Count group by enum including possible enum values that have 0 count - mysql

Question

I have a table of items. One of the fields is a category (represented by an enum). Some categories have zero items.
So I did this:
select category, count(*) as total from items group by category;
+------------+-------+
| category | total |
+------------+-------+
| one | 6675 |
+------------+-------+
I want to generate a table like this (where two is the other possible enum value):
+------------+-------+
| category | total |
+------------+-------+
| one | 6675 |
+------------+-------+
| two | 0 |
+------------+-------+
How do I do this with an mysql SQL query?

Answer 1

Enum datatype is generally preferred for those cases where possible options (values) are not too many (prefer <= 10), and you are not going to add new options in future (atleast not very frequently). So, a good use-case for Enum is gender: (m, f, n). In your case, it would be generally better to have a Master table of all possible Categories, instead of using Enum for them. Then it is easier to do a LEFT JOIN from the Master table.
However, as asked by you:
A solution uses the enum type to generate the table, and includes 0
entries
Works for all MySQL/MariaDB versions:
We will need to get the list of all possible Enum values from INFORMATION_SCHEMA.COLUMNS:
SELECT
SUBSTRING(COLUMN_TYPE, 6, CHAR_LENGTH(COLUMN_TYPE) - 6) AS enum_values
FROM
information_schema.COLUMNS
WHERE
TABLE_NAME = 'items' -- your table name
AND
COLUMN_NAME = 'category' -- name of the column
AND
TABLE_SCHEMA = 'your_db' -- name of the database (schema)
But then, this query will give you all the enum values in comma-separated string, like below:
'one','two','three','four'
Now, we will need to convert this string into multiple rows. To achieve that, we can use a Sequence (Number series) table. You can define a permanent table in your database storing integers ranging from 1 to 100 (you may find this table helpful in many other cases as well) (OR, another approach is to use a Derived Table - check this to get an idea: https://stackoverflow.com/a/58052199/2469308).
CREATE TABLE seq (n tinyint(3) UNSIGNED NOT NULL, PRIMARY KEY(n));
INSERT INTO seq (n) VALUES (1), (2), ...... , (99), (100);
Now, we will do a JOIN between "enum values string" and seq table, based on the position of comma, to extract enum values into different rows. Note that instead of just using , (comma) to extract enum values, we would use ',' (to avoid cases when there might be a comma inside the value string). String operations utilizing Substring_Index(), Trim(), Char_Length() etc functions can be used to extract enum values. You can check this answer to get a general idea about this technique:
Schema (View on DB Fiddle)
CREATE TABLE items
(id INT UNSIGNED NOT NULL AUTO_INCREMENT PRIMARY KEY,
category ENUM('one','two','three','four'),
item_id INT UNSIGNED) ENGINE=InnoDB;
INSERT INTO items (category, item_id)
VALUES ('one', 1),
('two', 2),
('one', 2),
('one', 3);
CREATE TABLE seq (n tinyint(3) UNSIGNED NOT NULL,
PRIMARY KEY(n));
INSERT INTO seq (n) VALUES (1),(2),(3),(4),(5);
Query #1
SELECT Trim(BOTH '\'' FROM Substring_index(Substring_index(e.enum_values,
'\',\'',
seq.n),
'\',\'', -1)) AS cat
FROM (SELECT Substring(column_type, 6, Char_length(column_type) - 6) AS
enum_values
FROM information_schema.columns
WHERE table_name = 'items'
AND column_name = 'category'
AND table_schema = 'test') AS e
JOIN seq
ON ( Char_length(e.enum_values) - Char_length(REPLACE(e.enum_values,
'\',\'',
''))
) / 3 >= seq.n - 1
| cat |
| ----- |
| one |
| two |
| three |
| four |
Now, the hard part is done. All we need to do is do a LEFT JOIN from this subquery (having all category enum values) to your items table, to get Count per category.
The final query follows (View on DB Fiddle):
SELECT all_cat.cat AS category,
Count(i.item_id) AS total
FROM (SELECT Trim(BOTH '\'' FROM Substring_index(
Substring_index(e.enum_values,
'\',\'',
seq.n),
'\',\'', -1)) AS cat
FROM (SELECT Substring(column_type, 6, Char_length(column_type) - 6)
AS
enum_values
FROM information_schema.columns
WHERE table_name = 'items'
AND column_name = 'category'
AND table_schema = 'test') AS e
JOIN seq
ON ( Char_length(e.enum_values) - Char_length(
REPLACE(e.enum_values,
'\',\'',
''))
) / 3 >= seq.n - 1) AS all_cat
LEFT JOIN items AS i
ON i.category = all_cat.cat
GROUP BY all_cat.cat
ORDER BY total DESC;
Result
| category | total |
| -------- | ----- |
| one | 3 |
| two | 1 |
| three | 0 |
| four | 0 |

