neo4jclient：我们如何在一次查询中 return 节点标签和属性？

Question

        var pubs = client.Cypher
            .Match("(s)")
            .Where("s:APublisher OR s:BPublisher OR s:CPublisher OR s:DPublisher ")
                           .Return(s => s.As<Publisher>())
                           .Results.ToList<Publisher>();

通过上述查询，返回节点的所有属性并为每个节点创建 "Publisher" 个对象。

如何获取节点标签，并将其设置为新创建的 "Publisher" 对象的 "Labels" 属性？

谢谢..

Answer 1

您需要使用适当的函数提取标签并将它们分配给您的对象标签属性。试试像

var pubs = client.Cypher
    .Match("(s)")
    .Where("s:APublisher OR s:BPublisher OR s:CPublisher OR s:DPublisher ")
    .Return(s => new {
        Pub = s.As<Publisher>()
        PubLabels = s.Labels() })
    .Results
    .ToList();

var pubsList = new List<Publisher>(pubs.Count);
for each (var pub in pubs)
{
    var publisher = pub.Pub;
    publisher.Labels = pub.PubLabels;
    pubsList.Add(publisher);
}

neo4jclient：我们如何在一次查询中 return 节点标签和属性？

neo4jclient: how do we return node label and properties in one query?

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