nickf
nickf

Reputation: 546055

Is it possible to perform a bitwise group function?

I have a field in a table which contains bitwise flags. Let's say for the sake of example there are three flags: 4 => read, 2 => write, 1 => execute and the table looks like this*:

  user_id  |  file  |  permissions
-----------+--------+---------------
        1  |  a.txt |  6    ( <-- 6 = 4 + 2 = read + write)
        1  |  b.txt |  4    ( <-- 4 = 4 = read)
        2  |  a.txt |  4
        2  |  c.exe |  1    ( <-- 1 = execute)

I'm interested to find all users who have a particular flag set (eg: write) on ANY record. To do this in one query, I figured that if you OR'd all the user's permissions together you'd get a single value which is the "sum total" of their permissions:

  user_id  |  all_perms
-----------+-------------
        1  |  6        (<-- 6 | 4 = 6)
        2  |  5        (<-- 4 | 1 = 5)

*My actual table isn't to do with files or file permissions, 'tis but an example

Is there a way I could perform this in one statement? The way I see it, it's very similar to a normal aggregate function with GROUP BY:

SELECT user_id, SUM(permissions) as all_perms
FROM permissions
GROUP BY user_id

...but obviously, some magical "bitwise-or" function instead of SUM. Anyone know of anything like that?

(And for bonus points, does it work in oracle?)

Upvotes: 23

Views: 7989

Answers (5)

Josh Bode
Josh Bode

Reputation: 3742

You would need to know the possible permission components (1, 2 and 4) apriori (thus harder to maintain), but this is pretty simple and would work:

SELECT user_id,
       MAX(BITAND(permissions, 1)) +
       MAX(BITAND(permissions, 2)) +
       MAX(BITAND(permissions, 4)) all_perms
FROM permissions
GROUP BY user_id

Upvotes: 1

manji
manji

Reputation: 47978

MySQL:

SELECT user_id, BIT_OR(permissions) as all_perms
FROM permissions
GROUP BY user_id

Upvotes: 22

Lieven Keersmaekers
Lieven Keersmaekers

Reputation: 58441

I'm interested to find all users who have a particular flag set (eg: write) on ANY record

What's wrong with simply

SELECT DISTINCT User_ID
FROM Permissions
WHERE permissions & 2 = 2

Upvotes: 0

cagcowboy
cagcowboy

Reputation: 30848

And you can do a bitwise or with...

FUNCTION BITOR(x IN NUMBER, y IN NUMBER)
RETURN NUMBER
AS
BEGIN
    RETURN x + y - BITAND(x,y);
END;

Upvotes: 2

nickf
nickf

Reputation: 546055

Ah, another one of those questions where I find the answer 5 minutes after asking... Accepted answer will go to the MySQL implementation though...

Here's how to do it with Oracle, as I discovered on Radino's blog

You create an object...

CREATE OR REPLACE TYPE bitor_impl AS OBJECT
(
  bitor NUMBER,

  STATIC FUNCTION ODCIAggregateInitialize(ctx IN OUT bitor_impl) RETURN NUMBER,

  MEMBER FUNCTION ODCIAggregateIterate(SELF  IN OUT bitor_impl,
                                       VALUE IN NUMBER) RETURN NUMBER,

  MEMBER FUNCTION ODCIAggregateMerge(SELF IN OUT bitor_impl,
                                     ctx2 IN bitor_impl) RETURN NUMBER,

  MEMBER FUNCTION ODCIAggregateTerminate(SELF        IN OUT bitor_impl,
                                         returnvalue OUT NUMBER,
                                         flags       IN NUMBER) RETURN NUMBER
)
/

CREATE OR REPLACE TYPE BODY bitor_impl IS
  STATIC FUNCTION ODCIAggregateInitialize(ctx IN OUT bitor_impl) RETURN NUMBER IS
  BEGIN
    ctx := bitor_impl(0);
    RETURN ODCIConst.Success;
  END ODCIAggregateInitialize;

  MEMBER FUNCTION ODCIAggregateIterate(SELF  IN OUT bitor_impl,
                                       VALUE IN NUMBER) RETURN NUMBER IS
  BEGIN
    SELF.bitor := SELF.bitor + VALUE - bitand(SELF.bitor, VALUE);
    RETURN ODCIConst.Success;
  END ODCIAggregateIterate;

  MEMBER FUNCTION ODCIAggregateMerge(SELF IN OUT bitor_impl,
                                     ctx2 IN bitor_impl) RETURN NUMBER IS
  BEGIN
    SELF.bitor := SELF.bitor + ctx2.bitor - bitand(SELF.bitor, ctx2.bitor);
    RETURN ODCIConst.Success;
  END ODCIAggregateMerge;

  MEMBER FUNCTION ODCIAggregateTerminate(SELF        IN OUT bitor_impl,
                                         returnvalue OUT NUMBER,
                                         flags       IN NUMBER) RETURN NUMBER IS
  BEGIN
    returnvalue := SELF.bitor;
    RETURN ODCIConst.Success;
  END ODCIAggregateTerminate;
END;
/

...and then define your own aggregate function

CREATE OR REPLACE FUNCTION bitoragg(x IN NUMBER) RETURN NUMBER
PARALLEL_ENABLE
AGGREGATE USING bitor_impl;
/

Usage:

SELECT user_id, bitoragg(permissions) FROM perms GROUP BY user_id

Upvotes: 9

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