sql - usar - uso de indices en oracle

Listar columnas con índices en PostgreSQL

Question

Me gustaría obtener las columnas en las que se encuentra un índice en PostgreSQL.

En MySQL puede usar SHOW INDEXES FOR table y ver la columna Column_name .

mysql> show indexes from foos;

+-------+------------+---------------------+--------------+-------------+-----------+-------------+----------+--------+------+------------+---------+
| Table | Non_unique | Key_name            | Seq_in_index | Column_name | Collation | Cardinality | Sub_part | Packed | Null | Index_type | Comment |
+-------+------------+---------------------+--------------+-------------+-----------+-------------+----------+--------+------+------------+---------+
| foos  |          0 | PRIMARY             |            1 | id          | A         |       19710 |     NULL | NULL   |      | BTREE      |         | 
| foos  |          0 | index_foos_on_email |            1 | email       | A         |       19710 |     NULL | NULL   | YES  | BTREE      |         | 
| foos  |          1 | index_foos_on_name  |            1 | name        | A         |       19710 |     NULL | NULL   |      | BTREE      |         | 
+-------+------------+---------------------+--------------+-------------+-----------+-------------+----------+--------+------+------------+---------+

¿Existe algo como esto para PostgreSQL?

Lo he intentado en el psql comando psql (con la opción -E para mostrar SQL) pero no muestra la información que estoy buscando.

Actualización: Gracias a todos los que agregaron sus respuestas. cope360 me dio exactamente lo que estaba buscando, pero varias personas intervinieron con enlaces muy útiles. Para referencia futura, consulte la documentación de pg_index (a través de Milen A. Radev ) y el muy útil artículo Extracción de información de META de PostgreSQL (a través de Michał Niklas ).

Answer 1

Similar a la respuesta aceptada, pero al haber dejado join en pg_attribute como join o query normal con pg_attribute, no da índices que son como:
create unique index unique_user_name_index on users (lower(name))

select 
    row_number() over (order by c.relname),
    c.relname as index, 
    t.relname as table, 
    array_to_string(array_agg(a.attname), '', '') as column_names 
from pg_class c
join pg_index i on c.oid = i.indexrelid and c.relkind=''i'' and c.relname not like ''pg_%'' 
join pg_class t on t.oid = i.indrelid
left join pg_attribute a on a.attrelid = t.oid and a.attnum = ANY(i.indkey) 
group by t.relname, c.relname order by c.relname;

Answer 2

# /di

La manera más fácil y más corta es /di , que listará todos los índices en la base de datos actual.

$ /di
                      List of relations
 Schema |            Name             | Type  |  Owner   |     Table     
--------+-----------------------------+-------+----------+---------------
 public | part_delivery_index         | index | shipper  | part_delivery
 public | part_delivery_pkey          | index | shipper  | part_delivery
 public | shipment_by_mandator        | index | shipper  | shipment_info
 public | shipment_by_number_and_size | index | shipper  | shipment_info
 public | shipment_info_pkey          | index | shipper  | shipment_info
(5 rows)

/di es el "hermano pequeño" del comando /d que enumerará todas las relaciones de la base de datos actual. Por lo tanto, /di ciertamente significa "muéstrame esta d atabases i ndex".

Typing /diS mostrará una lista de todos los índices utilizados en todo el sistema, lo que significa que también obtendrá todos los índices pg_catalog.

$ /diS
                                      List of relations
Schema   |                   Name                    | Type  |  Owner   |          Table
------------+-------------------------------------------+-------+----------+-------------------------
 pg_catalog | pg_aggregate_fnoid_index                  | index | postgres | pg_aggregate
 pg_catalog | pg_am_name_index                          | index | postgres | pg_am
 pg_catalog | pg_am_oid_index                           | index | postgres | pg_am
 pg_catalog | pg_amop_fam_strat_index                   | index | postgres | pg_amop
 pg_catalog | pg_amop_oid_index                         | index | postgres | pg_amop
 pg_catalog | pg_amop_opr_fam_index                     | index | postgres | pg_amop
 pg_catalog | pg_amproc_fam_proc_index                  | index | postgres | pg_amproc
 pg_catalog | pg_amproc_oid_index                       | index | postgres | pg_amproc
 pg_catalog | pg_attrdef_adrelid_adnum_index            | index | postgres | pg_attrdef
--More-- 

Con estos dos comandos, puede agregar un + después para obtener incluso más información, como el tamaño del espacio en disco que necesita el índice y una descripción, si está disponible.

