SQL група по урок: Брой, сума, средно и обяснение на клаузи

В GROUP BYклаузата е мощен, но понякога труден изявление да си помисля.

Дори осем години по-късно, всеки път, когато използвам a GROUP BY, трябва да спра и да помисля какво всъщност прави.

В тази статия ще разгледаме как да изградим GROUP BYклауза, какво прави тя с вашата заявка и как можете да я използвате, за да извършвате агрегиране и да събирате прозрения за вашите данни.

Ето какво ще разгледаме:

  • Настройване на вашата база данни
  • Настройване на примерни данни (създаване на продажби)
  • Как работи една GROUP BYработа?
  • GROUP BYКлаузи за писане
  • Обобщените ( COUNT, SUM, AVG)
  • Работа с множество групи
  • Използване на функции в GROUP BY
  • Филтриране на групи с HAVING
  • Агрегира с имплицитно групиране

Настройване на вашата база данни

Преди да можем да напишем нашите заявки, трябва да настроим нашата база данни.

За тези примери ще използваме PostgreSQL, но показаните тук заявки и концепции лесно ще бъдат преведени във всяка друга съвременна система от бази данни (като MySQL, SQL Server и т.н.).

За да работим с нашата база данни PostgreSQL, можем да използваме psql - интерактивната програма за команден ред PostgreSQL. Ако имате друг клиент на база данни, с който ви е приятно да работите, това също е добре.

За начало нека създадем нашата база данни. След като PostgreSQL вече е инсталиран, можем да createdb изпълним командата на нашия терминал, за да създадем нова база данни. Обадих се на моята fcc:

$ createdb fcc 

След това нека стартираме интерактивната конзола с помощта на командата psqlи се свържем с базата данни, която току-що направихме с помощта на \c :

$ psql psql (11.5) Type "help" for help. john=# \c fcc You are now connected to database "fcc" as user "john". fcc=# 
Забележка: Почистих psqlизхода в тези примери, за да улесня четенето, така че не се притеснявайте, ако изходът, показан тук, не е точно това, което сте виждали във вашия терминал.

Препоръчвам ви да следвате тези примери и да изпълните тези заявки за себе си. Ще научите и запомните много повече, като работите през тези примери, вместо просто да ги четете.

Настройване на данните (създаване на продажби)

За нашите примери ще използваме таблица, която съхранява записите за продажбите на различни продукти в различни магазини.

Ще наречем тази таблица salesи тя ще бъде просто представяне на продажбите в магазина: името на местоположението, името на продукта, цената и времето, когато е продадена.

Ако изграждахме тази таблица в реално приложение, щяхме да настроим чужди ключове за други таблици (като locationsили products). Но за илюстриране на GROUP BYконцепциите просто ще използваме прости TEXTколони.

Нека създадем таблицата и да вмъкнем някои данни за продажбите:

CREATE TABLE sales( location TEXT, product TEXT, price DECIMAL, sold_at TIMESTAMP ); INSERT INTO sales(location, product, price, sold_at) VALUES ('HQ', 'Coffee', 2, NOW()), ('HQ', 'Coffee', 2, NOW() - INTERVAL '1 hour'), ('Downtown', 'Bagel', 3, NOW() - INTERVAL '2 hour'), ('Downtown', 'Coffee', 2, NOW() - INTERVAL '1 day'), ('HQ', 'Bagel', 2, NOW() - INTERVAL '2 day'), ('1st Street', 'Bagel', 3, NOW() - INTERVAL '2 day' - INTERVAL '1 hour'), ('1st Street', 'Coffee', 2, NOW() - INTERVAL '3 day'), ('HQ', 'Bagel', 3, NOW() - INTERVAL '3 day' - INTERVAL '1 hour'); 

Имаме три местоположения: HQ , Downtown и 1st Street.

Имаме два продукта, кафе и багел , и ние вмъкваме тези продажби с различни sold_atстойности, за да представим артикулите, продавани в различни дни и часове.

Има някои продажби днес, някои вчера, а други от вчера.

Как работи една GROUP BYработа?

За да илюстрираме как GROUP BYработи клаузата, нека първо поговорим чрез пример.

Представете си, че имахме стая, пълна с хора, които са родени в различни страни.

Ако искахме да намерим средната височина на хората в стаята за всяка държава, първо бихме помолили тези хора да се разделят на групи въз основа на тяхната държава на раждане.

След като бяха разделени в групите си, можехме да изчислим средната височина в тази група.

