Why am I getting Impossible WHERE noticed after reading const tables in explain query?

I have a unique compound key like fr(fromid,toid) in the table, when I run the query with explain I get the following result:

Impossible WHERE noticed after reading const tables


The query I ran:

explain SELECT rid FROM relationship WHERE fromid=78 AND toid=60


Any help?

EDIT1:
When I use the below query:

explain SELECT rid FROM relationship WHERE fromid=60 and toid=78 AND is_approved='s'  OR is_approved='f' OR is_approved='t'


I see USING WHERE instead of the previous message, but when I use the below query:

explain SELECT rid FROM relationship WHERE fromid=60 and toid=78 AND (is_approved='s'  OR is_approved='f' OR is_approved='t')


I again get the first impossible ... message! What these parenthesis do here?

EDIT2:

CREATE TABLE relationship (
rid int(10) unsigned NOT NULL AUTO_INCREMENT,
fromid mediumint(8) unsigned NOT NULL,
toid mediumint(8) unsigned NOT NULL,
type tinyint(3) unsigned NOT NULL,
is_approved char(1) NOT NULL,
PRIMARY KEY (rid),
UNIQUE KEY fromid (fromid,toid),
KEY toid (toid),
CONSTRAINT relationship_ibfk_1 FOREIGN KEY (fromid) REFERENCES user (uid) ON DELETE CASCADE ON UPDATE CASCADE,
CONSTRAINT relationship_ibfk_2 FOREIGN KEY (toid) REFERENCES user (uid) ON DELETE CASCADE ON UPDATE CASCADE
) ENGINE=InnoDB


EDIT3:
As mysql site say:

Impossible WHERE noticed after reading const tables

MySQL has read all const (and system) tables and notice that the WHERE clause is always false.

But in the query I get the result I want, the WHERE part is not false. Is there someone who could explain this and shed some light on the subject?

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What is the table definition including datatypes? Your second set of queries are not equivalent BTW. And has higher precedence than Or so the equivalent parenthesised version is WHERE (fromid=60 and toid=78 AND is_approved='s') OR is_approved='f' OR is_approved='t' –  Martin Smith Jul 20 '12 at 12:13
What does SELECT COUNT(1) FROM relationship WHERE fromid=78 AND toid=60; return ??? –  RolandoMySQLDBA Jul 20 '12 at 14:51
@RolandoMySQLDBA, there will be using index in extra instead of impossible... –  john.locke Jul 20 '12 at 15:34
@MartinSmith you should submit your comment as the answer !!! –  RolandoMySQLDBA Jul 20 '12 at 15:44

You are getting the message

Impossible WHERE noticed after reading const tables

This is documented in the page you already linked.

MySQL has read all const (and system) tables and notice that the WHERE clause is always false

const tables are defined as

The table has at most one matching row, which is read at the start of the query. ... const is used when you compare all parts of a PRIMARY KEY or UNIQUE index to constant values.

You have a UNIQUE KEY on (fromid,toid). The query on WHERE fromid=78 AND toid=60 can be satisfied by reading this unique index. From the message you are getting this must return no results.

Similarly the query WHERE fromid=60 and toid=78 AND (is_approved='s' OR is_approved='f' OR is_approved='t') can also use this index to locate the row of interest (though it still has a residual predicate to evaluate were any row to match).

Your other query is different

SELECT rid
FROM   relationship
WHERE  fromid = 60
AND toid = 78
AND is_approved = 's'
OR is_approved = 'f'
OR is_approved = 't'


AND has a higher precedence than Or, so this is the same as

SELECT rid
FROM   relationship
WHERE  ( ( fromid = 60 ) AND ( toid = 78 ) AND ( is_approved = 's' ) )
OR ( is_approved = 'f' )
OR ( is_approved = 't' )


This can no longer use that index and has different semantics in that it will return any rows where is_approved IN ('f','t') irrespective of what the values in the other columns are.

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So how should I say for example: if its fromid=12 AND toid=78 then check if is_approved='f' or is_approved='t' or is_approved='s' –  john.locke Jul 21 '12 at 4:51
@john.locke: That is your third query: WHERE fromid=60 AND toid=78 AND (is_approved='s' OR is_approved='f' OR is_approved='t') which can also be written as: WHERE fromid=60 AND toid=78 AND ( is_approved IN ('s', 'f', 't') ) –  ypercube Jul 21 '12 at 7:39
Exactly. Because no row matches the fromid=60 AND toid=78 part so no further check is needed (for the is_approved part). –  ypercube Jul 21 '12 at 7:41
You can't have row s that do. There is a unique constraint on (fromid,toid) so there would be a maximum of one surely? And from the message you say you are getting MySQL doesn't think that there is even one that does. Do you mean that you have some rows that match fromid=60 and some rows that match toid=78 but not necessarily the same rows? –  Martin Smith Jul 21 '12 at 7:49
Perhaps this is the AND-OR confusion. Maybe you want all rows that have fromid=60 and all rows that have toid=78 and then from those, keep only those that have either 's' or 'f' or t' is_approved? If yes, try this condition: WHERE (fromid=60 OR toid=78) AND (is_approved IN ('s', 'f', 't')) –  ypercube Jul 21 '12 at 8:52

Impossible WHERE noticed after reading const tables in explain query?

This error occurs due to invalid value being put on a column which are either primary key or unique key.

Try with a correct value in the where` clause .

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