Sunday, March 14, 2021

10 things in MySQL (that won’t work as expected)

(I just discovered cracked.com)

#10. Searching for a NULL

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SELECT  *
FROM    a
WHERE   a.column = NULL
In SQL, a NULL is never equal to anything, even another NULL. This query won't return anything and in fact will be thrown out by the optimizer when building the plan.
When searching for NULL values, use this instead:
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SELECT  *
FROM    a
WHERE   a.column IS NULL

#9. LEFT JOIN with additional conditions

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SELECT  *
FROM    a
LEFT JOIN
        b
ON      b.a = a.id
WHERE   b.column = 'something'
LEFT JOIN is like INNER JOIN except that it will return each record from a at least once, substituting missing fields from b with NULL values, if there are no actual matching records.
The WHERE condition, however, is evaluated after the LEFT JOIN so the query above checks column after it had been joined. And as we learned earlier, no NULL value can satisfy an equality condition, so the records from awithout corresponding record from b will unavoidably be filtered out.
Essentially, this query is an INNER JOIN, only less efficient.
To match only the records with b.column = 'something' (while still returning all records from a), this condition should be moved into ON clause:
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SELECT  *
FROM    a
LEFT JOIN
        b
ON      b.a = a.id
        AND b.column = 'something'

#8. Less than a value but not a NULL

Quite often I see the queries like this:
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SELECT  *
FROM    b
WHERE   b.column < 'something'
        AND b.column IS NOT NULL
&#91;/sourcecode&#93;
 
This is actually not an error: this query is valid and will do what's intended. However, <code>IS NOT NULL</code> here is redundant.
 
If <code>b.column</code> is a <code>NULL</code>, then <code>b.column < 'something'</code> will never be satisfied, since any comparison to <code>NULL</code> evaluates to a boolean <code>NULL</code> and does not pass the filter.
 
It is interesting that this additional <code>NULL</code> check is never used for <q>greater than</q> queries (like in <code>b.column > 'something'</code>).
 
This is because <code>NULL</code> go first in <code>ORDER BY</code> in <strong>MySQL</strong> and hence are incorrectly considered <q>less</q> than any other value by some people.
 
This query can be simplified:
 
 
SELECT  *
FROM    b
WHERE   b.column < 'something'
&#91;/sourcecode&#93;
 
and will still never return a <code>NULL</code> in <code>b.column</code>.
 
<h3 class="cracked">#7. Joining on NULL</h3>
 
<img src="https://explainextended.com/wp-content/uploads/2010/11/MG_3163-e1288839302867.jpg" alt="" title="Helicopter" width="700" height="467" class="aligncenter size-full wp-image-5105 noborder" />
 
 
SELECT  *
FROM    a
JOIN    b
ON      a.column = b.column
When column is nullable in both tables, this query won't return a match of two NULLs for the reasons described above: no NULLs are equal.
Here's a query to do that:
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SELECT  *
FROM    a
JOIN    b
ON      a.column = b.column
        OR (a.column IS NULL AND b.column IS NULL)
MySQL's optimizer treats this as an equijoin and provides a special join condition, ref_or_null.

#6. NOT IN with NULL values

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SELECT  a.*
FROM    a
WHERE   a.column NOT IN
        (
        SELECT column
        FROM    b
        )
This query will never return anything if there is but a single NULL in b.column. As with other predicates, both IN and NOT IN against NULL evaluate to NULL.
This should be rewritten using a NOT EXISTS:
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SELECT  a.*
FROM    a
WHERE   NOT EXISTS
        (
        SELECT NULL
        FROM    b
        WHERE   b.column = a.column
        )
Unlike INEXISTS always evaluates to either true or false.

#5. Ordering random samples

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SELECT  *
FROM    a
ORDER BY
        RAND(), column
LIMIT 10
This query attempts to select 10 random records ordered by column.
ORDER BY orders the output lexicographically: that is, the records are only ordered on the second expression when the values of the first expression are equal.
However, the results of RAND() are, well, random. It's infeasible that the values of RAND() will match, so ordering on column after RAND() is quite useless.
To order the randomly sampled records, use this query:
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SELECT  *
FROM    (
        SELECT  *
        FROM    mytable
        ORDER BY
                RAND()
        LIMIT 10
        ) q
ORDER BY
        column

#4. Sampling arbitrary record from a group

This query intends to select one column from each group (defined by grouper)
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SELECT  DISTINCT(grouper), a.*
FROM    a
DISTINCT is not a function, it's a part of SELECT clause. It applies to all columns in the SELECT list, and the parentheses here may just be omitted. This query may and will select the duplicates on grouper (if the values in at least one of the other columns differ).
Sometimes, it's worked around using this query (which relies on MySQL's extensions to GROUP BY):
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SELECT  a.*
FROM    a
GROUP BY
        grouper
Unaggregated columns returned within each group are arbitrarily taken.
At first, this appears to be a nice solution, but it has quite a serious drawback. It relies on the assumption that all values returned, though taken arbitrarily from the group, will still belong to one record.
Though with current implementation is seems to be so, it's not documented and can be changed in any moment (especially if MySQL will ever learn to apply index_union after GROUP BY). So it's not safe to rely on this behavior.
This query would be easy to rewrite in a cleaner way if MySQL supported analytic functions. However, it's still possible to make do without them, if the table has a PRIMARY KEY defined:
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SELECT  a.*
FROM    (
        SELECT  DISTINCT grouper
        FROM    a
        ) ao
JOIN    a
ON      a.id =
        (
        SELECT  id
        FROM    a ai
        WHERE   ai.grouper = ao.grouper
        LIMIT 1
        )

#3. Sampling first record from a group

This is a variation of the previous query:
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SELECT  a.*
FROM    a
GROUP BY
        grouper
ORDER BY
        MIN(id) DESC
Unlike the previous query, this one attempts to select the record holding the minimal id.
Again: it is not guaranteed that the unaggregated values returned by a.* will belong to a record holding MIN(id) (or even to a single record at all).
Here's how to do it in a clean way:
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SELECT  a.*
FROM    (
        SELECT  DISTINCT grouper
        FROM    a
        ) ao
JOIN    a
ON      a.id =
        (
        SELECT  id
        FROM    a ai
        WHERE   ai.grouper = ao.grouper
        ORDER BY
                ai.grouper, ai.id
        LIMIT 1
        )
This query is just like the previous one but with ORDER BY added to ensure that the first record in id order will be returned.

#2. IN and comma-separated list of values

This query attempts to match the value of column against any of those provided in a comma-separated string:
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SELECT  *
FROM    a
WHERE   column IN ('1, 2, 3')
This does not work because the string is not expanded in the IN list.
Instead, if column column is a VARCHAR, it is compared (as a string) to the whole list (also as a string), and of course will never match. If column is of a numeric type, the list is cast into the numeric type as well (and only the first item will match, at best).
The correct way to deal with this query would be rewriting it as a proper IN list
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SELECT  *
FROM    a
WHERE   column IN (1, 2, 3)
, or as an inline view:
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SELECT  *
FROM    (
        SELECT  1 AS id
        UNION ALL
        SELECT  2 AS id
        UNION ALL
        SELECT  3 AS id
        ) q
JOIN    a
ON      a.column = q.id
, but this is not always possible.
To work around this without changing the query parameters, one can use FIND_IN_SET:
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SELECT  *
FROM    a
WHERE   FIND_IN_SET(column, '1,2,3')
This function, however, is not sargable and a full table scan will be performed on a.

#1. LEFT JOIN with COUNT(*)

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SELECT  a.id, COUNT(*)
FROM    a
LEFT JOIN
        b
ON      b.a = a.id
GROUP BY
        a.id
This query intends to count number of matches in b for each record in a.
The problem is that COUNT(*) will never return a 0 in such a query. If there is no match for a certain record in a, the record will be still returned and counted.
COUNT should be made to count only the actual records in b. Since COUNT(*), when called with an argument, ignores NULLs, we can pass b.a to it. As a join key, it can never be a null in an actual match, but will be if there were no match:
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SELECT  a.id, COUNT(b.a)
FROM    a
LEFT JOIN
        b
ON      b.a = a.id
GROUP BY
        a.id
P.S. In case you were wondering: no, the pictures don't have any special meaning. I just liked them.