将“其他”数据存储到 MySQL 数据库中的最佳方式

Question

I have a form with radio buttons which stores the value (which is the ID) in my MySQL database along with the necessary information from the user.

INSERT INTO table (user_id, name, address, prefer_id) VALUES ('', ?, ?, ?);

So when I try to fetch the data, I use LEFT JOIN , to get the necessary description from table2 :

SELECT a.name, a.address, b.prefer_desc FROM table a
LEFT JOIN table2 b ON a.prefer_id = b.prefer_id
WHERE a.user_id = ?

But I have created an other option, in case the option the user prefers is not in the list. A textbox will appear when the user selects Other in the list of radio buttons so they can type-in freely their preferred data. Check this fiddle to see an example.

The first logic that I've thinked of is to create a separate table which stores the typed-in data of the user.

other_tb:

 other_id | user_id | typed_in |
----------+---------+----------+
    1     |     1   |   cake   |
    2     |     3   |   pizza  |

So when I fetch the data, I use php 's if() condition if the prefer_id is 4 (or other), and if it does, I will use another SELECT query to get the other data in other_tb table.

SELECT typed_in FROM other_tb WHERE user_id = ?

Is there a way to do all of this in a single query?

OR

Is this the best option, or is there a right or better way in this kind of situation?

Answer 1

try this

SELECT 
    a.name, 
    a.address, 
    IF(a.prefer_id=4,(SELECT typed_in FROM other_tb WHERE user_id = a.user_id),b.prefer_desc) as prefer_desc
FROM table a
LEFT JOIN table2 b 
    ON a.prefer_id = b.prefer_id
WHERE a.user_id = ?

Answer 2

If you are using the second table as you described, and you want to keep the same format of the output and assuming the records in other_tb are unique for each user_id , you could use something like this:

SELECT a.name, a.address, 
       CASE 
           WHERE a.prefer_id = 4 THEN c.typed_in
           ELSE b.prefer_desc
       END CASE AS prefer_desc
FROM table a
LEFT JOIN table2   b ON a.prefer_id = b.prefer_id
LEFT JOIN other_tb c ON a.user_id   = c.user_id
WHERE a.user_id = ?

I think this can also be written as:

SELECT a.name, a.address, 
       CASE a.prefer_id
           WHERE 4 THEN c.typed_in
           ELSE b.prefer_desc
       END CASE AS prefer_desc
FROM table a
LEFT JOIN table2   b ON a.prefer_id = b.prefer_id
LEFT JOIN other_tb c ON a.user_id   = c.user_id
WHERE a.user_id = ?

Answer 3

There are two approaches that I was able to practice depending on the situation:

1. Allow users to add more options so that their answer is still indexed to table2 . So when fetching data, you may still use LEFT JOIN to get the user's preferred option without any other condition. But this method will allow users to see the options that was added by other users.
2. If the options are "fixed" and only the administrator is allowed to add options to table2 , have another column to table1 , which stores the other answer of the user (let say other_option column). This method allows the user to still freely state their option without adding options to table2 .

Here is table1 :

+---------+------+---------+-----------+--------------+
| user_id | name | address | prefer_id | other_option |
+---------+------+---------+-----------+--------------+

What I do is I put 0 to prefer_id if the user prefers to answer other_option . So your program expects that if the prefer_id is 0, you have an input in other_option .