Комментарии к публикации
Выполнение запросов в MySQL

У вас неточность в разделе «Соединение таблиц»:

У поля date указан тип данных DATE, но при этом ниже в это поле вставляются данные в формате DATETIME, TIMESTAMP.

P.S. Можно, чтоб вместо мейла ник отображался? А то ник поставил уже после публикации комментария.

В пояснении к одному из запросов пропущена звездочка

У вас:
"Часть запроса users_data., product сообщает о необходимости..."

Как должно быть, если я все верно понял:
"Часть запроса users_data.*, product сообщает о необходимости..."

Спасибо еще раз за курс!
Обратите внимание, в разделе Сортировка, для команды

SELECT * FROM users ORDER by name;

приведен пример (картинка) для команды

SELECT * FROM users WHERE NOT name = "Ольга";

который был указан выше.

UPDATE users SET name="Ваня" WHERE name = "Иван";
DELETE FROM users WHERE email = "[email protected]";
SELECT * FROM users WHERE name LIKE "%т%" or name LIKE "%с%";
SELECT name FROM users GROUP BY name HAVING COUNT(name)=1;

Делюсь аптечкой от ERROR 1055 при попытке группировки (у кого первоначально установлен sql_mode = only_full_group_by):

SET GLOBAL sql_mode=(SELECT REPLACE(@@sql_mode, 'ONLY_FULL_GROUP_BY',''));

Мне помогло...

UPDATE users SET name='Ваня' WHERE name='Иван';
DELETE FROM users WHERE name='[email protected]';
SELECT * FROM users WHERE name LIKE '%т%' OR name LIKE '%с%';
SELECT name FROM users GROUP BY name HAVING COUNT(name) = 1;

Задание №1:

UPDATE users SET name = 'Ваня' WHERE name = 'Иван';

Задание №2:

DELETE FROM users WHERE email = '[email protected]';

Задание №3:

SELECT * FROM users WHERE name LIKE '%т%';

Задание №4:

SELECT name, COUNT(name) FROM users GROUP BY name HAVING COUNT(name)=1;

1) UPDATE users SET name = "Ваня" WHERE name = "Иван";
2) DELETE FROM users WHERE email = "[email protected]";
3) SELECT * FROM users WHERE name LIKE "%т%" OR name LIKE "%с%";
4) SELECT name FROM users GROUP BY name HAVING COUNT(name)=1;

UPDATE users SET name = "Ваня" WHERE name = "Иван";
DELETE FROM users WHERE email = "[email protected]";
SELECT * FROM users WHERE name LIKE "%т%" OR name LIKE "%с%";
SELECT name, COUNT(name) FROM users GROUP BY name HAVING COUNT(name)=1;

UPDATE users SET name = "Ваня" WHERE name = "Иван";
DELETE FROM users WHERE email = "[email protected]";
SELECT name FROM users WHERE name LIKE "%т%" or name LIKE "%с%";
SELECT name, COUNT(name) FROM users GROUP BY name HAVING COUNT(name)=1;

Добрый день, на самом деле сервер по умолчанию не позволяет такую вольность (и это наверно правильно):
SELECT id, email, name FROM users GROUP BY name;
на такой запрос появляется ошибка (реальный запрос SELECT marka, plate_nr FROM trucks GROUP BY marka; и ответ сервера):
ERROR 1055 (42000): Expression #2 of SELECT list is not in GROUP BY clause and contains nonaggregated column 'test.trucks.plate_nr' which is not functionally dependent on columns in GROUP BY clause; this is incompatible with sql_mode=only_full_group_by
чтобы не было ошибки надо либо менять настройки сервера либо писать:
SELECT ANY_VALUE(id), ANY_VALUE(email), name FROM users GROUP BY name;
проверено на MySQL 8.0.20, Debian.

Д/З

UPDATE users SET name='Ваня' WHERE name='Иван';
DELETE FROM users WHERE email='[email protected]';
SELECT * FROM users WHERE name LIKE "%т%" OR name LIKE "%с%";
SELECT name FROM users GROUP BY name HAVING COUNT(name)=1;

UPDATE users SET name="Ваня" WHERE name="Иван";
DELETE FROM users WHERE email="[email protected]";
SELECT * FROM users WHERE name LIKE "%т%" OR name LIKE "%с%";
SELECT name, COUNT(name) FROM users GROUP BY name HAVING COUNT(name)=1;

UPDATE users SET name = "Ваня" WHERE name = "Иван";

DELETE FROM users WHERE email = "[email protected]";

SELECT * FROM users WHERE name LIKE "%т%" OR name LIKE "%с%";

SELECT name, COUNT(name) FROM users GROUP BY name HAVING COUNT(name)=1;

Первое задание

UPDATE users SET name = "Ваня" WHERE name = "Иван"

Второе задание

DELETE FROM `users` WHERE id = 8

Третье задание

SELECT * FROM `users` WHERE name LIKE "%т%" OR "%c%"

