py2neo: Graph.find_one 有多个 key/values

Question

我在查找 py2neo 和 find_one (http://py2neo.org/2.0/essentials.html)

时遇到了一些问题

我想要的 Cypher 是：

MATCH (p:Person) WHERE p.name='Alice' AND p.age=22 RETURN p

比如说，如果有多个 key/value 集（例如，如果图中有多个 'Alice'）。

我的问题是我不知道该给什么graph.find_one，有效代码是：

graph.find_one('Person', 'name', 'Alice')

我想要的是（这不起作用！）：

graph.find_one('Person', {'name': 'Alice', 'age': 22})

一个可能的（不好的）解决方案是制作一个 graph.find，然后遍历结果属性并查找年龄，但我不喜欢该解决方案。

奖励： graph.find 可以做年龄 > 25 的事情吗？

编辑：新 "solution"

find_person = "MATCH (p:Person) WHERE p.name = {N} AND p.age = {A} RETURN p"

>>> tx = graph.cypher.begin()
>>> tx.append(find_person, {'N': 'Alice', 'A': 22})
>>> res = tx.process()
>>> print(res[0][0][0])
(n423:Person {age:22,name:"Lisa"})

我不喜欢的是我错过了 Note 对象，（而且我不完全理解 RecordListList，以及如何导航它 nicley）

Answer 1

如果您查看 source code you will find that unfortunately find and find_one don't support that type of queries. You should directly use the Cypher interface：

d = {'name': 'Alice', 'age' : 22}

# quote string values
d = {k:"'{}'".format(v) if isinstance(v, basestring) else v 
                     for k,v in d.items()}

cond = ' AND '.join("p.{}={}".format(prop, value) for prop, value in d.items())

query = "MATCH (p:Person) {condition} RETURN p"
query = query.format(condition=cond)
# "MATCH (p:Person) p.age=22 AND p.name='Alice' RETURN p"
results = graph.cypher.execute(query)

Answer 2

根据@elyase 的回答和原来的 py2neo.Graph.find，我制作了这段代码。请随时发表评论和改进.. :-)

def find_dict(graph, label, key_value=None, limit=None):
    """ Iterate through a set of labelled nodes, optionally filtering
    by property key/value dictionary
    """
    if not label:
        raise ValueError("Empty label")
    from py2neo.cypher.lang import cypher_escape
    if key_value is None:
        statement = "MATCH (n:%s) RETURN n,labels(n)" % cypher_escape(label)
    else:
        # quote string values
        d = {k: "'{}'".format(v) if isinstance(v, str) else v
             for k, v in key_value.items()}

        cond = ""
        for prop, value in d.items():
            if not isinstance(value, tuple):
                value = ('=', value)

            if cond == "":
                cond += "n.{prop}{value[0]}{value[1]}".format(
                    prop=prop,
                    value=value,
                )
            else:
                cond += " AND n.{prop}{value[0]}{value[1]}".format(
                    prop=prop,
                    value=value,
                )

        statement = "MATCH (n:%s ) WHERE %s RETURN n,labels(n)" % (
            cypher_escape(label), cond)
    if limit:
        statement += " LIMIT %s" % limit
    response = graph.cypher.post(statement)
    for record in response.content["data"]:
        dehydrated = record[0]
        dehydrated.setdefault("metadata", {})["labels"] = record[1]
        yield graph.hydrate(dehydrated)
    response.close()


def find_dict_one(graph, label, key_value=None):
    """ Find a single node by label and optional property. This method is
    intended to be used with a unique constraint and does not fail if more
    than one matching node is found.
    """
    for node in find_dict(graph, label, key_value, limit=1):
        return node

使用find_dict_one：

>>> a = find_dict_one(graph, 'Person', {'name': 'Lisa', 'age': 23})
>>>     print(a)
(n1:Person {age:23,name:"Lisa"})

将 find_dict 与元组一起使用：

>>> a = find_dict(graph, 'Person', {'age': ('>', 21)}, 2)    >>> for i in a:
>>>     print(i)
(n2:Person {age:22,name:"Bart"})
(n1:Person {age:23,name:"Lisa"})

使用不带元组的 find_dict：

>>> a = find_dict(graph, 'Person', {'age': 22}, 2)    >>> for i in a:
>>>     print(i)
(n2:Person {age:22,name:"Bart"})

py2neo: Graph.find_one 有多个 key/values

py2neo: Graph.find_one with multiple key/values

python

neo4j

py2neo