Answer 2

Here is some fun with MySQL 8.0 and JSON_TABLE():
select c.category, count(i.category) as total
from information_schema.COLUMNS s
join json_table(
replace(replace(replace(trim('enum' from s.COLUMN_TYPE),'(','['),')',']'),'''','"'),
'$[*]' columns (category varchar(50) path '$')
) c
left join items i on i.category = c.category
where s.TABLE_SCHEMA = 'test' -- replace with your db/schema name
and s.TABLE_NAME = 'items'
and s.COLUMN_NAME = 'category'
group by c.category
It converts the ENUM type definition from information_schema to a JSON array, which is then converted by JSON_TABLE() to a table, which you then can use for a LEFT JOIN.
See demo on db-fiddle
Note: The categories should not contain any characters from ()[]'".
But seriously – Just create the categories table. There are more reasons to do that. For example you might want to render a drop-down menu with all possible categories. That would be simple with
select category from categories

Answer 3

I would say that it's basically bad practice to encode your enumerations into the script. Therefore, create a table with the enumerations present (and their relative keys), then it's a simple case of grouping a left joined query...
SELECT
cat.enum_name,
COUNT(data.id) AS total
FROM
category_table cat
LEFT JOIN
data_table data
ON cat.cate_id = data.cat_id
GROUP BY
cat.enum_name

Answer 4

using in-select subquery
select cat.categoryname
(
select count(*) -- count total
from items as i
where i.category = cat.category -- connect
) as totalcount
from cat
order by cat.categoryname

Answer 5

You can make a fictive dataset of the different categories and do a left join with your original table as shown below.
SELECT A.category, count(*) total FROM
(SELECT 'one' as Category
UNION ALL
SELECT 'two' as Category) A
LEFT JOIN items B
ON A.Category=B.Category
GROUP BY B.Category;
If you would prefer to get a list of all the category dynamically, then save them in another table (say All_category_table) then do a join as shown below:
SELECT A.category, count(*) total FROM
(SELECT Category FROM All_category_table) A
LEFT JOIN items B
ON A.Category=B.Category
GROUP BY B.Category;

