If I write a query:
SELECT TOP(10)
FROM table
ORDER BY name
will TSQL first take the top 10 elements, then order those, or will it ORDER BY the full table then return the top 10 of the entire table?
TSQL must order the elements then take the TOP N of those ordered elements. Ordering, to be efficient, should require an index so as to not do a full table sort. That is, with an index, the sorting doesn't happen on every query, so TOP N can take advantage of that fact.
Your question can be explained more visually and directly in SSMS's explain plan UI.
As to the title of your question, slightly different from the content: ORDERBY is not "optimized" for TOP per se, but it is a requirement. This is noted by the commenters to your question.
Related
I will be having queries on my database with where clauses similar to this:
SELECT * FROM table WHERE a = 'string_value' AND b = 'other_string_value' AND t > <timestamp>
and less often to this:
SELECT * FROM table WHERE a = 'string_value' AND t > <timestamp>
I have created a multicolumn index on a, b and t on that order. However I am not sure if it will be optimal for my second -less frequent- query.
Will this index do an index scan on b or skip it and move to the t index immediately? (To be honest Im not sure how index scans work exactly). Should I create a second multi-column index on a and t only for the second query?
The docs state that
'the index is most efficient when there are constraints on the leading (leftmost) columns'
But in the example it doesn't highlight my case where the 'b' equality column is missing in the where clause.
The 2nd query will be much less effective with the btree index on (a,b,t) because the absence of b means t cannot be used efficiently (it can still be used as an in-index filter, but that is not nearly as good as being used as a start/stop point). An index on (a,t) will be able to support the 2nd query much more efficiently.
But that doesn't mean you have to create that index as well. Indexes take space and must be maintained, so are far from free. It might be better to just live with less-than-optimal plans for the 2nd query, since that query is used "less often". On the other hand, you did bother to post about it, so maybe "less often" is still pretty often. So you might be better off just to build the extra index and spend your time worrying about something else.
A btree index can be thought of like a phonebook, which is sorted on last name, then first name, then middle name. Your first query is like searching for "people named Mary Smith with a middle name less than Cathy" You can use binary search to efficiently find the first "Mary Smith", then you scan through those until the middle name is > 'Cathy', and you are done. Compare that to "people surnamed Smith with a middle name less than Cathy". Now you have to scan all the Smith's. You can't stop at the first middle name > Cathy, because any change in first name resets the order of the middle names.
Given that b only has 10 distinct values, you could conceivably use the (a,b,t) index in a skip scan quite efficiently. But PostgreSQL doen't yet implement skip scans natively. You can emulate them, but that is fragile, ugly, a lot of work, and easy to screw up. Nothing you said here makes me think it would be worthwhile to do.
I have an ltree column containing a tree with a depth of 3. I'm trying to write a query that can select all children at a specific depth (level 1 = get all parents, 2 = get all children, 3 = get all grandchildren). I know this is pretty straightforward with n_level:
SELECT path FROM hierarchies
WHERE
nlevel(path) = 1
LIMIT 1000;
I have 200,000 dummy records and it's pretty fast (~170 ms). However, this query uses a sequential scan. I think it'd be better to write it in a way that takes advantage of the ltree operators supported by the GiST index. Frustratingly, I can't seem to wrap my brain around them, and I haven't found a similar question on SO or DBA (besides this one on finding leaves)
Any advice is appreciated!
The only index that could support your query is a simple b-tree index on an expression.
create index on hierarchies((nlevel(path)))
Note however that it is quite possible for the planner to choose a sequential scan anyway, exemplary in the case the number of rows with level 1 is much more than other levels.
I have two tables in which I have data coming from two different sources. One of the field of each table contains the title of a movie, but for some reason out of my control, the titles are not always exactly the same.
So I use the ts_vector to get rid of all the minor differences (stop words, plurals and so on).
See an example here: http://sqlfiddle.com/#!17/5ccbc/3
My problem is how to compare the two ts_vector without taking into account the numberic values, but just the text content. If I compare directly the two fields, I only get the exact match between values, including position of each word. The only solution I have found is using the strip() function, that remove positions and weights from tsvector, leaving only the text content.
I was wondering if there is a fastest way to compare ts_vectors.
