SQL学习-2.8 子查询和临时表格

3.练习-首个子查询

在写带有子查询的句子时，为了阅读起来更容易，应该要按照正确的缩进方式，将内查询整体进行缩进。

问题：

对现有的这个的数据库进行分析，那些渠道平均每天发送给Patch&Posey的流量最大

解法：首先建立该问题的内查询：统计每个渠道每天的访问次数(count(*)),然后基于内查询的结果集进行外查询：计算访问次数的平均值avg().

SELECT channel,AVG(event_count) AS avg_event_count
FROM(SELECT DATE_TRUNC('day',occurred_at) as day,channel,COUNT(*) AS event_countFROM web_eventsGROUP BY 1,2) subGROUP BY 1ORDER BY 2 DESC;# 结果展示
channel avg_event_count
direct  4.8964879852125693
organic 1.6672504378283713
facebook    1.5983471074380165
adwords 1.5701906412478336
twitter 1.3166666666666667
banner  1.2899728997289973

首先，我们需要按照日期和渠道分组。然后按事件数（第三列）排序，这样可以快速得出第一个问题的答案。
SELECT DATE_TRUNC('day',occurred_at) AS day,
channel, COUNT(*) as events
FROM web_events
GROUP BY 1,2
ORDER BY 3 DESC;可以看出，要获得这一结果，提供了整个原始表格。查询的附加部分包括 *，并且我们需要为表格设置别名。此外，是在 SELECT 语句中（而不是 FROM）中提供表格。
SELECT *
FROM (SELECT DATE_TRUNC('day',occurred_at) AS day,
channel, COUNT(*) as events
FROM web_events
GROUP BY 1,2
ORDER BY 3 DESC) sub;最后，我们在以下语句中能够获得显示每个渠道一天的平均事件数的表格。
SELECT channel, AVG(events) AS average_events
FROM (SELECT DATE_TRUNC('day',occurred_at) AS day,
channel, COUNT(*) as events
FROM web_events
GROUP BY 1,2) sub
GROUP BY channel
ORDER BY 2 DESC;

7. 练习更多关于子查询的内容

子查询（第二部分）

在你写的第一个子查询中，你编写了一个子查询来创建表格，然后可以在 FROM 语句中查询该表格。但是，如果只返回一个值，则可以在逻辑语句中使用该值，例如 WHERE、HAVING，甚至 SELECT，该值可以嵌套在 CASE 语句中。

在下一页面中，我们将讲解这个示例，然后你将尝试自己回答一些问题。

专家提示

注意，在条件语句中编写子查询时，不能包含别名。这是因为该子查询会被当做单个值（或者对于 IN 情况是一组值），而不是一个表格。

同时注意，这里的查询对应的是单个值。如果我们返回了整个列，则需要使用 IN 来执行逻辑参数。如果我们要返回整个表格，则必须为该表格使用别名，并对整个表格执行其他逻辑。

1.
SELECT DATE_TRUNC('month',MIN(occurred_at)) as min_month
FROM orders;
# result
min_month
2013-12-01T00:00:00.000Z2.
SELECT *
FROM orders
WHERE DATE_TRUNC('month',occurred_at) =(SELECT DATE_TRUNC('month',MIN(occurred_at)) as min_monthFROM orders)
ORDER BY occurred_at DESC;
# result
仅仅返回了2013-12的所有订单的信息。

