Question

我打算将我的整个图形（两个节点都有关系和＆＃34;独立的节点）导出到Gephi中。为了实现它，我当前执行两个查询：

// export relationships
match path = (n)--() 
with collect(path) as paths
call apoc.gephi.add(null, 'workspace1', paths, '', ['attr1', 'attr2']) yield nodes, relationships, time
return nodes, relationships, time

// export independent nodes
match path = (p) 
where not (p)--()
with collect(path) as paths
call apoc.gephi.add(null, 'workspace1', paths, '', ['attr1', 'attr2']) yield nodes, relationships, time
return nodes, relationships, time

我尝试用以下单个查询替换它们：

match path = (n)-[*0..]-() 
with collect(path) as paths
call apoc.gephi.add(null, 'workspace1', paths, '', ['attr1', 'attr2']) yield nodes, relationships, time
return nodes, relationships, time

不幸的是，查询永远不会完成并且有效地使用DoS-es Neo4j（导致Neo4j端的CPU和RAM消耗很高并使其无响应）。我也试图用[*0..10]限制关系深度，但它没有帮助。

使用单个查询导出数据的正确方法是什么？

Answer 1

我会尝试以下情况......

public function getInputFilterSpecification()
{
    return [
        'username' => [
            'name' => 'username',
            'required' => true,
            'filters' => [
                'name' => 'StringToLower',
                'name'=>'StripTags',
                'name'=>'StringTrim' 
            ],
            validators => [
                [
                    'name'=>'Regex',
                    'options'=> [
                        'pattern' => '/^[a-z0-9_.-]{1,50}+$/', 
                        'pattern_js' => '^[a-zA-Z0-9_\.\-]{1,50}$' 
                    ]
                ]
            ] 
        ],
        'password' => [
            'name' => 'password',
            'required' => true,
            'filters' => [
                array('name'=>'StripTags'),
                array('name'=>'StringTrim')
            ],
            'validators' => [
                [
                    'name'=>'Regex',
                    'options'=> [
                    'pattern' => '/^[a-z0-9_.-]{6,25}+$/', 
                    'pattern_js' => '^[a-zA-Z0-9_\.\-]{6,25}$' 
                    ]
                ]
            ]
        ]
    ];
}

所以我们已经添加了关系的方向并限制为只有1跳。这样我们就可以删除导出重复项并加快导出速度。

如何将_all_节点和所有关系从Neo4j导出到Gephi？

1 个答案: