How can I write equivalent of following in mongo? I need to ignore some characters(spaces, hyphen) from a particular column before conditions are checked. For the sake of putting an example of mysql I am just removing space.
select * from TABLE
where REPLACE('name', ' ', '') = 'TEST'
So if name column has " T E S T" that should match.
You can try with $where operator in your query:
{$where: "this.name.replace(/[ -]/g,'') == 'TEST'"}
or:
{$where: "this.name.match(/T[ -]*E[ -]*S[ -]*T/)"}
or directly a $regex:
{name: /T[ -]*E[ -]*S[ -]*T/}
More info about $where $regex operators.
Related
I try to create advanced search to my database.
I want to do something like that: if the user type for search = overf**w
and I have in my database an cloumn that his value = overflow - show him.
this my code:
$name = str_replace('*', '_', $name);
SELECT name FROM table WHERE name LIKE CONCAT('%', ?, '%')
its not working, I dont know what the problem.
You can't use LIKE in this situation, you need to use REGEXP() to do a wildcard search. Replace * or ** with .*. To only return names that starts with the given value use ^ at the beginning of the regular expression
SELECT name
FROM actors
WHERE name REGEXP('^overf.*w')
I don't know php but your $name parameter should be set like this (in pseudo code)
$name = '^' + replace($name, '**', '.*')
I have a table with an integer column. It has 12 records numbered 1000 to 1012. Remember, these are ints.
This query returns, as expected, 12 results:
select count(*) from proposals where qd_number::text like '%10%'
as does this:
SELECT COUNT(*) FROM "proposals" WHERE (lower(first_name) LIKE '%10%' OR qd_number::text LIKE '%10%' )
but this query returns 2 records:
SELECT COUNT(*) FROM "proposals" WHERE (lower(first_name) || ' ' || qd_number::text LIKE '%10%' )
which implies using || in concatenated where expressions is not equivalent to using OR. Is that correct or am I missing something else here?
You probably have nulls in first_name. For these records (lower(first_name) || ' ' || qd_number::text results in null, so you don't find the numbers any longer.
using || in concatenated where expressions is not equivalent to using ORIs that correct or am I missing something else here?
That is correct.
|| is the string concatenation operator in SQL, not the OR operator.
I got stuck during database migration on PostgreSQL and need your help.
I have two tables that I need to join: drzewa_mateczne.migracja (data I need to migrate) and ibl_as.t_adres_lesny (dictionary I need to join with migracja).
I need to join them on replace(drzewa_mateczne.migracja.adresy_lesne, ' ', '') = replace(ibl_as.t_adres_lesny.adres, ' ', ''). However my data is not very regular, so I want to join it on first good match with the dictionary.
I've created the following query:
select
count(*)
from
drzewa_mateczne.migracja a
where
length(a.adresy_lesne) > 0
and replace(a.adresy_lesne, ' ', '') = (select substr(replace(al.adres, ' ', ''), 1, length(replace(a.adresy_lesne, ' ', ''))) from ibl_as.t_adres_lesny al limit 1)
The query doesn't return any rows.
It does successfully join empty rows if ran without
length(a.adresy_lesne) > 0
The two following queries return rows (as expected):
select replace(adres, ' ', '')
from ibl_as.t_adres_lesny
where substr(replace(adres, ' ', ''), 1, 16) = '16-15-1-13-180-c'
limit 1
select replace(adresy_lesne, ' ', ''), length(replace(adresy_lesne, ' ', ''))
from drzewa_mateczne.migracja
where replace(adresy_lesne, ' ', '') = '16-15-1-13-180-c'
I'm suspecting that there might be a problem in sub-query inside the 'where' clause in my query. If you guys could help me resolve this issue, or at least point me in the right direction, I'd be very greatful.
Thanks in advance,
Jan
You can largely simplify to:
SELECT count(*)
FROM drzewa_mateczne.migracja a
WHERE a.adresy_lesne <> ''
AND EXISTS (
SELECT 1 FROM ibl_as.t_adres_lesny al
WHERE replace(al.adres, ' ', '')
LIKE (replace(a.adresy_lesne, ' ', '') || '%')
)
a.adresy_lesne <> '' does the same as length(a.adresy_lesne) > 0, just faster.