9.更多的子查询

1.
SELECT r.name region_name, s.name rep_name,MAX(total_amt_usd)
FROM region r
JOIN Sales_reps s
ON r.id = s.region_id
JOIN accounts a
ON a.Sales_rep_id = s.id
JOIN orders o
ON a.id = o.account_id
GROUP BY 1,2
# result 50 results
region_name rep_name    max
Midwest Carletta Kosinski   8836.64
Midwest Charles Bidwell 18784.55
Midwest Chau Rowles 11049.08
Midwest Cliff Meints    48675.90
Midwest Cordell Rieder  45465.23
Midwest Delilah Krum    11093.97
Midwest Julie Starr 13947.39
Midwest Kathleen Lalonde    8776.53
Midwest Sherlene Wetherington   13211.32
Northeast   Akilah Drinkard 4642.17
Northeast   Ayesha Monica   27902.70
Northeast   Cara Clarke 9021.14
Northeast   Debroah Wardle  13226.23
Northeast   Elba Felder 9078.63
Northeast   Elna Condello   40928.70
Northeast   Ernestine Pickron   92991.05
Northeast   Eugena Esser    81243.96
Northeast   Gianna Dossey   12079.83
Northeast   Julia Behrman   84099.62
Northeast   Lavera Oles 9711.46
Northeast   Michel Averette 23989.98
Northeast   Nakesha Renn    7382.16
Northeast   Necole Victory  93106.81
Northeast   Renetta Carew   43421.28
Northeast   Retha Sears 31329.24
Northeast   Samuel Racine   14054.42
Northeast   Shawanda Selke  29699.28
Northeast   Sibyl Lauria    16998.48
Northeast   Silvana Virden  24885.98
Northeast   Tia Amato   77285.75
Southeast   Babette Soukup  11264.85
Southeast   Calvin Ollison  10116.04
Southeast   Derrick Boggess 24000.14
Southeast   Dorotha Seawell 95005.82
Southeast   Earlie Schleusner   19195.48
Southeast   Maren Musto 17859.13
Southeast   Moon Torian 20855.55
Southeast   Nelle Meaux 22460.24
Southeast   Saran Ram   28073.65
Southeast   Vernita Plump   24111.28
West    Arica Stoltzfus 107533.55
West    Brandie Riva    93505.69
West    Dawna Agnew 232207.07
West    Elwood Shutt    82163.71
West    Georgianna Chisholm 15849.34
West    Hilma Busick    21032.39
West    Marquetta Laycock   22694.65
West    Maryanna Fiorentino 112875.18
West    Micha Woodford  16631.12
West    Soraya Fulton   93547.84
………………

上面对于第一个问题的解答是错误的，这样的结果是每个区域的每个销售代表的所有记录中的最大值；首先没有进行销售额求和，也没有做到统同一区域下面的销售代表的总销售额的比较，从而取最大销售额的销售代表。、

SELECT region_name, MAX(total_amt) total_amt
FROM(SELECT s.name rep_name, r.name region_name, SUM(o.total_amt_usd) total_amt
FROM sales_reps s
JOIN accounts a
ON a.sales_rep_id = s.id
JOIN orders o
ON o.account_id = a.id
JOIN region r
ON r.id = s.region_id
GROUP BY 1, 2) inner1
GROUP BY 1;

以上才是第一题的正确答案和相应的结果返回。

2.
SELECT region_name,total_amt
FROM(SELECT r.name region_name, SUM(o.total_amt_usd) total_amtFROM sales_reps sJOIN accounts aON a.sales_rep_id = s.idJOIN orders oON o.account_id = a.idJOIN region rON r.id = s.region_idGROUP BY 1ORDER BY 2) sub
LIMIT 1;
#result
region_name total_amt
Midwest 3013486.51# 以下是官网给出的答案，稍显复杂，但逻辑要更严谨一点
对于具有最高销售额 (total_amt_usd) 的区域，总共下了多少个订单？我写的第一个查询是获取每个区域的 total_amt_usd。
SELECT r.name region_name, SUM(o.total_amt_usd) total_amt
FROM sales_reps s
JOIN accounts a
ON a.sales_rep_id = s.id
JOIN orders o
ON o.account_id = a.id
JOIN region r
ON r.id = s.region_id
GROUP BY r.name;然后，我们仅从该表格中获取销售额最高的区域。可以通过两种方法来获取，一种是使用子查询后的最大值，另一种是按降序排序，然后获取最高值。
SELECT MAX(total_amt)
FROM (SELECT r.name region_name, SUM(o.total_amt_usd) total_amt
FROM sales_reps s
JOIN accounts a
ON a.sales_rep_id = s.id
JOIN orders o
ON o.account_id = a.id
JOIN region r
ON r.id = s.region_id
GROUP BY r.name) sub;最终，我们要获取具有该区域销售额的总订单量：
SELECT r.name, SUM(o.total) total_orders
FROM sales_reps s
JOIN accounts a
ON a.sales_rep_id = s.id
JOIN orders o
ON o.account_id = a.id
JOIN region r
ON r.id = s.region_id
GROUP BY r.name
HAVING SUM(o.total_amt_usd) = (
SELECT MAX(total_amt)
FROM (SELECT r.name region_name, SUM(o.total_amt_usd) total_amt
FROM sales_reps s
JOIN accounts a
ON a.sales_rep_id = s.id
JOIN orders o
ON o.account_id = a.id
JOIN region r
ON r.id = s.region_id
GROUP BY r.name) sub);结果就是 Northeast，总订单为 1230378 个。

