Monday, October 14, 2013

10-things-in-mysql-that-wont-work-as-expected

http://explainextended.com/2010/11/03/10-things-in-mysql-that-wont-work-as-expected/ Reference from this guys, it seems useful 10. Searching for a NULL view sourceprint? 1. SELECT * 2. FROM a 3. 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: view sourceprint? 1. SELECT * 2. FROM a 3. WHERE a.column IS NULL #9. LEFT JOIN with additional conditions view sourceprint? 1. SELECT * 2. FROM a 3. LEFT JOIN 4. b 5. ON b.a = a.id 6. WHERE b.column = 'something' A 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 a without 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: view sourceprint? 1. SELECT * 2. FROM a 3. LEFT JOIN 4. b 5. ON b.a = a.id 6. AND b.column = 'something' #8. Less than a value but not a NULL Quite often I see the queries like this: view sourceprint? 1. SELECT * 2. FROM b 3. WHERE b.column < 'something' 4. AND b.column IS NOT NULL This is actually not an error: this query is valid and will do what’s intended. However, IS NOT NULL here is redundant. If b.column is a NULL, then b.column < 'something' will never be satisfied, since any comparison to NULL evaluates to a boolean NULL and does not pass the filter. It is interesting that this additional NULL check is never used for greater than queries (like in b.column > 'something'). This is because NULL go first in ORDER BY in MySQL and hence are incorrectly considered less than any other value by some people. This query can be simplified: view sourceprint? 1. SELECT * 2. FROM b 3. WHERE b.column < 'something' and will still never return a NULL in b.column. #7. Joining on NULL view sourceprint? 1. SELECT * 2. FROM a 3. JOIN b 4. 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: view sourceprint? 1. SELECT * 2. FROM a 3. JOIN b 4. ON a.column = b.column 5. 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 view sourceprint? 1. SELECT a.* 2. FROM a 3. WHERE a.column NOT IN 4. ( 5. SELECT column 6. FROM b 7. ) 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: view sourceprint? 1. SELECT a.* 2. FROM a 3. WHERE NOT EXISTS 4. ( 5. SELECT NULL 6. FROM b 7. WHERE b.column = a.column 8. ) Unlike IN, EXISTS always evaluates to either true or false. #5. Ordering random samples view sourceprint? 1. SELECT * 2. FROM a 3. ORDER BY 4. RAND(), column 5. 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: view sourceprint? 01. SELECT * 02. FROM ( 03. SELECT * 04. FROM mytable 05. ORDER BY 06. RAND() 07. LIMIT 10 08. ) q 09. ORDER BY 10. column #4. Sampling arbitrary record from a group This query intends to select one column from each group (defined by grouper) view sourceprint? 1. SELECT DISTINCT(grouper), a.* 2. 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): view sourceprint? 1. SELECT a.* 2. FROM a 3. GROUP BY 4. 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: view sourceprint? 01. SELECT a.* 02. FROM ( 03. SELECT DISTINCT grouper 04. FROM a 05. ) ao 06. JOIN a 07. ON a.id = 08. ( 09. SELECT id 10. FROM a ai 11. WHERE ai.grouper = ao.grouper 12. LIMIT 1 13. ) #3. Sampling first record from a group This is a variation of the previous query: view sourceprint? 1. SELECT a.* 2. FROM a 3. GROUP BY 4. grouper 5. ORDER BY 6. 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: view sourceprint? 01. SELECT a.* 02. FROM ( 03. SELECT DISTINCT grouper 04. FROM a 05. ) ao 06. JOIN a 07. ON a.id = 08. ( 09. SELECT id 10. FROM a ai 11. WHERE ai.grouper = ao.grouper 12. ORDER BY 13. ai.grouper, ai.id 14. LIMIT 1 15. ) 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: view sourceprint? 1. SELECT * 2. FROM a 3. 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 view sourceprint? 1. SELECT * 2. FROM a 3. WHERE column IN (1, 2, 3) , or as an inline view: view sourceprint? 01. SELECT * 02. FROM ( 03. SELECT 1 AS id 04. UNION ALL 05. SELECT 2 AS id 06. UNION ALL 07. SELECT 3 AS id 08. ) q 09. JOIN a 10. 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: view sourceprint? 1. SELECT * 2. FROM a 3. 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(*) view sourceprint? 1. SELECT a.id, COUNT(*) 2. FROM a 3. LEFT JOIN 4. b 5. ON b.a = a.id 6. GROUP BY 7. 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: view sourceprint? 1. SELECT a.id, COUNT(b.a) 2. FROM a 3. LEFT JOIN 4. b 5. ON b.a = a.id 6. GROUP BY 7. a.id P.S. In case you were wondering: no, the pictures don't have any special meaning. I just liked them.