Text 如何在Neo4j中将选择性查询结果连接为字符串？ 初始情况

我编写了一个有效的密码查询，它返回四个不同的数量

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;

（因为amountB的值为0，所以在结果中忽略它。）

我对数百万行使用这个Cyper查询。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返回字符串形式的数量呢？你能给我一个如何解决这个挑战的建议吗？非常感谢你为我指明了正确的方向

---

解决问题的方法/初步结果 塞弗声明：

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;

仍然开放：

• 由于值0，正在筛选amountB

您想使用该功能吗

匹配
具有
将（不同类型A）计数为amountA，
将（不同类型B）计数为数量B，
将（不同类型C）计数为数量C，
将（不同类型的）计数为amountD
//重新格式化为列表
具有
[{name:'amountA'，value:amountA}，{name:'amountB'，value:amountB}，{name:'amountC'，value:amountC}，{name:'amountD'，value:amountD}]作为数量
//过滤掉0
使用过滤器（数量中的x，其中x.value>0）作为数量
//将列表转换为字符串
回来
减少（结果=数量[0]。名称+“：”+数量[0]。值，尾部的n（数量）|结果+“，“+n.名称+”：“+n.值）作为格式化数量；作为格式化的数量；

请注意，如果所有值都为0（null+string=null），则返回null

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;

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

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;