对于第三个问题，分析：题意是首先根据得到标准纸销售量最大的客户的名字、标准纸的购买数、所有类型的纸的股买数；而最后的结果是返回所有客户中总购买数大于该客户的总体购买数的客户名称。

首先，我们要得出购买标准纸张数量 (standard_qty) 最多的客户。以下查询获取了该客户，以及总消费：
SELECT a.name account_name, SUM(o.standard_qty) total_std, SUM(o.total) total
FROM accounts a
JOIN orders o
ON o.account_id = a.id
GROUP BY 1
ORDER BY 2 DESC
LIMIT 1;现在，我将使用上述信息获取总消费更高的所有客户：
SELECT a.name
FROM orders o
JOIN accounts a
ON a.id = o.account_id
GROUP BY 1
HAVING SUM(o.total) > (SELECT total
FROM (SELECT a.name act_name, SUM(o.standard_qty) tot_std, SUM(o.total) total
FROM accounts a
JOIN orders o
ON o.account_id = a.id
GROUP BY 1
ORDER BY 2 DESC
LIMIT 1) sub);上述查询列出了具有更多订单的客户列表。我们还可以使用另一个简单的子查询获取数量。
SELECT COUNT(*)
FROM (SELECT a.name
FROM orders o
JOIN accounts a
ON a.id = o.account_id
GROUP BY 1
HAVING SUM(o.total) > (SELECT total
FROM (SELECT a.name act_name, SUM(o.standard_qty) tot_std, SUM(o.total) total
FROM accounts a
JOIN orders o
ON o.account_id = a.id
GROUP BY 1
ORDER BY 2 DESC
LIMIT 1) inner_tab)
) counter_tab;

对于问题四

SELECT a.name, w.channel,COUNT(*)
FROM accounts a
JOIN web_events w
ON a.id = w.account_id AND a.id =
(SELECT a.id FROM (
SELECT a.id,a.name,SUM(total_amt_usd)
FROM accounts a
JOIN orders o
ON a.id = o.account_id
GROUP BY 1,2
ORDER BY 3 DESC
LIMIT 1) sub)
GROUP BY 1,2
ORDER BY 3 DESC;
# 上面的结果是错误的，其返回了所有相应的客户的在一些渠道上的events。与下面的主要的区别在于：子查询部分的结果中的第一个字段没有进行重命名，下面的部分进行了a_id的重命名。SELECT a.name, w.channel,COUNT(*)
FROM accounts a
JOIN web_events w
ON a.id = w.account_id AND a.id =
(SELECT a_id FROM (
SELECT a.id as a_id,a.name,SUM(total_amt_usd)
FROM accounts a
JOIN orders o
ON a.id = o.account_id
GROUP BY 1,2
ORDER BY 3 DESC
LIMIT 1) sub)
GROUP BY 1,2
ORDER BY 3 DESC;