Answer 6

This answer is applicable for when you do not have another table holding the possible category values.
Let's say you have a table called real_table with a not null & value constrained column category. In this column you know you can theoretically encounter 5 different values: 'CATEGORY_0', 'CATEGORY_1', 'CATEGORY_2', 'CATEGORY_3', 'CATEGORY_4':
CREATE TABLE real_table
(
id VARCHAR(255) NOT NULL
PRIMARY KEY,
category VARCHAR(255) NOT NULL
CONSTRAINT category_in CHECK (
category in ('CATEGORY_0',
'CATEGORY_1',
'CATEGORY_2',
'CATEGORY_3',
'CATEGORY_4')
)
);
But your actual data set in the table does not include any row with value 'CATEGORY_0'. So when you run a query such as:
SELECT real_table.category AS category, COUNT(*) AS cnt
FROM real_table
GROUP BY real_table.category;
you will see, that you get result like this:
category
cnt
CATEGORY_1
150
CATEGORY_2
20
CATEGORY_3
12
CATEGORY_4
1
Hmm, the 'CATEGORY_0' is omitted. Not good.
Since your categories are not backed by another table, then you must create an artificial dataset of the possible categories that looks as below:
SELECT 'CATEGORY_0' AS category_entry
UNION ALL
SELECT 'CATEGORY_1' AS category_entry
UNION ALL
SELECT 'CATEGORY_2' AS category_entry
UNION ALL
SELECT 'CATEGORY_3' AS category_entry
UNION ALL
SELECT 'CATEGORY_4' AS category_entry;
You can use this in your original query as a table to do a right join on:
SELECT all_categories.category_entry AS category,
COUNT(real_table.id) AS cnt -- important to count some non-null value, such as PK of the real_table
FROM real_table
RIGHT JOIN
(SELECT 'CATEGORY_0' AS category_entry -- not present in any row in table 'all_categories'
UNION ALL
SELECT 'CATEGORY_1' AS category_entry
UNION ALL
SELECT 'CATEGORY_2' AS category_entry
UNION ALL
SELECT 'CATEGORY_3' AS category_entry
UNION ALL
SELECT 'CATEGORY_4' AS category_entry) all_categories
ON real_table.category = all_categories.category_entry
GROUP BY all_categories.category_entry;
Now when you run the query, you should get the desired output:
category
cnt
CATEGORY_0
0
CATEGORY_1
150
CATEGORY_2
20
CATEGORY_3
12
CATEGORY_4
1
The 'CATEGORY_0' is now included with zero cnt. Nice.
Now let's say that the category column is not not null constrained and can also possibly include some other unexpected category values (e.g. 'CATEGORY_66'):
CREATE TABLE real_table
(
id VARCHAR(255) NOT NULL
PRIMARY KEY,
category VARCHAR(255) -- nullable and no constraint for valid values
);
We would like to include these null and unexpected category counts in the result set as well.
Then we must prepare the artificial dataset of the possible categories differently:
SELECT DISTINCT all_categories.category_entry
FROM (SELECT 'CATEGORY_0' AS category_entry -- not present in any row in table 'all_categories'
UNION ALL
SELECT 'CATEGORY_1' AS category_entry
UNION ALL
SELECT 'CATEGORY_2' AS category_entry
UNION ALL
SELECT 'CATEGORY_3' AS category_entry
UNION ALL
SELECT 'CATEGORY_4' AS category_entry
UNION ALL
SELECT DISTINCT category
FROM real_table AS category_entry) all_categories;
and use it as before:
SELECT distinct_categories.category_entry AS category,
COUNT(real_table.id) AS cnt -- important to count some non-null value, such as PK of the real_table
FROM real_table
RIGHT JOIN
(SELECT DISTINCT all_categories.category_entry
FROM (SELECT 'CATEGORY_0' AS category_entry -- not present in any row in table 'all_categories'
UNION ALL
SELECT 'CATEGORY_1' AS category_entry
UNION ALL
SELECT 'CATEGORY_2' AS category_entry
UNION ALL
SELECT 'CATEGORY_3' AS category_entry
UNION ALL
SELECT 'CATEGORY_4' AS category_entry
UNION ALL
SELECT DISTINCT category
FROM real_table AS category_entry) all_categories) distinct_categories
ON real_table.category = distinct_categories.category_entry
GROUP BY distinct_categories.category_entry;
Now when you run the query, the output should also include counts for additional categories and null categories
category
cnt
CATEGORY_0
0
CATEGORY_1
150
CATEGORY_2
20
CATEGORY_3
12
CATEGORY_4
1
CATEGORY_66
13
10
Both unexpected 'CATEGORY_66' (with 13 entries) as well as null category (with 10 entries) are now included in the result set
I cannot vouch for the performance of the provided queries - somebody more experienced might weigh in on that?