$ /di+
                                 List of relations
 Schema |            Name             | Type  |  Owner   |     Table     | Size  | Description 
--------+-----------------------------+-------+----------+---------------+-------+-------------
 public | part_delivery_index         | index | shipper  | part_delivery | 16 kB | 
 public | part_delivery_pkey          | index | shipper  | part_delivery | 16 kB | 
 public | shipment_by_mandator        | index | shipper  | shipment_info | 19 MB | 
 public | shipment_by_number_and_size | index | shipper  | shipment_info | 19 MB | 
 public | shipment_info_pkey          | index | shipper  | shipment_info | 53 MB | 
(5 rows)

En psql puede encontrar ayuda fácilmente sobre los comandos que escriben /? .

Answer 3

¿Qué tal una solución simple?

SELECT 
  t.relname table_name,
  ix.relname index_name,
  indisunique,
  indisprimary, 
  regexp_replace(pg_get_indexdef(indexrelid), ''.*/((.*)/)'', ''/1'') columns
FROM pg_index i
JOIN pg_class t ON t.oid = i.indrelid
JOIN pg_class ix ON ix.oid = i.indexrelid
WHERE t.relname LIKE ''test%''

`

Answer 4

Algunos datos de muestra ...

create table test (a int, b int, c int, constraint pk_test primary key(a, b));
create table test2 (a int, b int, c int, constraint uk_test2 unique (b, c));
create table test3 (a int, b int, c int, constraint uk_test3b unique (b), constraint uk_test3c unique (c), constraint uk_test3ab unique (a, b));

Use la función pg_get_indexdef :

select pg_get_indexdef(indexrelid) from pg_index where indrelid = ''test''::regclass;

                    pg_get_indexdef
--------------------------------------------------------
 CREATE UNIQUE INDEX pk_test ON test USING btree (a, b)
(1 row)


select pg_get_indexdef(indexrelid) from pg_index where indrelid = ''test2''::regclass;
                     pg_get_indexdef
----------------------------------------------------------
 CREATE UNIQUE INDEX uk_test2 ON test2 USING btree (b, c)
(1 row)


select pg_get_indexdef(indexrelid) from pg_index where indrelid =''test3''::regclass;
                      pg_get_indexdef
------------------------------------------------------------
 CREATE UNIQUE INDEX uk_test3b ON test3 USING btree (b)
 CREATE UNIQUE INDEX uk_test3c ON test3 USING btree (c)
 CREATE UNIQUE INDEX uk_test3ab ON test3 USING btree (a, b)
(3 rows)

Answer 5

Aquí hay una función que envuelve la respuesta de cope360:

CREATE OR REPLACE FUNCTION getIndices(_table_name varchar)
  RETURNS TABLE(table_name varchar, index_name varchar, column_name varchar) AS $$
  BEGIN
    RETURN QUERY
    select
    t.relname::varchar as table_name,
    i.relname::varchar as index_name,
    a.attname::varchar as column_name
from
    pg_class t,
    pg_class i,
    pg_index ix,
    pg_attribute a
where
    t.oid = ix.indrelid
    and i.oid = ix.indexrelid
    and a.attrelid = t.oid
    and a.attnum = ANY(ix.indkey)
    and t.relkind = ''r''
    and t.relname = _table_name
order by
    t.relname,
    i.relname;
  END;
  $$ LANGUAGE plpgsql;

Uso:

select * from getIndices(''<my_table>'')

Answer 6

Buena respuesta de @cope360, convertida para usar la sintaxis de unión.

select t.relname as table_name
     , i.relname as index_name
     , array_to_string(array_agg(a.attname), '', '') as column_names
from pg_class t
join pg_index ix
on t.oid = ix.indrelid
join pg_class i
on i.oid = ix.indexrelid
join pg_attribute a
on a.attrelid = t.oid
and a.attnum = ANY(ix.indkey)
where t.relkind = ''r''
and t.relname like ''test%''
group by t.relname
       , i.relname
order by t.relname
       , i.relname
;

Answer 7

Combinado con código de otros y creado una vista:

CREATE OR REPLACE VIEW view_index AS 
SELECT
     n.nspname  as "schema"
    ,t.relname  as "table"
    ,c.relname  as "index"
    ,pg_get_indexdef(indexrelid) as "def"
FROM pg_catalog.pg_class c
    JOIN pg_catalog.pg_namespace n ON n.oid        = c.relnamespace
    JOIN pg_catalog.pg_index i ON i.indexrelid = c.oid
    JOIN pg_catalog.pg_class t ON i.indrelid   = t.oid
WHERE c.relkind = ''i''
    and n.nspname not in (''pg_catalog'', ''pg_toast'')
    and pg_catalog.pg_table_is_visible(c.oid)
ORDER BY
     n.nspname
    ,t.relname
    ,c.relname;

Answer 8

Crea algunos datos de prueba ...

