Returning values even the result are empty - php

Question

I have a table that is similar below.
| user_id | point_1 | point_2 | point_3
453123 1234 32 433
321543 1 213 321
My query is something like this:
$query = "SELECT * FROM my_table WHERE user_id = 12345 OR user_id = 987654321"
Obviously, this will return nothing since user_id 12345 OR user_id 987654321 do not exist on the table.
But I still want to return something like the one below :
| user_id | point_1 | point_2 | point_3
12345 0 0 0
987654321 0 0 0

Answer 1

You could use an inline view as a rowsource for your query. To return a zero in place of a NULL (which would be returned by the outer join when no matching row is found in my_table, you can use the IFNULL function.
e.g.
SELECT s.user_id
, IFNULL(t.point_1,0) AS point_1
, IFNULL(t.point_2,0) AS point_2
, IFNULL(t.point_3,0) AS point_3
FROM ( SELECT 12345 AS user_id
UNION ALL SELECT 987654321
) s
LEFT
JOIN my_table t
ON t.user_id = s.user_id
NOTE: If datatype of user_id column my_table is character, then I'd enclose the literals in the inline view in single quotes. e.g. SELECT '12345' AS user_id. If the characterset of the column doesn't match your client characterset, e.g. database column is latin1, and client characterset is UTF8, you'd want to force the character strings to be a compatible (coercible) characterset... SELECT _latin1'12345' AS user_id

Answer 2

You can't get the result you want using only a select statement. Only rows that exist somewhere will be returned.
The only way I can think to do this is to insert the query values into a temp table and then outer join against that for your query.
So the basic process would be:
create table temp1 (user_id integer);
insert into temp1 values (987654321); -- repeat as needed for query.
select t.user_id, m.* from temp1 t left outer join my_table m on m.user_id = t.user_id;
drop table temp1;
This isn't very efficient though.

Answer 3

Your desired result resembles the result of an OUTER JOIN - when some records exist only in one table and not the other, an OUTER JOIN will show all of the rows from one of the joined tables, filling in missing fields from the other table with NULL values.
To solve your particular problem purely in SQL, you could create a second table that contains a single field with all of the user_id values that you want to be able to show in your result. Something like:
+-----------+
| user_id |
+-----------+
| 1 |
+-----------+
| 2 |
+-----------+
| 3 |
+-----------+
| ... |
+-----------+
| 12344 |
+-----------+
| 12345 |
+-----------+
| 12346 |
+-----------+
| ... |
+-----------+
And so on. If this second table is named all_ids, you could then get your desired result by modifying your query as follows (exact syntax may vary by database implementation):
SELECT
*
FROM
all_ids AS i
LEFT OUTER JOIN
my_table AS t ON i.user_id = t.user_id
WHERE
i.user_id = 12345
OR i.user_id = 987654321;
This should produce the following result set:
+-----------+----------+----------+----------+
| user_id | point_1 | point_2 | point_3 |
+-----------+----------+----------+----------+
| 12345 | NULL | NULL | NULL |
+-----------+----------+----------+----------+
| 987654321 | NULL | NULL | NULL |
+-----------+----------+----------+----------+
It should be noted that this table full of IDs could take up a significant amount of disk space. An integer column in MySQL can hold 4,294,967,296 4-byte values, or 16 GB of data sitting around purely for your convenience in displaying some other data you don't have. So unless you need some smaller range or set of IDs available, or have disk space coming out your ears, this approach simply may not be practical.
Personally, I would not ask the database to do this in the first place. Essentially it's a display issue; you already get all the information you need from the fact that certain rows were not returned by your query. I would solve the display issue outside of the database, which in your case means filling in those zeroes with PHP.