如何使用变量作为 SELECT 语句的列名

Question

I am unable to use a previously defined variable representing a column name in a SELECT statement and need some help.

TABLE:

main | i1 | i2 ||..........| i24  | i25 |
.
. 
17 | 121 | 123| .........
19 | 123 | 212| .........   | 4832 | 56392 |

From a previous SELECT/SET statement @indent variable equaled 25. As I can't use an integer as a column name the columns are i1 - i25 in the tableindex. However when I attempt to use the @VARIABLE in the SELECT statment I get the following:

SET @colindex1=(CONCAT(i, @indent);    
SELECT @colindex1 FROM tableindex WHERE main=19;

RESULTS

| @colindex1 |
|  i25  |

SHOULD BE THE VALUE OF THE CELL:

| i25  |
| 56392|

So the next attach was to use a DECLARE statement.

DECLARE @colindex1 VARCHAR;

But I got the following error:

ERROR 1064 (42000): You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'declare @colindex1 INT' at line 1

I need to combine my integer (25) value with the letter (i) for the column "i25' and then use it a SELECT statement. However I am failing at this point.

Can someone help

BASED ON ANSWER HERE IS A TEST USING THE SPECIFIED SYNTAX AND ERROR:

SET @indent = 1+24; 
SET @colindent1 = (SELECT CONCAT('i', @indent)); 
set @stmt = (SELECT @colindent1 FROM tableindex WHERE main=19);
 PREPARE thestmt FROM @stmt; 

ERROR 1064 (42000): ..near 'i25' at line 1 AAAAHH! WHAT AM I DOING WRONG?

HOWEVER:

set @stmt = (SELECT i25 FROM tableindex WHERE main=19);
Query OK, 0 rows affected (0.00 sec)

Answer 1

To use variables to represent columns (or tables), you need to use PREPARE/EXECUTE. Try something like this:

    SET @colindex1=(CONCAT(i, @indent);    
    SET @stmt = 'SELECT @colindex1 FROM tableindex WHERE main=19';
    PREPARE thestmt FROM @stmt;
    EXECUTE thestmt;

Answer 2

Keep it simple:

SET @stmt = CONCAT('SELECT i', @indent,' FROM tableindex WHERE main=19';
PREPARE thestmt FROM @stmt;
EXECUTE thestmt;