I am still new to writing SQL queries.
I am trying to find all rows that have a specific pattern match that is a combination of a constant_string_1, followed by any combination of integers, followed by another constant_string_2.
So for ex I am trying to find all rows that match:
Roger_117788_Maryland
Roger_188_Maryland
Roger_211_Maryland
I have tried:
select * from cust where cust_string ilike 'Roger_[0-9]_Maryland'
However that query doesn't generates 0 results. I have also tried using [:digit:] and \d but neither has worked. Not sure what I am doing wrong. Any help is appreciated.
Thanks
You can work this with just "%". Ilike may cause issues, so you can use "like" after transforming your column to lower case to match all cases.
select * from cust where lower(cust_string) like 'roger%maryland';
The following worked. I ended up using regexp_instr, but it was the multiple-digit match was where my query was not giving the results.
select * from cust where regexp_instr(cust_string, '^(Roger_)[0-9]{1,10}__Maryland$') > 0'
This did two things. One it achieved the result set above, i.e. Roger_188_Maryland, etc. But it also removed any multiple occurrences of the patterns, so for ex it eliminated a multi-match like Roger_188_Roger_211_Maryland. Thanks everyone for your help.
Related
Given a table name table and a string column named column, I want to search for the word word in that column in the following way: exact matches be on top, followed by prefix matches and finally postfix matches.
Currently I got the following solutions:
Solution 1:
select column
from (select column,
case
when column like 'word' then 1
when column like 'word%' then 2
when column like '%word' then 3
end as rank
from table) as ranked
where rank is not null
order by rank;
Solution 2:
select column
from table
where column like 'word'
or column like 'word%'
or column like '%word'
order by case
when column like 'word' then 1
when column like 'word%' then 2
when column like '%word' then 3
end;
Now my question is which one of the two solutions are more efficient or better yet, is there a solution better than both of them?
Your 2nd solution looks simpler for the planner to optimize, but it is possible that the first one gets the same plan as well.
For the Where, is not needed as it is covered by ; it might confuse the DB to do 2 checks instead of one.
But the biggest problem is the third one as this has no way to be optimized by an index.
So either way, PostgreSQL is going to scan your full table and manually extract the matches. This is going to be slow for 20,000 rows or more.
I recommend you to explore fuzzy string matching and full text search; looks like that is what you're trying to emulate.
Even if you don't want the full power of FTS or fuzzy string matching, you definitely should add the extension "pgtrgm", as it will enable you to add a GIN index on the column that will speedup LIKE '%word' searches.
https://www.postgresql.org/docs/current/pgtrgm.html
And seriously, have a look to FTS. It does provide ranking. If your requirements are strict to what you described, you can still perform the FTS query to "prefilter" and then apply this logic afterwards.
There are tons of introduction articles to PostgreSQL FTS, here's one:
https://www.compose.com/articles/mastering-postgresql-tools-full-text-search-and-phrase-search/
And even I wrote a post recently when I added FTS search to my site:
https://deavid.wordpress.com/2019/05/28/sedice-adding-fts-with-postgresql-was-really-easy/
I have two table in my database , agridata and geoname. I am trying to find out geoid column for names in agridata like below
select geonameid , name from geoname where name in (select distinct district_name from agridata );
I want to do a fuzzy match of the names as exact names are not in database. How to go about it ?
You can use a variety of matching algorithms (see here), but I'm not 100% sure they will work with an in clause. I'd imagine you really want to use a soundex join e.g.
select distinct g.geonameid, g.name from geoname g join agridata a on soundex(a.name) = g.name
or similar.
If you've got a huge match set to deal with, you may want to consider using some kind of search index such as ElasticSearch/Solr.
Use extension for PostgreSQL called pg_trgm, implementation of trigram matching.
"We can measure the similarity of two strings by counting the number of trigrams they share. This simple idea turns out to be very effective for measuring the similarity of words in many natural languages"
I used it, it's very fast and gives great results.
I have a UDF in my database which basically tries to get a station (e.g. bus/train) based on some input data (geographic/name/type). Inside this function i try to check if there are any rows matching the given values:
SELECT
COUNT(s.id)
INTO
firsttry
FROM
geographic.stations AS s
WHERE
ST_DWithin(s.the_geom,plocation,0.0017)
AND
s.name <-> pname < 0.8
AND
s.type ~ stype;
The firsttry variable now contains the value 1. If i use the following (slightly extended) SELECT statement i get no results:
RETURN query SELECT
s.id, s.name, s.type, s.the_geom,
similarity(
regexp_replace(s.name::text,'(Hauptbahnhof|Hbf)','Hbf'),
regexp_replace(pname::text,'(Hauptbahnhof|Hbf)','Hbf')
)::double precision AS sml,
st_distance(s.the_geom,plocation) As dist from geographic.stations AS s
WHERE ST_DWithin(s.the_geom,plocation,0.0017) and s.name <-> pname < 0.8
AND s.type ~ stype
ORDER BY dist asc,sml desc LIMIT 1;
the parameters are as follows:
stype = '^railway'
pname = 'Amsterdam Science Park'
plocation = ST_GeomFromEWKT('SRID=4326;POINT(4.9492530 52.3531670)')
the tuple i need to be returned is:
id name type geom (displayed as ST_AsText)
909658;"Amsterdam Sciencepark";"railway_station";"POINT(4.9482893 52.352904)"
The same UDF returns quite well for a lot of other stations, but this is one (of more) which just won't work. Any suggestions?