create table test (a int, b int, c int, constraint pk_test primary key(a, b));
create table test2 (a int, b int, c int, constraint uk_test2 unique (b, c));
create table test3 (a int, b int, c int, constraint uk_test3b unique (b), constraint uk_test3c unique (c),constraint uk_test3ab unique (a, b));

Listar índices y columnas indexadas:

select
    t.relname as table_name,
    i.relname as index_name,
    a.attname as column_name
from
    pg_class t,
    pg_class i,
    pg_index ix,
    pg_attribute a
where
    t.oid = ix.indrelid
    and i.oid = ix.indexrelid
    and a.attrelid = t.oid
    and a.attnum = ANY(ix.indkey)
    and t.relkind = ''r''
    and t.relname like ''test%''
order by
    t.relname,
    i.relname;

 table_name | index_name | column_name
------------+------------+-------------
 test       | pk_test    | a
 test       | pk_test    | b
 test2      | uk_test2   | b
 test2      | uk_test2   | c
 test3      | uk_test3ab | a
 test3      | uk_test3ab | b
 test3      | uk_test3b  | b
 test3      | uk_test3c  | c

Enrolle los nombres de las columnas:

select
    t.relname as table_name,
    i.relname as index_name,
    array_to_string(array_agg(a.attname), '', '') as column_names
from
    pg_class t,
    pg_class i,
    pg_index ix,
    pg_attribute a
where
    t.oid = ix.indrelid
    and i.oid = ix.indexrelid
    and a.attrelid = t.oid
    and a.attnum = ANY(ix.indkey)
    and t.relkind = ''r''
    and t.relname like ''test%''
group by
    t.relname,
    i.relname
order by
    t.relname,
    i.relname;

 table_name | index_name | column_names
------------+------------+--------------
 test       | pk_test    | a, b
 test2      | uk_test2   | b, c
 test3      | uk_test3ab | a, b
 test3      | uk_test3b  | b
 test3      | uk_test3c  | c

Answer 9

Cuando se juega con índices, el orden en que se construyen las columnas en el índice es tan importante como las propias columnas.

La siguiente consulta enumera todos los índices de una tabla determinada y todas sus columnas de forma ordenada.

SELECT
  table_name,
  index_name,
  string_agg(column_name, '','')
FROM (
       SELECT
         t.relname AS table_name,
         i.relname AS index_name,
         a.attname AS column_name,
         (SELECT i
          FROM (SELECT
                  *,
                  row_number()
                  OVER () i
                FROM unnest(indkey) WITH ORDINALITY AS a(v)) a
          WHERE v = attnum)
       FROM
         pg_class t,
         pg_class i,
         pg_index ix,
         pg_attribute a
       WHERE
         t.oid = ix.indrelid
         AND i.oid = ix.indexrelid
         AND a.attrelid = t.oid
         AND a.attnum = ANY (ix.indkey)
         AND t.relkind = ''r''
         AND t.relname LIKE ''tablename''
       ORDER BY table_name, index_name, i
     ) raw
GROUP BY table_name, index_name

Answer 10

Este comando muestra la vista de las tablas variables, índices y restricciones también

=# /d table_name;

Ejemplo:

testannie=# /d dv.l_customer_account;

Answer 11

La información sin procesar está en pg_index .

Answer 12

No creo que esta versión exista aún en este hilo: proporciona tanto la lista de nombres de columna junto con el ddl para el índice.

CREATE OR REPLACE VIEW V_TABLE_INDEXES AS

SELECT
     n.nspname  as "schema"
    ,t.relname  as "table"
    ,c.relname  as "index"
    ,i.indisunique AS "is_unique"
    ,array_to_string(array_agg(a.attname), '', '') as "columns"
    ,pg_get_indexdef(i.indexrelid) as "ddl"
FROM pg_catalog.pg_class c
    JOIN pg_catalog.pg_namespace n ON n.oid        = c.relnamespace
    JOIN pg_catalog.pg_index i ON i.indexrelid = c.oid
    JOIN pg_catalog.pg_class t ON i.indrelid   = t.oid
    JOIN pg_attribute a ON a.attrelid = t.oid AND a.attnum = ANY(i.indkey)
WHERE c.relkind = ''i''
      and n.nspname not in (''pg_catalog'', ''pg_toast'')
      and pg_catalog.pg_table_is_visible(c.oid)
GROUP BY
    n.nspname
    ,t.relname
    ,c.relname
    ,i.indisunique
    ,i.indexrelid
ORDER BY
    n.nspname
    ,t.relname
    ,c.relname;

Descubrí que los índices que usan funciones no se vinculan con los nombres de las columnas, por lo que ocasionalmente se encuentra una lista de índices, por ejemplo, un nombre de columna cuando en realidad se usa 3.

