[英]Accessing Connected Vertices in OrientDB Javascript Functions
我正在尝试使用OrientDB studio中的“函数”中的内置功能将未使用的Workstation分组。 获取这些顶点的查询工作正常,但是我试图避免使用Traverse,因为它非常慢-太慢而无法用于生产中。 与其遍历每个免费电台并将其与所有邻居分组在一起,不如将每个分组的“命名”与集合中最小的@rid保持在一起。
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;
我的问题来自访问顶点的邻居。 它可以正确确定if语句中是否有邻居return freeDesks[j].getRecord().field("out_NEIGHBOUR_OF")
但我希望能够获得每个邻居的@rid,因此可以将@rid分为几类。
freeDesks[j].getRecord().field("out_NEIGHBOUR_OF")
返回边缘记录,但是我似乎无法使用field()方法获取“ in”或“ out”字段(在此对象上找不到)或将其作为数组[]访问。
[
{
"@type": "d",
"@rid": "#34:18176",
"@version": 6,
"@class": "NEIGHBOUR_OF",
"out": "#16:13",
"in": "#16:1408",
"@fieldTypes": "out=x,in=x"
}
]
您能否帮助获得邻居@rids的列表/数组,以便我可以用其余代码遍历它们?
干杯!
一个简单的示例,以了解您可以做什么:
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
定义此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:
然后您可以:
select getNeighbours("#12:0")
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