Is it possible to do a SELECT
statement with a predetermined order, ie. selecting IDs 7,2,5,9 and 8 and returning them in that order, based on nothing more than the ID field?
Both these statements return them in the same order:
SELECT id FROM table WHERE id in (7,2,5,9,8)
SELECT id FROM table WHERE id in (8,2,5,9,7)
The SQL ORDER BY Keyword The ORDER BY keyword is used to sort the result-set in ascending or descending order. The ORDER BY keyword sorts the records in ascending order by default. To sort the records in descending order, use the DESC keyword.
By default, SQL Server sorts out results using ORDER BY clause in ascending order.
SELECT column-list FROM table_name [WHERE condition] [ORDER BY column1, column2, .. columnN] [ASC | DESC]; You can use more than one column in the ORDER BY clause. Make sure whatever column you are using to sort that column should be in the column-list.
SQL queries initiated by using a SELECT statement support the ORDER BY clause. The result of the SELECT statement is sorted in an ascending or descending order.
I didn't think this was possible, but found a blog entry here that seems to do the type of thing you're after:
SELECT id FROM table WHERE id in (7,2,5,9,8)
ORDER BY FIND_IN_SET(id,"7,2,5,9,8");
will give different results to
SELECT id FROM table WHERE id in (7,2,5,9,8)
ORDER BY FIND_IN_SET(id,"8,2,5,9,7");
FIND_IN_SET
returns the position of id
in the second argument given to it, so for the first case above, id
of 7 is at position 1 in the set, 2 at 2 and so on - mysql internally works out something like
id | FIND_IN_SET
---|-----------
7 | 1
2 | 2
5 | 3
then orders by the results of FIND_IN_SET
.
Your best bet is:
ORDER BY FIELD(ID,7,2,4,5,8)
...but it's still ugly.
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