Sử dụng hàm CONCAT_WS(). Tạo bảng demo38:
mysql> CREATE TABLE demo38 (
-> user_id INT,
-> user_first_name VARCHAR(20),
-> user_last_name VARCHAR(20),
-> user_date_of_birth DATE
-> );
Chèn dữ liệu:
mysql> INSERT INTO demo38 VALUES(10,'John','Smith','1990-10-01');
mysql> INSERT INTO demo38 VALUES(11,'David','Miller','1994-01-21');
mysql> INSERT INTO demo38 VALUES(11,'John','Doe','1992-02-01');
mysql> INSERT INTO demo38 VALUES(12,'Adam','Smith','1996-11-11');
mysql> INSERT INTO demo38 VALUES(13,'Chris','Brown','1997-03-10');
Hiển thị dữ liệu:
mysql> SELECT * FROM demo38;
Kết quả:
+---------+-----------------+----------------+--------------------+
| user_id | user_first_name | user_last_name | user_date_of_birth |
+---------+-----------------+----------------+--------------------+
| 10 | John | Smith | 1990-10-01 |
| 11 | David | Miller | 1994-01-21 |
| 11 | John | Doe | 1992-02-01 |
| 12 | Adam | Smith | 1996-11-11 |
| 13 | Chris | Brown | 1997-03-10 |
+---------+-----------------+----------------+--------------------+
Truy vấn nối chuỗi và lọc theo user_id IN (11,13):
mysql> SELECT CONCAT_WS('/', user_first_name, user_last_name, 'the date of birth year is=', DATE_FORMAT(user_date_of_birth, '%Y')) AS Output
-> FROM demo38
-> WHERE user_id IN (11,13);
Kết quả:
+----------------------------------------------+
| Output |
+----------------------------------------------+
| David/Miller/the date of birth year is=/1994 |
| John/Doe/the date of birth year is=/1992 |
| Chris/Brown/the date of birth year is=/1997 |
+----------------------------------------------+