在 OrientDb 中查找和删除重复的边

Question

假设我们有 Vertex User 和 Edge FriendsWith。 FriendsWith 可以双向出入（通常是出入或在两个用户之间）。

重复是指从一个用户到另一个用户的外出或进入超过一次（外出和进入一起不被视为重复）

有没有办法找到重复的边缘并删除它们？

更新添加图片说明问题

谢谢。

Answer 1

这是我的 javascript 函数：

var g=orient.getGraph();
var C=g.command('sql','select from FriendsWith');
var arr = new Array(C.length);
var toRemove = new Array();

for(i=0;i<C.length;i++){
  var found = false;
  for (x = 0; x < i+1 && !found; x++) {
  if (arr[x] === C[i].getProperty("out").getId()+" "+C[i].getProperty("in").getId()) {
    found = true;
    toRemove.push(C[i].getId());
    }
  }
  arr[i] = C[i].getProperty("out").getId()+" "+C[i].getProperty("in").getId();
}

for(a=0;a<toRemove.length;a++){
  var C=g.command('sql','delete edge '+toRemove[a]);
}

希望对您有所帮助。 再见

Answer 2

你可以试试这个功能

var g=orient.getGraph();
var friends=g.command("sql","select from FriendsWith");
var paths=[];
for(i=0;i<friends.length;i++){
    paths.push(friends[i]);
}
for(i=0;i<paths.length;i++){
    var myEdge=paths[i];
    var vIn=myEdge.getProperty("in").getId();
    var vOut=myEdge.getProperty("out").getId();
    for(j=0;j<paths.length;j++){
        if(i<j){
            var edge=paths[j];
            var vInCopy=edge.getProperty("in").getId();
            var vOutCopy=edge.getProperty("out").getId();
            if((vIn==vInCopy && vOut==vOutCopy) || (vIn==vOutCopy && vOut==vInCopy)){
                g.command("sql","delete edge FriendsWith where @rid="+edge.getId());
                paths.splice(j, 1);
                j--;
            } 
        }
    }
}

之前 后

Answer 3

我创建了一个小型数据库来尝试您的案例。这是我的代码：

create class User extends V
create class follows extends E

create property User.id integer
create property User.name String

create vertex User set id=1, name="Paul"
create vertex User set id=2, name="John"
create vertex User set id=3, name="Mark"
create vertex User set id=4, name="Robert"

create edge follows from (select from User where id=1) to (select from User where id=2)
create edge follows from (select from User where id=2) to (select from User where id=1)
create edge follows from (select from User where id=1) to (select from User where id=3)
create edge follows from (select from User where id=2) to (select from User where id=3)
create edge follows from (select from User where id=3) to (select from User where id=2)
create edge follows from (select from User where id=3) to (select from User where id=4)

图表：

然后我创建了一个简单的 Javascript 函数，一旦发现重复的边缘，它就会删除方向边缘。

输入：ridA（你开始@rid）

代码：

var g=orient.getGraph();
var outF=g.command('sql','select expand(out("follows")) from '+ridA);
var inF=g.command('sql','select expand(in("follows")) from '+ridA);
for(x=0;x<outF.length;x++){
  var ridOut=outF[x].getId();
  for(y=0;y<inF.length;y++){
    var ridIn=inF[y].getId();
    if(ridOut==ridIn){
      g.command('sql','delete edge follows from '+ridIn+' to '+ridA);
    }
  }
}

编辑：

例如，如果您尝试从顶点 #12:1 中删除重复的边，则在启动该函数后，两条“follow”方向边将被删除。