Возможно, вы захотите использовать синтаксис INSERT INTO ... SELECT
:
INSERT INTO final_table
SELECT id, name, email, pic
FROM data
JOIN pics ON (pics.msg_id = data.id);
Пример:
Ваши текущие данные:
CREATE TABLE data (id int, name varchar(20), email varchar(100));
INSERT INTO data VALUES (1, 'name1', 'a@example.com');
INSERT INTO data VALUES (2, 'name2', 'b@example.com');
INSERT INTO data VALUES (3, 'name3', 'c@example.com');
INSERT INTO data VALUES (4, 'name4', 'd@example.com');
CREATE TABLE pics (msg_id int, pic varchar(100));
INSERT INTO pics VALUES (1, 'pic1.jpg');
INSERT INTO pics VALUES (1, 'pic2.jpg');
INSERT INTO pics VALUES (2, 'pic3.jpg');
INSERT INTO pics VALUES (2, 'pic4.jpg');
INSERT INTO pics VALUES (3, 'pic5.jpg');
Ваш новый стол:
CREATE TABLE final_table (
id int, name varchar(20), email varchar(100), pic varchar(100)
);
INSERT INTO final_table
SELECT id, name, email, pic
FROM data
LEFT JOIN pics ON (pics.msg_id = data.id);
Результат:
SELECT * FROM final_table;
+------+-------+---------------+----------+
| id | name | email | pic |
+------+-------+---------------+----------+
| 1 | name1 | a@example.com | pic1.jpg |
| 1 | name1 | a@example.com | pic2.jpg |
| 2 | name2 | b@example.com | pic3.jpg |
| 2 | name2 | b@example.com | pic4.jpg |
| 3 | name3 | c@example.com | pic5.jpg |
| 4 | name4 | d@example.com | NULL |
+------+-------+---------------+----------+
6 rows in set (0.00 sec)