Neo4j Cypher 查询:WHERE 子句中不考虑括号?
Neo4j Cypher query: Parenthesis not considered in WHERE clause?
我正在尝试 运行 查询,例如:
MATCH (n:Type1)-[:relation1]->(:Type1)<-[:relation2]-(:Type2) WHERE
((n.prop1="A" and n.prop2 in [42, 34]) OR
(n.prop1="B" and n.prop2 in [67]) OR
(n.prop1="C" and n.prop2 in [51, 43]) OR
(n.prop1="D" and n.prop2 in [56, 42]))
RETURN DISTINCT id(n) as node_id, n.prop3 as node_prop3, n.prop1 as prop1, n.prop2 as prop2;
这给出了混合结果,即 prop1="D"
与 prop2=51
匹配,这显然不符合预期条件。就好像括号在条件的形成中没有发言权。
当我在 WHERE 子句中 运行 查询 w/o 括号时,结果是相同的。
虽然,如果我查询类似:
MATCH (n:Type1)-[:relation1]->(:Type1)<-[:relation2]-(:Type2) WHERE
((n.prop1="A" and n.prop2=42) OR
(n.prop1="A" and n.prop2=34) OR
(n.prop1="B" and n.prop2=67) OR
(n.prop1="C" and n.prop2=51) OR
(n.prop1="C" and n.prop2=43) OR
(n.prop1="D" and n.prop2=56) OR
(n.prop1="D" and n.prop2=42))
RETURN DISTINCT id(n) as node_id, n.prop3 as node_prop3, n.prop1 as prop1, n.prop2 as prop2;
在将示例 2 中的单独条件替换为示例 1 中的 IN
以避免重复比较 prop1
的字符串时,我是否遗漏了什么?
我相信 IN
如您所愿。
设置示例数据。
(:Type1 {prop1:"A", prop2:34}),
(:Type1 {prop1:"B", prop2:67}),
(:Type1 {prop1:"C", prop2:51}),
(:Type1 {prop1:"C", prop2:43}),
(:Type1 {prop1:"D", prop2:56}),
(:Type1 {prop1:"D", prop2:42}),
(:Type1 {prop1:"D", prop2:51}),
(:Type1 {prop1:"B", prop2:42});
MATCH (t:Type1)
MERGE (t)-[:relation1]->(:Type1)<-[:relation2]-(:Type2);
现在 运行 您的查询。
MATCH (n:Type1)-[:relation1]->(:Type1)<-[:relation2]-(:Type2) WHERE
((n.prop1="A" and n.prop2 in [42, 34]) OR
(n.prop1="B" and n.prop2 in [67]) OR
(n.prop1="C" and n.prop2 in [51, 43]) OR
(n.prop1="D" and n.prop2 in [56, 42]))
RETURN DISTINCT id(n) as node_id, n.prop3 as node_prop3, n.prop1 as prop1, n.prop2 as prop2;
语句returns七条记录,结果不包含prop1=D,prop2=51的节点
如果省略括号,请参阅 this answer 了解布尔运算符优先级的信息。
我正在尝试 运行 查询,例如:
MATCH (n:Type1)-[:relation1]->(:Type1)<-[:relation2]-(:Type2) WHERE
((n.prop1="A" and n.prop2 in [42, 34]) OR
(n.prop1="B" and n.prop2 in [67]) OR
(n.prop1="C" and n.prop2 in [51, 43]) OR
(n.prop1="D" and n.prop2 in [56, 42]))
RETURN DISTINCT id(n) as node_id, n.prop3 as node_prop3, n.prop1 as prop1, n.prop2 as prop2;
这给出了混合结果,即 prop1="D"
与 prop2=51
匹配,这显然不符合预期条件。就好像括号在条件的形成中没有发言权。
当我在 WHERE 子句中 运行 查询 w/o 括号时,结果是相同的。
虽然,如果我查询类似:
MATCH (n:Type1)-[:relation1]->(:Type1)<-[:relation2]-(:Type2) WHERE
((n.prop1="A" and n.prop2=42) OR
(n.prop1="A" and n.prop2=34) OR
(n.prop1="B" and n.prop2=67) OR
(n.prop1="C" and n.prop2=51) OR
(n.prop1="C" and n.prop2=43) OR
(n.prop1="D" and n.prop2=56) OR
(n.prop1="D" and n.prop2=42))
RETURN DISTINCT id(n) as node_id, n.prop3 as node_prop3, n.prop1 as prop1, n.prop2 as prop2;
在将示例 2 中的单独条件替换为示例 1 中的 IN
以避免重复比较 prop1
的字符串时,我是否遗漏了什么?
我相信 IN
如您所愿。
设置示例数据。
(:Type1 {prop1:"A", prop2:34}),
(:Type1 {prop1:"B", prop2:67}),
(:Type1 {prop1:"C", prop2:51}),
(:Type1 {prop1:"C", prop2:43}),
(:Type1 {prop1:"D", prop2:56}),
(:Type1 {prop1:"D", prop2:42}),
(:Type1 {prop1:"D", prop2:51}),
(:Type1 {prop1:"B", prop2:42});
MATCH (t:Type1)
MERGE (t)-[:relation1]->(:Type1)<-[:relation2]-(:Type2);
现在 运行 您的查询。
MATCH (n:Type1)-[:relation1]->(:Type1)<-[:relation2]-(:Type2) WHERE
((n.prop1="A" and n.prop2 in [42, 34]) OR
(n.prop1="B" and n.prop2 in [67]) OR
(n.prop1="C" and n.prop2 in [51, 43]) OR
(n.prop1="D" and n.prop2 in [56, 42]))
RETURN DISTINCT id(n) as node_id, n.prop3 as node_prop3, n.prop1 as prop1, n.prop2 as prop2;
语句returns七条记录,结果不包含prop1=D,prop2=51的节点
如果省略括号,请参阅 this answer 了解布尔运算符优先级的信息。