P.S. The use of the <-> operator is coming from the pg_trgm module.
Some ideas on how to troubleshoot this:
Break your troubleshooting into steps. Start with the simplest query possible. No aggregates, just joins and no filters. Then add filters. Then add order by, then add aggregates. Look at exactly where the change occurs.
Try reindexing the database.
One possibility that occurs to me based on this is that it could be a corrupted index used in the second query but not the first. I have seen corrupted indexes in the past and usually they throw errors but at least in theory they should be able to create a problem like this.
If this is correct, your query will suddenly return rows if you remove the ORDER BY clause.
If you have a corrupted index, then you need to pay close attention to hardware. Is the RAM ECC? Is the processor overheating? How are you disks doing?
A second possibility is that there is a typo on a join condition of filter statement. Normally this is something I would suspect first but it is easy enough to weed out index problems to start there. If removing the ORDER BY doesn't change things, then chances are it is a typo. If you can't find a typo, then try reindexing.
I'm attempting to construct a LIKE operator in my query on DB2 that is checking if a varchar is just two digits. I've looked online and it seems like DB2 does not support a character range i.e. [0-9]. I've tried LIKE '[0-9][0-9]' and I didn't get an error from DB2, but no rows showed up in my result set from that query when I can see rows that exactly match this through looking at a SELECT * of the same table.
Is there anyway I can replicate this in DB2 if it is indeed true? Is my syntax for the LIKE wrong? Thanks in advance.
The TRANSLATE function is more appropriate for validating an expression that contains a limited number of valid values.
WHERE TRANSLATE( yourExpressionOrColumn, '000000000', '123456789') = '00'
Found it. No you cannot and there are no symbols that can represent an OR in LIKE.
I have a Postgres table with more than 8 million rows. Given the following two ways of doing the same query via DBD::Pg, I get wildly different results.
$q .= '%';
## query 1
my $sql = qq{
SELECT a, b, c
FROM t
WHERE Lower( a ) LIKE '$q'
};
my $sth1 = $dbh->prepare($sql);
$sth1->execute();
## query 2
my $sth2 = $dbh->prepare(qq{
SELECT a, b, c
FROM t
WHERE Lower( a ) LIKE ?
});
$sth2->execute($q);
query 2 is at least an order of magnitude slower than query 1... seems like it is not using the indexes, while query 1 is using the index.
Would love hear why.
With LIKE expressions, b-tree indexes can only be used if the search pattern is left-anchored, i.e. terminated with %. More details in the manual.
Thanks to #evil otto for the link. This link to the current version.
Your first query provides this essential information at prepare time, so the query planner can use a matching index.
Your second query does not provide any information about the pattern at prepare time, so the query planner cannot use any indexes.
I suspect that in the first case the query compiler/optimizer detects that the clause is a constant, and can build an optimal query plan. In the second it has to compile a more generic query because the bound variable can be anything at run-time.
Are you running both test cases from same file using same $dbh object?
I think reason of increasing speed in second case is that you using prepared statement which is already parsed(but maybe I wrong:)).
Ahh, I see - I will drop out after this comment since I don't know Perl. But I would trust that the editor is correct in highlighting the $q as a constant. I'm guessing that you need to concatenate the value into the string, rather than just directly referencing the variable. So, my guess is that if + is used for string concatenation in perl, then use something like:
my $sql = qq{
SELECT a, b, c
FROM t
WHERE Lower( a ) LIKE '
} + $q + qq{'};
(Note: unless the language is tightly integrated with the database, such as Oracle/PLSQL, you usually have to create a completely valid SQL string before submitting to the database, instead of expecting the compiler to 'interpolate'/'Substitute' the value of the variable.)
I would again suggest that you get the COUNT() of the statements, to make sure that you are comparing apple to apples.
I don't know Postgres at all, but I think in Line 7 (WHERE Lower( a ) LIKE '$q'
), $q is actually a constant. It looks like your editor thinks so too, since it is highlighted in red. You probably still need to use the ? for the variable.
To test, do a COUNT(*), and make sure they match - I could be way offbase.