对于问题五

SELECT AVG(total_amt)
FROM(SELECT a.id,a.name,SUM(total_amt_usd) AS total_amt
FROM accounts a
JOIN orders o
ON a.id = o.account_id
GROUP BY 1,2
ORDER BY 3 DESC
LIMIT 10) sub;avg
304846.969000000000# 官方答案
首先，我们需要找出总消费 (total_amt_usd) 在前十名的客户。
SELECT a.id, a.name, SUM(o.total_amt_usd) tot_spent
FROM orders o
JOIN accounts a
ON a.id = o.account_id
GROUP BY a.id, a.name
ORDER BY 3 DESC
LIMIT 10;现在计算这十个客户的平均消费。
SELECT AVG(tot_spent)
FROM (SELECT a.id, a.name, SUM(o.total_amt_usd) tot_spent
FROM orders o
JOIN accounts a
ON a.id = o.account_id
GROUP BY a.id, a.name
ORDER BY 3 DESC
LIMIT 10) temp;

对于问题六：

SELECT AVG(avg_amt)
FROM
(SELECT o.account_id,AVG(o.total_amt_usd) avg_amt
FROM orders o
GROUP BY 1
HAVING AVG(o.total_amt_usd) >(SELECT AVG(o.total_amt_usd) avg_total_amtFROM orders oJOIN accounts aON o.account_id =  a.id)
) subA# 官网给出的答案
首先，算出所有客户的总消费 (total_amt_usd) 平均值：
SELECT AVG(o.total_amt_usd) avg_all
FROM orders o
JOIN accounts a
ON a.id = o.account_id;然后，只获取高于这一平均值的客户。
SELECT o.account_id, AVG(o.total_amt_usd)
FROM orders o
GROUP BY 1
HAVING AVG(o.total_amt_usd) > (SELECT AVG(o.total_amt_usd) avg_all
FROM orders o
JOIN accounts a
ON a.id = o.account_id);最后，算出这些值的平均值。
SELECT AVG(avg_amt)
FROM (SELECT o.account_id, AVG(o.total_amt_usd) avg_amt
FROM orders o
GROUP BY 1
HAVING AVG(o.total_amt_usd) > (SELECT AVG(o.total_amt_usd) avg_all
FROM orders o
JOIN accounts a
ON a.id = o.account_id)) temp_table;

12+13 练习利用with实现的公用表达式(CTE)

问题：你需要算出每个渠道每天的平均事件数。解决方案：SELECT channel, AVG(events) AS average_events
FROM (SELECT DATE_TRUNC('day',occurred_at) AS day,
channel, COUNT(*) as events
FROM web_events
GROUP BY 1,2) sub
GROUP BY channel
ORDER BY 2 DESC;
我们使用 WITH 语句重新编写查询。注意：你可以获取内部查询：SELECT DATE_TRUNC('day',occurred_at) AS day,
channel, COUNT(*) as events
FROM web_events
GROUP BY 1,2
我们在此部分放入 WITH 语句。注意，在下面我们将表格的别名设为 events：WITH events AS (
SELECT DATE_TRUNC('day',occurred_at) AS day,
channel, COUNT(*) as events
FROM web_events
GROUP BY 1,2)
现在，我们可以像对待数据库中的任何其他表格一样使用这个新创建的 events 表格：WITH events AS (
SELECT DATE_TRUNC('day',occurred_at) AS day,
channel, COUNT(*) as events
FROM web_events
GROUP BY 1,2)SELECT channel, AVG(events) AS average_events
FROM events
GROUP BY channel
ORDER BY 2 DESC;
对于上述示例，我们只需一个额外的表格，但是想象下我们要创建第二个表格来从中获取数据。我们可以按照以下方式来创建额外的表格并从中获取数据：WITH table1 AS (
SELECT *
FROM web_events),table2 AS (
SELECT *
FROM accounts)SELECT *
FROM table1
JOIN table2
ON table1.account_id = table2.id;
然后，你可以按照相同的方式使用 WITH 语句添加越来越多的表格。底部的练习将确保你掌握了这些新查询的所有必要组成部分。例如
WITH table1 AS (SELECT *FROM web_events),table2 AS (SELECT *FROM accounts)SELECT *
FROM table1
JOIN table2
ON table1.account_id = table2.id;

