don't have a Locale CURRENTLY_LOCATED_ON them
var freeDesks = db.query("SELECT FROM Workpoint WHERE @rid NOT IN (SELECT @rid FROM (SELECT EXPAND(OUT('CURRENTLY_LOCATED_ON').OUT('LOCATED_ON')) FROM Person) WHERE @class = 'Workpoint')");    

//Iterate through all vacant Workpoints
for (var j=0; j < freeDesks.length; j++){
    var baseNodeRid = freeDesks[j].getRecord().getIdentity().toString();                    // The RID of the Workpoint
    var baseNodeNumber = parseFloat(baseNodeRid.replace("#", "").replace(":","."));         // The RID converted to a number for comparisons. The lower RID takes precedence
    var baseSanitized = baseNodeRid.replace(":","-")                                        // Keys cannot contain colon so they are replaced with a dash    

    if (freeDesks[j].getRecord().field("out_NEIGHBOUR_OF") == null ) {
        // Desks without neighbours can be put in a group on their own
        groups[baseSanitized] = new Array();
        groups[baseSanitized].push(baseNodeRid);
        mappedDesks[baseSanitized] = baseSanitized;
    } else {
        //Iterate over all the desk's neighbours
        for (var n = 0; n < freeDesks[j].getRecord().field("out_NEIGHBOUR_OF").length; n++){          

            //Convert the neighbour's RID to a number too
            var nSanitized = n.replace(":","-");

            if (parseFloat(n.replace("#", "").replace(":",".")) > baseNodeNumber ){
                //The neighbour's node group is larger than the current one. This needs to be merged into the group with the smaller rid    

                //Move the desks from the neighbour's group into the base's group. If it has none then do nothing
                var nGroup = groups[mappedDesks[nSanitized]]
                if ( nGroup != null) {
                    groups[baseSanitized] = groups[baseSanitized].concat(nGroup);

                    //Change the mapping of each moved desk to the base's
                    for (var g = 0; g < nGroup.length; g++){
                        mappedDesks[nGroup[g]] = baseSanitized;          
                    }
                }    



                //Delete the reference to the old group
                delete groups[mappedDesks[nSanitized]];    

                //Update the mappings for the desks dealt with
                mappedDesks[nSanitized] = baseSanitized;
                mappedDesks[baseSanitized] = baseSanitized;    

            } else {
                // The neighbour is lower than the current desk so the desk should be merged into the neighbour's group
                mappedDesks[baseSanitized] = nSanitized;
                groups[nSanitized].push(baseNodeRid);
            }
        }
    }    
}    

return groups;

我的问题来自于访问一个顶点的邻居。它正确地确定是否有if语句邻居return freeDesks[j].getRecord().field("out_NEIGHBOUR_OF")但我希望能够得到每个邻居的@rid这样我就可以在@rids成组进行排序。

freeDesks[j].getRecord().field("out_NEIGHBOUR_OF")返回边缘记录，但我不似乎能够得到“中”或使用该字段（）方法“走出去”领域（不是这个对象上找到）或访问它作为数组[]。

[
    {
        "@type": "d",
        "@rid": "#34:18176",
        "@version": 6,
        "@class": "NEIGHBOUR_OF",
        "out": "#16:13",
        "in": "#16:1408",
        "@fieldTypes": "out=x,in=x"
    }
]

你能帮助是让邻居列表/阵列@rids这样我就可以在他们重复的代码的其他人呢？

干杯!

一个简单的例子来理解你可以做什么：

create class Person extends V    
create class IsNeighbour extends E

create vertex Person set name = 'P1'              // 12:0
create vertex Person set name = 'P2'              // 12:1
create vertex Person set name = 'P3'              // 12:2

create edge IsNeighbour from #12:0 to #12:1
create edge IsNeighbour from #12:0 to #12:2

这个定义JavaScript函数：

var gdb = orient.getGraphNoTx();
var v = gdb.command("sql", "select from " + personRID);
var neighbours = v[0].getRecord().field("out_IsNeighbour").iterator();
var result = [];

while(neighbours.hasNext()){
  var neighbour = neighbours.next();
  result.push(neighbour.field("in"));
}

return result;

像这样：