Четвертое задание

SELECT DISTINCT name FROM `users`

UPDATE users SET name = "Ваня" WHERE name = "Иван"
DELETE FROM users WHERE email = "[email protected]"
SELECT * FROM users WHERE name LIKE "%т%" OR name LIKE "%с%"
SELECT name, COUNT(name)FROM users GROUP BY name HAVING COUNT(name) = 1

UPDATE users SET name = 'Ваня' WHERE name = 'Иван';
DELETE FROM users WHERE email = '[email protected]';
SELECT * FROM users WHERE (name LIKE '%т%') OR (name LIKE '%с%');
SELECT name FROM users GROUP BY name HAVING count(email) = 1;

1. Напишите запрос, который обновит у всех пользователей имя с "Иван" на "Ваня".

   UPDATE users SET name = 'Ваня' WHERE name = 'Иван';

2. Напишите запрос, который удалит из базы пользователя с почтой [email protected]

   DELETE FROM users WHERE email = “[email protected]”;

3. Напишите запрос, который выведет все записи, у которых в имени есть хотя бы одна из букв: "т" или "с"

   SELECT * FROM users WHERE name LIKE "%т%" OR name LIKE "%с%";

4. Выведите имена, встречающиеся в таблице только один раз.

   SELECT name FROM users GROUP BY name HAVING COUNT(name)=1;

Исправил 1 -е.
1.

UPDATE users SET name="Ваня" WHERE name = "Иван";

update users
    set name='Ваня'
    where name='Иван';

delete from users 
    where email='[email protected]';

select * from users 
    where name Like '%т%' or name like '%с%';

select name from users group by name having count(name)=1;

UPDATE users SET name = "Ваня" WHERE name = "Иван";

DELETE FROM users WHERE email = "[email protected]";

SELECT * FROM users WHERE name LIKE "%т%" OR name LIKE "%с%";

SELECT name FROM users GROUP BY name HAVING COUNT(name) = 1;

update users set name = 'Ваня' where name = 'Иван';
delete from users where email = '[email protected]';
select * from users where name like '%т%' or name like '%с%';
select name from users group by name having count(name) = 1;

UPDATE users SET name = "Ваня" WHERE name = "Иван";
DELETE FROM users WHERE email="[email protected]";
SELECT * FROM users WHERE name LIKE "%т%" OR name LIKE "%с%";
SELECT name, COUNT(name) FROM users GROUP BY name HAVING COUNT(name)=1;

Задание 1:
update users set name = "Ваня" where name = "Иван";
Задание 2:
delete from users where email = "[email protected]";
Задание 3:
select * from users where name like "%т%" or "%с%";
Задание 4:
select name from users group by name having count(name) = 1;

UPDATE users 
SET name = 'Ваня' 
WHERE
name = 'Иван'

DELETE 
FROM
    users 
WHERE
    email = '[email protected]'

SELECT
    * 
FROM
    users 
WHERE
    name LIKE '%т%' 
    OR name LIKE '%с%'

SELECT name,

       COUNT( name ) 
FROM
    users 
GROUP BY
name 
HAVING
    COUNT( name )=1

1) UPDATE users SET name = "Ваня" WHERE name = "Иван";
2) DELETE FROM users WHERE email = "[email protected]";
3) SELECT * FROM users WHERE name LIKE "%т%" or name LIKE "%с%";
4) SELECT name FROM users GROUP BY name HAVING COUNT(name)=1;

1. UPDATE users SET name = "Ваня" WHERE name = "Иван";
2. DELETE FROM users WHERE email = "[email protected]";
3. SELECT * FROM users WHERE name LIKE "%т%" OR name LIKE "%с%";
4. SELECT name FROM users GROUP BY name HAVING COUNT(name) = 1;

Напишите запрос, который обновит у всех пользователей имя с "Иван" на "Ваня".

UPDATE `users` SET `name`= "Ваня" WHERE name LIKE "%Иван%";

Напишите запрос, который удалит из базы пользователя с почтой "[email protected]"

DELETE FROM users WHERE id = 8;

Напишите запрос, который выведет все записи, у которых в имени есть хотя бы одна из букв: "т" или "с"

SELECT * FROM users WHERE name LIKE "%т%" OR "%с%";

Выведите имена, встречающиеся в таблице только один раз.

SELECT name, COUNT(name) FROM users GROUP BY name HAVING COUNT(name)= 1;

UPDATE users SET name="Ваня" WHERE name = "Иван";
DELETE FROM users WHERE email = "[email protected]";
SELECT * FROM users WHERE name LIKE "%т%" or name LIKE "%с%";
SELECT name FROM users GROUP BY name HAVING COUNT(name)=1;

UPDATE users SET name = "Ваня" WHERE name = "Иван";
DELETE FROM users WHERE email = "[email protected]";
SELECT * FROM users WHERE name LIKE "%т%" OR name LIKE "%с%";
SELECT name, COUNT(name) FROM users GROUP BY name HAVING COUNT(name)=1;