Ето как GROUP BYработи клаузата. Първо дефинираме как искаме да групираме редовете заедно - след това можем да извършим изчисления или агрегиране на групите.

Няколко групи

Можем да групираме данните в толкова групи или подгрупи, колкото искаме.

Например, след като помолим хората да се разделят на групи въз основа на техните страни по рождение, бихме могли да кажем на всяка от тези групи държави да се разделят допълнително на групи въз основа на цвета на очите им.

Правейки това, имаме групи хора въз основа на комбинацията от тяхната държава на раждане и цвета на очите им.

Сега бихме могли да намерим средната височина във всяка от тези по-малки групи и бихме получили по-специфичен резултат: средна височина на държава за цвят на очите .

GROUP BYклаузи често се използват за ситуации, в които можете да използвате фразата за нещо или за всяко нещо :

  • Средна височина за страна на раждане
  • Общ брой хора за всяка комбинация от цветове на очите и косата
  • Общи продажби на продукт

GROUP BYКлаузи за писане

А GROUP BYклауза е много лесно да се напише-ние просто използвайте ключовите думи GROUP BYи след това определете областта (ите), което искаме да се групират по:

SELECT ... FROM sales GROUP BY location;

Тази проста заявка групира salesданните ни по locationколоната.

Направихме групирането - но какво влагаме в нашето SELECT?

Очевидното нещо, което трябва да изберете, е нашето - ние се locationгрупираме по него, така че поне искаме да видим името на групите, които сме направили:

SELECT location FROM sales GROUP BY location; 

Резултатът са нашите три места:

 location ------------ 1st Street HQ Downtown (3 rows) 

If we look at our raw table data (SELECT * FROM sales;), we'll see that we have four rows with a location of HQ, two rows with a location of Downtown, and two rows with a location of 1st Street:

 product | location | price | sold_at ---------+------------+-------+---------------------------- Coffee | HQ | 2 | 2020-09-01 09:42:33.085995 Coffee | HQ | 2 | 2020-09-01 08:42:33.085995 Bagel | Downtown | 3 | 2020-09-01 07:42:33.085995 Coffee | Downtown | 2 | 2020-08-31 09:42:33.085995 Bagel | HQ | 2 | 2020-08-30 09:42:33.085995 Bagel | 1st Street | 3 | 2020-08-30 08:42:33.085995 Coffee | 1st Street | 2 | 2020-08-29 09:42:33.085995 Bagel | HQ | 3 | 2020-08-29 08:42:33.085995 (8 rows) 

By grouping on the location column, our database takes these inputs rows and identifies the unique locations among them—these unique locations serve as our "groups."

But what about the other columns in our table?

If we try to select a column like product that we didn't group by...

SELECT location, product FROM sales GROUP BY location; 

...we run into this error:

ERROR: column "sales.product" must appear in the GROUP BY clause or be used in an aggregate function 

The problem here is we've taken eight rows and squished or distilled them down to three.

We can't just return the rest of the columns like normal—we had eight rows, and now we have three.

What do we do with the remaining five rows of data? Which of the eight rows' data should be displayed on these three distinct location rows?

There's not a clear and definitive answer here.

To use the rest of our table data, we also have to distill the data from these remaining columns down into our three location groups.

This means that we have to aggregate or perform a calculation to produce some kind of summary information about our remaining data.

Aggregations (COUNT, SUM, AVG)

Once we've decided how to group our data, we can then perform aggregations on the remaining columns.

These are things like counting the number of rows per group, summing a particular value across the group, or averaging information within the group.

To start, let's find the number of sales per location.

Since each record in our sales table is one sale, the number of sales per location would be the number of rows within each location group.

To do this we'll use the aggregate function COUNT() to count the number of rows within each group:

SELECT location, COUNT(*) AS number_of_sales FROM sales GROUP BY location; 

We use COUNT(*) which counts all of the input rows for a group.

(COUNT() also works with expressions, but it has slightly different behavior.)

Here's how the database executes this query:

  • FROM sales — First, retrieve all of the records from the sales table
  • GROUP BY location — Next, determine the unique location groups
  • SELECT ... — Finally, select the location name and the count of the number of rows in that group

We also give this count of rows an alias using AS number_of_sales to make the output more readable. It looks like this:

 location | number_of_sales ------------+----------------- 1st Street | 2 HQ | 4 Downtown | 2 (3 rows) 

The 1st Street location has two sales, HQ has four, and Downtown has two.

Here we can see how we've taken the remaining column data from our eight independent rows and distilled them into useful summary information for each location: the number of sales.

