Postgresql 从2个查询中减去结果

我有两个查询，它们给出相同的结果（相同的列、行），但数据不同。我需要将查询#1的一列的结果减去查询#2的同一列

这是我的疑问

QUERY 1

SELECT 
COUNT(DISTINCT CASE WHEN customer.ref IN ('emp', 'oth') THEN customer.id END) as visit_count, locations.name
FROM visits 
LEFT JOIN customer ON customer.id = visits.id
LEFT JOIN locations ON customer.loc_id = locations.id
WHERE visits.group_id = 1
GROUP BY locations.name, locations.id;

RESULTS
visit_count  |   location
---------------------------
      5      |    loc1
      1      |    loc2
      1      |    loc3

QUERY 2

SELECT COUNT(customer.id) as visit_count, locations.name FROM locations 
LEFT JOIN visits ON locations.name = visits.location_check
LEFT JOIN customer ON visits.cust_id = customer.id AND customer.group_id = 1
WHERE locations.group_id = 1 AND locations.active = true
GROUP BY locations.location_name, locations.id;

RESULTS
visit_count  |   location
---------------------------
      10      |    loc1
      15      |    loc2
      18      |    loc3

结果是一样的，我只需要从第二个结果中减去第一个结果。所以我想要的结果是：

visit_count  |   location
---------------------------
      5      |    loc1
      14     |    loc2
      17     |    loc3

---

有没有一种方法可以将它们连接在一起并从另一列中减去一列？

这是最简单的答案：

SELECT b.visit_count-a.visit_count, a.location
FROM
(
  SELECT 
  COUNT(DISTINCT CASE WHEN customer.ref IN ('emp', 'oth') THEN customer.id END) as visit_count, locations.name
  FROM visits 
  LEFT JOIN customer ON customer.id = visits.id
  LEFT JOIN locations ON customer.loc_id = locations.id
  WHERE visits.group_id = 1
  GROUP BY locations.name, locations.id;
) as a
(
  SELECT COUNT(customer.id) as visit_count, locations.name FROM locations 
  LEFT JOIN visits ON locations.name = visits.location_check
  LEFT JOIN customer ON visits.cust_id = customer.id AND customer.group_id = 1
  WHERE locations.group_id = 1 AND locations.active = true
  GROUP BY locations.location_name, locations.id;
) as b
WHERE a.location=b.location;

这显然只是您提供的内容的复制/粘贴，但似乎有一些细节需要解决，比如一个查询中的

customer.id=visions.id

，另一个查询中的

customer.id=visions.cust\u id

您似乎希望获得每个位置的总访问次数，减去

customer.ref

为

emp

或

oth

的访问次数。如果是这样的话，我想你只是想要这样的东西：

SELECT count(*), l.name as location
FROM visits v
JOIN customer c ON c.id=v.id
JOIN locations l ON l.id=c.loc_id
WHERE v.group_id = 1
AND l.active
AND c.ref NOT IN ('emp','oth')
GROUP BY l.name;

当然，正确的查询取决于您的表结构，因为列名似乎不一致，所以我们在这里并不真正了解表结构

---

披露：我为

工作这是最直接的答案：

SELECT b.visit_count-a.visit_count, a.location
FROM
(
  SELECT 
  COUNT(DISTINCT CASE WHEN customer.ref IN ('emp', 'oth') THEN customer.id END) as visit_count, locations.name
  FROM visits 
  LEFT JOIN customer ON customer.id = visits.id
  LEFT JOIN locations ON customer.loc_id = locations.id
  WHERE visits.group_id = 1
  GROUP BY locations.name, locations.id;
) as a
(
  SELECT COUNT(customer.id) as visit_count, locations.name FROM locations 
  LEFT JOIN visits ON locations.name = visits.location_check
  LEFT JOIN customer ON visits.cust_id = customer.id AND customer.group_id = 1
  WHERE locations.group_id = 1 AND locations.active = true
  GROUP BY locations.location_name, locations.id;
) as b
WHERE a.location=b.location;

这显然只是您提供的内容的复制/粘贴，但似乎有一些细节需要解决，比如一个查询中的

customer.id=visions.id

，另一个查询中的

customer.id=visions.cust\u id

您似乎希望获得每个位置的总访问次数，减去

customer.ref

为

emp

或

oth

的访问次数。如果是这样的话，我想你只是想要这样的东西：

SELECT count(*), l.name as location
FROM visits v
JOIN customer c ON c.id=v.id
JOIN locations l ON l.id=c.loc_id
WHERE v.group_id = 1
AND l.active
AND c.ref NOT IN ('emp','oth')
GROUP BY l.name;

当然，正确的查询取决于您的表结构，因为列名似乎不一致，所以我们在这里并不真正了解表结构

---

披露：我为

工作，您可以将两个查询都转换为子查询，

连接

结果并在外部查询中进行计算。但可能有更有效的解决方案；为了让我们提供建议，您需要在原始表中显示数据，而不仅仅是现有查询的结果。好的，您可以将两个查询都转换为子查询，

join

结果并在外部查询中进行计算。但可能有更有效的解决方案；为了让我们提供建议，您需要在原始表中显示数据，而不仅仅是现有查询的结果。感谢您的解释！这真的帮助了我！：）谢谢你的解释！这真的帮助了我！：）