# 针对上面利用子查询做的结果，利用公共表达式重新做了优化。

以下是使用 WITH 条件重新编写的之前问题的答案。这些查询通常更容易读懂。

提供每个区域拥有最高销售额 (total_amt_usd) 的销售代表的姓名。

WITH t1 AS (
SELECT s.name rep_name, r.name region_name, SUM(o.total_amt_usd) total_amt
FROM sales_reps s
JOIN accounts a
ON a.sales_rep_id = s.id
JOIN orders o
ON o.account_id = a.id
JOIN region r
ON r.id = s.region_id
GROUP BY 1,2
ORDER BY 3 DESC),
t2 AS (
SELECT region_name, MAX(total_amt) total_amt
FROM t1
GROUP BY 1)
SELECT t1.rep_name, t1.region_name, t1.total_amt
FROM t1
JOIN t2
ON t1.region_name = t2.region_name AND t1.total_amt = t2.total_amt;

对于具有最高销售额 (total_amt_usd) 的区域，总共下了多少个订单？

WITH t1 AS (
SELECT r.name region_name, SUM(o.total_amt_usd) total_amt
FROM sales_reps s
JOIN accounts a
ON a.sales_rep_id = s.id
JOIN orders o
ON o.account_id = a.id
JOIN region r
ON r.id = s.region_id
GROUP BY r.name),
t2 AS (
SELECT MAX(total_amt)
FROM t1)
SELECT r.name, SUM(o.total) total_orders
FROM sales_reps s
JOIN accounts a
ON a.sales_rep_id = s.id
JOIN orders o
ON o.account_id = a.id
JOIN region r
ON r.id = s.region_id
GROUP BY r.name
HAVING SUM(o.total_amt_usd) = (SELECT * FROM t2);

对于购买标准纸张数量 (standard_qty) 最多的客户（在作为客户的整个时期内），有多少客户的购买总数依然更多？

WITH t1 AS (
SELECT a.name account_name, SUM(o.standard_qty) total_std, SUM(o.total) total
FROM accounts a
JOIN orders o
ON o.account_id = a.id
GROUP BY 1
ORDER BY 2 DESC
LIMIT 1),
t2 AS (
SELECT a.name
FROM orders o
JOIN accounts a
ON a.id = o.account_id
GROUP BY 1
HAVING SUM(o.total) > (SELECT total FROM t1))
SELECT COUNT(*)
FROM t2;

对于（在作为客户的整个时期内）总消费 (total_amt_usd) 最多的客户，他们在每个渠道上有多少 web_events？

WITH t1 AS (
SELECT a.id, a.name, SUM(o.total_amt_usd) tot_spent
FROM orders o
JOIN accounts a
ON a.id = o.account_id
GROUP BY a.id, a.name
ORDER BY 3 DESC
LIMIT 1)
SELECT a.name, w.channel, COUNT(*)
FROM accounts a
JOIN web_events w
ON a.id = w.account_id AND a.id =  (SELECT id FROM t1)
GROUP BY 1, 2
ORDER BY 3 DESC;

对于总消费前十名的客户，他们的平均终身消费 (total_amt_usd) 是多少?

WITH t1 AS (
SELECT a.id, a.name, SUM(o.total_amt_usd) tot_spent
FROM orders o
JOIN accounts a
ON a.id = o.account_id
GROUP BY a.id, a.name
ORDER BY 3 DESC
LIMIT 10)
SELECT AVG(tot_spent)
FROM t1;

比所有客户的平均消费高的企业平均终身消费 (total_amt_usd) 是多少？

WITH t1 AS (
SELECT AVG(o.total_amt_usd) avg_all
FROM orders o
JOIN accounts a
ON a.id = o.account_id),
t2 AS (
SELECT o.account_id, AVG(o.total_amt_usd) avg_amt
FROM orders o
GROUP BY 1
HAVING AVG(o.total_amt_usd) > (SELECT * FROM t1))
SELECT AVG(avg_amt)
FROM t2;