SUM

In a similar way, instead of counting the number of rows in a group, we could sum information within the group—like the total amount of money earned from those locations.

To do this we'll use the SUM() function:

SELECT location, SUM(price) AS total_revenue FROM sales GROUP BY location; 

Instead of counting the number of rows in each group we sum the dollar amount of each sale, and this shows us the total revenue per location:

 location | total_revenue ------------+--------------- 1st Street | 5 HQ | 9 Downtown | 5 (3 rows) 

Average (AVG)

Finding the average sale price per location just means swapping out the SUM() function for the AVG() function:

SELECT location, AVG(price) AS average_revenue_per_sale FROM sales GROUP BY location; 

Working with multiple groups

So far we've been working with just one group: location.

What if we wanted to sub-divide that group even further?

Similar to the "birth countries and eye color" scenario we started with, what if we wanted to find the number of sales per product per location?

To do this all we need to do is add the second grouping condition to our GROUP BY statement:

SELECT ... FROM sales GROUP BY location, product;

By adding a second column in our GROUP BY we further sub-divide our location groups into location groups per product.

Тъй като сега също групираме по productколона, вече можем да го върнем в нашата SELECT!

(Ще хвърля някои ORDER BYклаузи за тези заявки, за да улесня четенето на изхода.)

SELECT location, product FROM sales GROUP BY location, product ORDER BY location, product; 

Разглеждайки резултата от новото ни групиране, можем да видим нашите уникални комбинации местоположение / продукт:

 location | product ------------+--------- 1st Street | Bagel 1st Street | Coffee Downtown | Bagel Downtown | Coffee HQ | Bagel HQ | Coffee (6 rows) 

Сега, когато имаме свои групи, какво искаме да направим с останалите данни от колоните?

Е, можем да намерим броя продажби на продукт на място, като използваме същите обобщени функции като преди:

SELECT location, product, COUNT(*) AS number_of_sales FROM sales GROUP BY location, product ORDER BY location, product; 
 location | product | number_of_sales ------------+---------+----------------- 1st Street | Bagel | 1 1st Street | Coffee | 1 Downtown | Bagel | 1 Downtown | Coffee | 1 HQ | Bagel | 2 HQ | Coffee | 2 (6 rows) 
Като упражнение за четеца ™: намерете общите приходи (сума) на всеки продукт на място.

Използване на функции в GROUP BY

След това нека се опитаме да намерим общия брой продажби на ден .

If we follow a similar pattern as we did with our locations and group by our sold_at column...

SELECT sold_at, COUNT(*) AS sales_per_day FROM sales GROUP BY sold_at ORDER BY sold_at; 

...we might expect to have each group be each unique day—but instead we see this:

 sold_at | sales_per_day ----------------------------+--------------- 2020-08-29 08:42:33.085995 | 1 2020-08-29 09:42:33.085995 | 1 2020-08-30 08:42:33.085995 | 1 2020-08-30 09:42:33.085995 | 1 2020-08-31 09:42:33.085995 | 1 2020-09-01 07:42:33.085995 | 1 2020-09-01 08:42:33.085995 | 1 2020-09-01 09:42:33.085995 | 1 (8 rows) 

It looks like our data isn't grouped at all—we get each row back individually.

But, our data is actually grouped! The problem is each row's sold_at is a unique value—so every row gets its own group!

The GROUP BY is working correctly, but this is not the output we want.

The culprit is the unique hour/minute/second information of the timestamp.

Each of these timestamps differ by hours, minutes, or seconds—so they are each placed in their own group.

We need to convert each of these date and time values into just a date:

  • 2020-09-01 08:42:33.085995 =>2020-09-01
  • 2020-09-01 09:42:33.085995 =>2020-09-01

Converted to a date, all of the timestamps on the same day will return the same date value—and will therefore be placed into the same group.

To do this, we'll cast the sold_at timestamp value to a date:

SELECT sold_at::DATE AS date, COUNT(*) AS sales_per_day FROM sales GROUP BY sold_at::DATE ORDER BY sold_at::DATE; 

In our GROUP BY clause we use ::DATE to truncate the timestamp portion down to the "day." This effectively chops off the hours/minutes/seconds of the timestamp and just returns the day.

In our SELECT, we also return this same expression and give it an alias to pretty up the output.

For the same reason we couldn't return product without grouping by it or performing some kind of aggregation on it, the database won't let us return just sold_at—everything in the SELECT must either be in the GROUP BY or some kind of aggregate on the resulting groups.

