Question

我是Couchbase DB服务器的新手，我试图通过一个查询来实现我对三个查询所做的工作，因为这样做效率不高。

我在同一存储桶中有三种不同的文档类型（x，y，z）；都具有类似的键：“区”，像这样：

文档x：

{
  "type": "x",
  "district": "Some district"
}

文档y：

{
  "type": "y",
  "district": "Some district"
}

文档z：

{
  "type": "z",
  "district": "Some district"
}

我目前已经在PHP中实现了以下伪代码：

$totalDistrictInX = "SELECT COUNT(x) FROM bucket WHERE type = 'x' AND district = 'Maboro';

$totalDistrictInY = "SELECT COUNT(x) FROM bucket WHERE type = 'y' AND district = 'Maboro';

$totalDistrictInZ = "SELECT COUNT(x) FROM bucket WHERE type = 'z' AND district = 'Maboro';

$totalCountOfMaboro = $totalDistrictInX + $totalDistrictInY + $totalDistrictInZ;

我无法使用JOIN查询，因为当前使用的Couchbase服务器低于5.50 which only supports joining documents between document key to document field and not between document fields。

是否有一种方法可以通过一个n1ql查询来实现？请提供任何帮助。

Answer 1

可以使用GROUP BY和COUNT组合来解决吗？

SELECT COUNT(x) FROM bucket WHERE district = 'Maboro' GROUP BY type

Documentation

Answer 2

使用不带分组依据的汇总查询来计算总数，控制要通过谓词计数的文档。

SELECT COUNT(1) AS cnt
FROM bucket
WHERE type IN ['x', 'y', 'z'] AND district = 'Maboro';

如果每种类型都需要计数，请使用GROUP BY

SELECT type, COUNT(1) AS cnt
FROM bucket
WHERE type IN ['x', 'y', 'z'] AND district = 'Maboro'
GROUP BY type;

如果要计算总数和单个类型，则将其计为数组

SELECT ARRAY_SUM(av[*].cnt) AS totalcnt, av AS details
LET av = (SELECT type, COUNT(1) AS cnt
          FROM bucket
          WHERE type IN ['x', 'y', 'z'] AND district = 'Maboro'
          GROUP BY type);

如何编写一个n1ql查询以求和同一存储桶中多个文档的值

2 个答案: