我有一个简单的场景,
我想选择Adrian和Permies之间的所有路径,
shortestPath函数给了我
Adrian -> Permies
但我无法得到另一条道路,
Adrian -> Devs -> Permies
答案 0 :(得分:1)
使用Java,您可以使用以下代码和以下类
OrientGraph g=new OrientGraph(currentPath);
AllPaths allPath = new AllPaths(g);
List<List<Vertex>> paths=allPath.getPaths("#11:0","#11:7");
AllPaths类
import java.util.ArrayList;
import java.util.List;
import com.tinkerpop.blueprints.Direction;
import com.tinkerpop.blueprints.Vertex;
import com.tinkerpop.blueprints.impls.orient.OrientGraph;
public class AllPaths {
private boolean stop=false;
private Vertex vertexFrom=null;
private List<Vertex> vertexPreviousStep=new ArrayList<Vertex>();
private List<Vertex> vertexCurrently=new ArrayList<Vertex>();
private List<List<Vertex>> paths=new ArrayList<List<Vertex>>();
private OrientGraph g;
public AllPaths(OrientGraph g) {
this.g=g;
}
protected List<List<Vertex>> getPaths(String ridFrom, String ridTo) {
if(!check(ridFrom,ridTo))
return paths;
vertexPreviousStep.add(vertexFrom);
List<Vertex> p=new ArrayList<Vertex>();
p.add(vertexFrom);
paths.add(p);
int step=1;
do{
stop=false;
for(Vertex v: vertexPreviousStep){
List<Vertex> toAdd=new ArrayList<Vertex>();
Iterable<Vertex> nodes = (Iterable<Vertex>) v.getVertices(Direction.OUT);
for(Vertex v1:nodes){
toAdd.add(v1);
if(!v1.getId().toString().equals(ridTo))
vertexCurrently.add(v1);
}
if(toAdd.size()!=0)
setPaths(v,toAdd,step);
}
change();
step++;
}while(stop==true);
return cleanPaths(ridTo);
}
private boolean check(String ridFrom,String ridTo) {
boolean findFrom=false,findTo=false;
for(Vertex v:g.getVertices()){
if(v.getId().toString().equals(ridFrom)){
findFrom=true;
vertexFrom=v;
}
else if(v.getId().toString().equals(ridTo))
findTo=true;
}
if(findFrom==false || findTo==false)
return false;
return true;
}
public void change(){
vertexPreviousStep.clear();
for(Vertex v:vertexCurrently)
vertexPreviousStep.add(v);
vertexCurrently.clear();
}
private void setPaths(Vertex previousVertex,List<Vertex> toAdd,int step) {
for(int i=0;i<paths.size();i++){
List<Vertex> list=paths.get(i);
Vertex last=list.get(list.size()-1);
if(last.getId().toString().equals(previousVertex.getId().toString()) && list.size()==step){
int j=0;
for(Vertex v1:toAdd){
boolean vertexFound=false;
for(Vertex v2:list){
if(v2.getId().toString().equals(v1.getId().toString()))
vertexFound=true;
}
if(vertexFound==false){
List<Vertex> listVertex=new ArrayList<Vertex>();
for(Vertex p:list)
listVertex.add(p);
listVertex.add(v1);
if(j==0){
stop=true;
paths.set(i,listVertex);
j++;
}
else
paths.add(listVertex);
}
}
}
}
}
public List<List<Vertex>> cleanPaths(String ridTo){
for(int i=0;i<paths.size();i++){
List<Vertex> list=paths.get(i);
if(!list.get(list.size()-1).getId().toString().equals(ridTo)){
paths.remove(i);
i--;
}
}
return paths;
}
}
答案 1 :(得分:0)
好吧,我将分享两种方式:
a)使用遍历但指定深度:
TRAVERSE * FROM (SELECT FROM Source WHERE field = 'value') MAXDEPTH 3
b)使用一些约束条款:
SELECT FROM (TRAVERSE * FROM (SELECT FROM Source WHERE field = 'value') while @rid <> (SELECT FROM Target WHERE field='value'))