The result is the sales per day that we originally wanted to see:

 date | sales_per_day ------------+--------------- 2020-08-29 | 2 2020-08-30 | 2 2020-08-31 | 1 2020-09-01 | 3 (4 rows) 

Filtering groups with HAVING

Next let's look at how to filter our grouped rows.

To do this, let's try to find days where we had more than one sale.

Without grouping, we would normally filter our rows by using a WHERE clause. For example:

SELECT * FROM sales WHERE product = 'Coffee'; 

With our groups, we may want to do something like this to filter our groups based on the count of rows...

SELECT sold_at::DATE AS date, COUNT(*) AS sales_per_day FROM sales WHERE COUNT(*) > 1 -- filter the groups? GROUP BY sold_at::DATE; 

Unfortunately, this doesn't work and we receive this error:

ERROR:  aggregate functions are not allowed in WHERE

Aggregate functions are not allowed in the WHERE clause because the WHERE clause is evaluated before the GROUP BY clause—there aren't any groups yet to perform calculations on.

But, there is a type of clause that allows us to filter, perform aggregations, and it is evaluated after the GROUP BY clause: the HAVING clause.

The HAVING clause is like a WHERE clause for your groups.

To find days where we had more than one sale, we can add a HAVING clause that checks the count of rows in the group:

SELECT sold_at::DATE AS date, COUNT(*) AS sales_per_day FROM sales GROUP BY sold_at::DATE HAVING COUNT(*) > 1; 

This HAVING clause filters out any rows where the count of rows in that group is not greater than one, and we see that in our result set:

 date | sales_per_day ------------+--------------- 2020-09-01 | 3 2020-08-29 | 2 2020-08-30 | 2 (3 rows) 

Just for the sake of completeness, here's the order of execution for all parts of a SQL statement:

  • FROM — Retrieve all of the rows from the FROM table
  • JOIN — Perform any joins
  • WHERE — Filter rows
  • GROUP BY - Form groups
  • HAVING - Filter groups
  • SELECT - Select the data to return
  • ORDER BY - Order the output rows
  • LIMIT - Return a certain number of rows

Aggregates with implicit grouping

The last topic we'll look at is aggregations that can be performed without a GROUP BY—or maybe better said they have an implicitgrouping.

These aggregations are useful in scenarios where you want to find one particular aggregate from a table—like the total amount of revenue or the greatest or least value of a column.

For example, we could find the total revenue across all locations by just selecting the sum from the entire table:

SELECT SUM(price) FROM sales; 
 sum ----- 19 (1 row) 

So far we've done $19 of sales across all locations (hooray!).

Another useful thing we could query is the first or last of something.

For example, what is the date of our first sale?

To find this we just use the MIN() function:

SELECT MIN(sold_at)::DATE AS first_sale FROM sales; 
 first_sale ------------ 2020-08-29 (1 row) 

(To find the date of the last sale just substitute MAX()for MIN().)

Using MIN / MAX

While these simple queries can be useful as a standalone query, they're often parts of filters for larger queries.

For example, let's try to find the total sales for the last day that we had sales.

One way we could write that query would be like this:

SELECT SUM(price) FROM sales WHERE sold_at::DATE = '2020-09-01'; 

This query works, but we've obviously hardcoded the date of 2020-09-01.

09/01/2020 may be the last date we had a sale, but it's not always going to be that date. We need a dynamic solution.

This can be achieved by combining this query with the MAX() function in a subquery:

SELECT SUM(price) FROM sales WHERE sold_at::DATE = ( SELECT MAX(sold_at::DATE) FROM sales ); 

In our WHERE clause we find the largest date in our table using a subquery: SELECT MAX(sold_at::DATE) FROM sales.

Then, we use this max date as the value we filter the table on, and sum the price of each sale.

Implicit grouping

I say that these are implicit groupings because if we try to select an aggregate value with a non-aggregated column like this...

SELECT SUM(price), location FROM sales; 

...we get our familiar error:

ERROR: column "sales.location" must appear in the GROUP BY clause or be used in an aggregate function 

GROUP BY is a tool

As with many other topics in software development, GROUP BY is a tool.

There are many ways to write and re-write these queries using combinations of GROUP BY, aggregate functions, or other tools like DISTINCT, ORDER BY, and LIMIT.

Understanding and working with GROUP BY's will take a little bit of practice, but once you have it down you'll find an entirely new batch of problems are now solvable to you!

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Thanks for reading!

John