You could create in index on the stripped vector:
create index on tbl1 (strip(ts_title));
create index on tbl2 (strip(ts_title));
But given that your query has to fetch every row of each table, it is unlikely this would serve much of a point. Doing a merge join between the precomputed stripped vectors could be faster, but probably not once you include the overhead of building and maintaining the indexes. If the real WHERE clause is more restrictive (selecting only a few rows from one or the other of the tables) then please share the real query.
I'd like to replace the following ABAP OpenSQL snippet (in the where clause of a much bigger statement) with an equivalent join.
... AND tf~tarifart = ( SELECT MAX( tf2~tarifart ) FROM ertfnd AS tf2 WHERE tf2~tariftyp = e1~tariftyp AND tf2~bis >= e1~bis AND tf2~ab <= e1~ab ) ...
My motivation: Query migration to ABAP CDS views (basically plain SQL with in comparison somewhat reduced expressiveness). Alas, correlated subqueries and EXISTS statements are not supported.
I googled a bit and found a possible solution (last post) here https://archive.sap.com/discussions/thread/3824523
However, the proposal
Selecting MAX(value)
Your scenarion using inner join to first CDS view
doesn't work in my case.
tf.bis (and tf.ab) need to be in the selection list of the new view to limit the rhs of the join (new view) to the correct time frames.
Alas, there could be multiple (non overlapping) sub time frames (contained within [tf.ab, tf.bis]) with the same tf.tarifart.
Since these couldn't be grouped together, this results in multiple rows on the rhs.
The original query does not have a problem with that (no join -> no Cartesian product).
I hope the following fiddle (working example) clears things up a bit: http://sqlfiddle.com/#!9/8d1f48/3
Given these constraints, to me it seems that an equivalent join is indeed impossible. Suggestions or even confirmations?
select doc_belzart,
doc_tariftyp,
doc_ab,
doc_bis,
max(tar_tarifart)
from
(
select document.belzart as doc_belzart,
document.tariftyp as doc_tariftyp,
document.ab as doc_ab,
document.bis as doc_bis,
tariff.tarifart as tar_tarifart,
tariff.tariftyp as tar_tariftyp,
tariff.ab as tar_ab,
tariff.bis as tar_bis
from dberchz1 as document
inner join ertfnd as tariff
on tariff.tariftyp = document.tariftyp and
tariff.ab <= document.ab and
tariff.bis >= document.bis
) as max_tariff
group by doc_belzart,
doc_tariftyp,
doc_ab,
doc_bis
Translated in English, you seem to want to determine the max applicable tariff for a set of documents.
I'd refactor this into separate steps:
Determine all applicable tariffs, meaning all tariffs that completely cover the document's time interval. This will become your first CDS view, and in my answer forms the sub-query.
Determine for all documents the max applicable tariff. This will form your second CDS view, and in my answer forms the outer query. This one has the MAX / GROUP BY to reduce the result set to one per document.
Consider I have a table T with fields a,b,c,d with two indices: first on a,b,c fields and second on a,b,d fields. Types of a,b,c and d are integer. Both indices are almost the same (on production they both have about 2Gb size, they have the same creation time and the same statistics of usage, table overall have about 60 millions rows).
I make two queries:
select * from T where a=... and b=... and c=...;
select * from T where a=... and b=... and d=...;
I expect that for the first query index on a,b,c fields is used and for the second index on a,b,d fields is used. However it's not the case and for both queries first index is used, but in second case with "filter"(I used expect analyze to gain this knowledge). For me such behavior is unacceptable, because in some circumstances number of entries in filter grows very fast and autovacuum/analyze (which actually helps the planner to use the right index) works too slow to prevent the unexpected latencies and downtime.
So my question is: how can I force postgresql not to use wrong index with filtering, but rather use the right index when all fields in query's 'where' and in that index match?
Finally I found the solution. It's not perfect and I will not mark it as the best one, however it works and could help someone.
What I have actually done is I have changed the indices, instead of a,b,c and a,b,d indices now I'm having c,a,b and d,a,b.
One problem appeared: I needed an index on 'a', because some queries rely on it. However when I add index solely on 'a', the problem from the first post appears again (the index on 'a' is used when planner thinks that its cost is lower than cost of c,a,b or d,a,b). So I decided to add new field to the table which is a copy (has the same data) of a field 'a', let's call it 'a1', and I added the index on this field. Now when I need to filter contents of 'a' somehow instead I'm filtering on 'a1' field. It's weird, but I couldn't find another solution.