在Neo4j中如何将选择性查询结果作为字符串连接起来？

Question

初始情况

我编写了一个有效的Cypher查询，该查询返回了四个不同的数量。

MATCH
  <complex satement>
WITH
  count(DISTINCT typeA) AS amountA,
  count(DISTINCT typeB) AS amountB,
  count(DISTINCT typeC) AS amountC,
  count(DISTINCT typeD) AS amountD
RETURN
  amountA, amountB, amountC, amountD;

目标解决方案

我现在想返回一个文本字符串，而不是四列表，其中所有四个量都串联在一起，包括描述性标签。但是，如果数量大于零，则数量只能是字符串的一部分。

╒════════════════════════════════════════════════════╕
│"formattedQuantities"                               │
╞════════════════════════════════════════════════════╡
│"amountA: 123456, amountC: 9876543, amountD: 2018"  │
└────────────────────────────────────────────────────┘

（由于amountB的值为0，因此在结果中将其省略。）

我将此Cyper查询用于数百万行。由于担心性能影响，我不想创建和调用自定义插件。

那么，如何使用Cypher和Neo4j以字符串形式返回数量？您能给我建议如何解决这个挑战吗？在此先感谢您为我指明了正确的方向！

---

解决问题/初步结果的方法

密码声明：

MATCH
<complex satement>
WITH
  count(DISTINCT typeA) AS amountA,
  count(DISTINCT typeB) AS amountB,
  count(DISTINCT typeC) AS amountC,
  count(DISTINCT typeD) AS amountD
WITH
  ['amountA: ', amountA, ', amountB: ', amountB, ', amountC: ', amountC, ', amountD: ', amountD] AS quantities
RETURN
  reduce(result = toString(head(quantities)), n IN tail(quantities) | result + n) AS formattedQuantities;

结果：

╒═════════════════════════════════════════════════════════════════╕
│"formattedQuantities"                                            │
╞═════════════════════════════════════════════════════════════════╡
│"amountA: 123456: 1, amountB: 0, amountC: 9876543, amountD: 2018"│
└─────────────────────────────────────────────────────────────────┘

仍然打开：

• 由于值为0而对amountB进行过滤

Answer 1

您要使用FILTER函数

MATCH
<complex satement>
WITH
  count(DISTINCT typeA) AS amountA,
  count(DISTINCT typeB) AS amountB,
  count(DISTINCT typeC) AS amountC,
  count(DISTINCT typeD) AS amountD
// Reformat to list
WITH
  [{name:'amountA', value:amountA}, {name:'amountB', value:amountB}, {name:'amountC', value:amountC}, {name:'amountD', value:amountD}] AS quantities
// Filter out 0's
WITH filter(x IN quantities WHERE x.value > 0) AS quantities
// Convert list to string
RETURN
  reduce(result = quantities[0].name + ": " + quantities[0].value, n IN tail(quantities) | result + ", " + n.name + ": " + n.value) AS formattedQuantities; AS formattedQuantities;

请注意，如果所有值均为0（null +字符串= null），则返回null