Question

我一直试图创建一个节点并将其链接到其他节点的列表，我得到了以下查询：

MATCH (s:Subject), (p:Programme {name: 'Bsc. Agriculture' })
Where s.name IN ['Physics (CAPE)', 'Biology (CAPE)', 'Chemistry (CAPE)']
Create (c: Combo {amt:1}), (c)-[:contains]->(s), (p)-[:requires]->(c) return *

但遗憾的是，组合节点创建了三次，每个组合节点都链接到一个Subject节点。有什么方法可以调整此查询，以便只创建一个Combo节点吗？

Answer 1

下面的查询使用aggregation function //bar//foo[count(./ancestor::quux) = count(./ancestor::bar/ancestor::quux)]为class ViewController: UIViewController, UICollectionViewDelegate, UICollectionViewDataSource, UICollectionViewDelegateFlowLayout { let cellId = "cellId" override func viewDidLoad() { super.viewDidLoad() setupCV() } var containerCollectionView: UICollectionView = { let layout = UICollectionViewLayout() let cv = UICollectionView(frame: .zero, collectionViewLayout: layout) cv.backgroundColor = UIColor.white return cv }() } extension ViewController { fileprivate func setupCV() { // these first two lines add the colelction view and apply constraints - they work and the CV is properly displayed inside the view controller self.view.addSubview(containerCollectionView) containerCollectionView.anchor(view.topAnchor, left: view.leftAnchor, bottom: view.bottomAnchor, right: view.rightAnchor, topConstant: 0, leftConstant: 0, bottomConstant: 0, rightConstant: 0, widthConstant: 0, heightConstant: 0) containerCollectionView.delegate = self containerCollectionView.dataSource = self containerCollectionView.register(UICollectionViewCell.self, forCellWithReuseIdentifier: cellId) } func collectionView(_ collectionView: UICollectionView, layout collectionViewLayout: UICollectionViewLayout, sizeForItemAt indexPath: IndexPath) -> CGSize { return CGSize(width: 150, height: 150) } func collectionView(_ collectionView: UICollectionView, cellForItemAt indexPath: IndexPath) -> UICollectionViewCell { let cell = collectionView.dequeueReusableCell(withReuseIdentifier: cellId, for: indexPath) cell.backgroundColor = UIColor.yellow return cell } func collectionView(_ collectionView: UICollectionView, numberOfItemsInSection section: Int) -> Int { return 3 } }生成一行数据（而不是三行），以确保只执行第一个COLLECT每p只有一次 - 因此每CREATE只产生一个p和Combo关系。然后requires子句创建所有必需的p关系。

FOREACH

Answer 2

您的查询为MATCH中找到的CREATE中使用的每个其他节点创建Combo节点。其中一个解决方案可能是您先创建Combo节点，然后在查询的其他部分使用它

Create (c: Combo {amt:1})
WITH c
MATCH (s:Subject), (p:Programme {name: 'Bsc. Agriculture' })
Where s.name IN ['Physics (CAPE)', 'Biology (CAPE)', 'Chemistry (CAPE)']
CREATE (c)-[:contains]->(s), (p)-[:requires]->(c) return *

Answer 3

如果使用MERGE而不是CREATE，它应该满足您的要求并避免创建重复的节点。但是这假设每个关系目标节点都由其所有唯一属性标识。试试这个......

MATCH (s:Subject), (p:Programme {name: 'Bsc. Agriculture' }) 
Where s.name IN ['Physics (CAPE)', 'Biology (CAPE)', 'Chemistry (CAPE)'] 
MERGE (c: Combo {amt:1}), (c)-[:contains]->(s), (p)-[:requires]->(c) return *

合并只有在元素不存在时才会创建元素。需要注意的是，有问题的元素必须与其所有参数（名称和属性）匹配。 https://graphaware.com/neo4j/2014/07/31/cypher-merge-explained.html还有更多内容。

您可以在一个查询中创建节点并将其链接到多个节点吗？

3 个答案: