我可以在 MySQL 中使用一个查询在表中插入（“挤入”）一行并在插入的行之后增加剩余值吗？

Question

I wonder if I can insert a row in a table and increase all remaining values after the inserted row?

Take look at the sample table below:

+----+----------+-----------+ 
| id | position | data      |
+----+----------+-----------+ 
| 1  |  1       | Paris     |
| 2  |  2       | London    |
| 3  |  3       | Berlin    |
| 4  |  4       | Madrid    |
+----+----------+-----------+ 

The id column is the primary key. The position column is used to sort the data in custom order. I now wonder if it's possible to create an INSERT/UPDATE-query in MySQL to "sqeeze in" a new city in the table, ie the position values need to be increased after the insert, like this:

+----+----------+-----------+ 
| id | position | data      |
+----+----------+-----------+ 
| 1  |  1       | Paris     |
| 2  |  2       | London    |
| 5  |  3       | New York  |  <-- inserted row... positions after are increased
| 3  |  4       | Berlin    |  <-- position +1
| 4  |  5       | Madrid    |  <-- position +1
+----+----------+-----------+ 

I first thought that I solve this in PHP by modifying each row in a loop. But I hope I can solve this with one single INSERT-query. Is that possible?

INSERT INTO table SET position=3, data="New York" {AND INCREASE THE REST of postion}

Answer 1

Rows in tables are not physically stored in-order of their primary-key values (the real answer, as ever, is "it's complicated" - I'll be happy to go into detail on this point if you like).

But w.r.t. your question, the solution is straightforward:

SET @insertAtPosition = 3;
SET @insertData = 'New York';

BEGIN TRANSACTION;

UPDATE tbl SET Position = Position + 1 WHERE Position >= @insertAtPosition;

INSERT INTO tbl ( Position, Data ) VALUES ( @insertAtPosition, @insertData );

COMMIT TRANSACTION;

---

Note that the use of BEGIN TRANSACTION and COMMIT TRANSACTION is important so that the Position values won't be updated without a new row being successfully inserted - if the INSERT (or the UPDATE ) fails for whatever reason, then the Position values will be unchanged, as though the attempt was never made.