Replace the correlated subquery with an EXISTS semi-join (to get only one match per row).
Replace the complex string construction with a simple LIKE expression.
More information on pattern matching and index support in these related answers:
PostgreSQL LIKE query performance variations
Difference between LIKE and ~ in Postgres
speeding up wildcard text lookups
What you're basically telling the database to do is to get you the count of rows from drzewa_mateczne.migracja that have a non-empty adresy_lesne field that is a prefix of the adres field of a semi-random ibl_as.t_adres_lesny row...
Lose the "limit 1" in the subquery and substitute the "=" with "in" and see if that is what you wanted...
Here is my query that results in a syntax error:
SELECT *
FROM account_invoice,sale_order
WHERE sale_order.name LIKE %account_invoice.origin%
The account_invoice.origin field contains the text of sale_order.name, plus other text as well, so I need to match sale_order.name string anywhere in the account_invoice.origin string.
I'm using PostgreSQL 8.4.
Try this
SELECT *
FROM account_invoice,sale_order
WHERE sale_order.name LIKE '%' || account_invoice.origin || '%'
% needs single quote because the pattern is a string.
|| is the operator for concatenation.
I am using PostgreSQL 9.1.4 with hstore and the PostgreSQL JDBC driver (9.1-901.jdbc4).
I am trying to use the contains operators (?, ?&, ?|) in a PreparedStatement, however the ? character is parsed as a variable placeholder. Can this character be escaped to send the correct operator in the query?
An example:
PreparedStatement stmt = conn.prepareStatement("SELECT a, b FROM table1 WHERE c ? 'foo' AND d = ?");
stmt.setInt(1, dValue);
stmt.executeQuery();
In this form the following example would raise an exception:
org.postgresql.util.PSQLException: No value specified for parameter 2.
Update:
After investigating the query parser in the pgjdbc driver this snippet seems to indicate that it is not possible to escape the ? character. The questions that remain are:
Is there anything in the JDBC spec which allows a ? to be escaped and be anything other than a parameter placeholder?
Is there any better work around for this issue than just using plain Statements with variables manually inserted into the query string?
Effectively, it looks like the java SQL parser is not hstore compliant.
But since the syntax c ? 'foo' is equivalent to exist(c, 'foo'), you can easily workaround this problem. Have a look at the following page to see what the verbose operators for hstore are.
Postgres hstore documentation
There is a discussion about this issue on pgsql-hackers mailing list: http://grokbase.com/t/postgresql/pgsql-hackers/1325c6ys9n/alias-hstores-to-so-that-it-works-with-jdbc
For now I like most this workaround which also supports indexes:
CREATE FUNCTION exist_inline (hstore, text) RETURNS bool AS $$ SELECT $1 ? $2; $$ LANGUAGE sql;
You can use this query to find the function backing an operator in PostgreSQL like this. In your example:
SELECT
oprname,
oprcode || '(' || format_type(oprleft, NULL::integer) || ', '
|| format_type(oprright, NULL::integer) || ')' AS function
FROM pg_operator
WHERE oprname LIKE '?%'
AND (SELECT oid FROM pg_type WHERE typname = 'hstore') IN (oprleft, oprright);
This produces:
|oprname|function |
|-------|--------------------------|
|? |exist(hstore, text) |
|?| |exists_any(hstore, text[])|
|?& |exists_all(hstore, text[])|
See also a related question about using JSON operators containing ?. Note that the function usage may not profit from the same indexing capability when using a GIN index on your HSTORE column.
If you'd like to add multiple key-value pairs using PreparedStatement then you can do:
PreparedStatement ps = c.prepareStatement(
"insert into xyz(id, data) values(?, hstore(?, ?))");
ps.setLong(1, 23456L);
ps.setArray(2, c.createArrayOf("text", new String[]{"name", "city"}));
ps.setArray(3, c.createArrayOf("text", new String[]{"Duke", "Valley"}));
This will insert: 23456, 'name=>Duke, city=>Valley'