Ejemplo:

CREATE INDEX ui1 ON table1 (coalesce(col1,''''),coalesce(col2,''''),col3)

La consulta devuelve solo ''col3'' como una columna en el índice, pero el DDL muestra el conjunto completo de columnas utilizadas en el índice.

Answer 13

Por favor, intente la consulta a continuación para profundizar en los índices requeridos

Consulta de la siguiente manera: lo he intentado personalmente y lo uso con frecuencia.

SELECT n.nspname as "Schema",
  c.relname as "Name",
  CASE c.relkind WHEN ''r'' THEN ''table'' WHEN ''v'' THEN ''view'' WHEN ''i'' 
THEN ''index'' WHEN ''S'' THEN ''sequence'' WHEN ''s'' THEN ''special'' END as "Type",
  u.usename as "Owner",
 c2.relname as "Table"
FROM pg_catalog.pg_class c
     JOIN pg_catalog.pg_index i ON i.indexrelid = c.oid
     JOIN pg_catalog.pg_class c2 ON i.indrelid = c2.oid
     LEFT JOIN pg_catalog.pg_user u ON u.usesysid = c.relowner
     LEFT JOIN pg_catalog.pg_namespace n ON n.oid = c.relnamespace
WHERE c.relkind IN (''i'','''')
      AND n.nspname NOT IN (''pg_catalog'', ''pg_toast'')
      AND pg_catalog.pg_table_is_visible(c.oid)
      AND c2.relname like ''%agg_transaction%'' --table name
      AND nspname = ''edjus'' -- schema name 
ORDER BY 1,2;

Answer 14

RESULTADO DE LA CONSULTA:

table |     column     |          type          | notnull |  index_name  | is_index | primarykey | uniquekey | default
-------+----------------+------------------------+---------+--------------+----------+-   -----------+-----------+---------
 nodes | dns_datacenter | character varying(255) | f       |              | f        | f          | f         |
 nodes | dns_name       | character varying(255) | f       | dns_name_idx | t        | f          | f         |
 nodes | id             | uuid                   | t       | nodes_pkey   | t        | t          | t         |
(3 rows)

CONSULTA:

SELECT  
c.relname AS table,
f.attname AS column,  
pg_catalog.format_type(f.atttypid,f.atttypmod) AS type,
f.attnotnull AS notnull,  
i.relname as index_name,
CASE  
    WHEN i.oid<>0 THEN ''t''  
    ELSE ''f''  
END AS is_index,  
CASE  
    WHEN p.contype = ''p'' THEN ''t''  
    ELSE ''f''  
END AS primarykey,  
CASE  
    WHEN p.contype = ''u'' THEN ''t'' 
    WHEN p.contype = ''p'' THEN ''t'' 
    ELSE ''f''
END AS uniquekey,
CASE
    WHEN f.atthasdef = ''t'' THEN d.adsrc
END AS default  FROM pg_attribute f  
JOIN pg_class c ON c.oid = f.attrelid  
JOIN pg_type t ON t.oid = f.atttypid  
LEFT JOIN pg_attrdef d ON d.adrelid = c.oid AND d.adnum = f.attnum  
LEFT JOIN pg_namespace n ON n.oid = c.relnamespace  
LEFT JOIN pg_constraint p ON p.conrelid = c.oid AND f.attnum = ANY (p.conkey)  
LEFT JOIN pg_class AS g ON p.confrelid = g.oid
LEFT JOIN pg_index AS ix ON f.attnum = ANY(ix.indkey) and c.oid = f.attrelid and c.oid = ix.indrelid 
LEFT JOIN pg_class AS i ON ix.indexrelid = i.oid 

WHERE c.relkind = ''r''::char  
AND n.nspname = ''public''  -- Replace with Schema name 
--AND c.relname = ''nodes''  -- Replace with table name, or Comment this for get all tables
AND f.attnum > 0
ORDER BY c.relname,f.attname;

Answer 15

Si desea conservar el orden de las columnas en el índice, esta es una forma (muy fea) de hacerlo:

select table_name,
    index_name,
    array_agg(column_name)
from (
    select
        t.relname as table_name,
        i.relname as index_name,
        a.attname as column_name,
        unnest(ix.indkey) as unn,
        a.attnum
    from
        pg_class t,
        pg_class i,
        pg_index ix,
        pg_attribute a
    where
        t.oid = ix.indrelid
        and i.oid = ix.indexrelid
        and a.attrelid = t.oid
        and a.attnum = ANY(ix.indkey)
        and t.relkind = ''r''
        and t.relnamespace = <oid of the schema you''re interested in>
    order by
        t.relname,
        i.relname,
        generate_subscripts(ix.indkey,1)) sb
where unn = attnum
group by table_name, index_name

el orden de las columnas se almacena en la columna pg_index.indkey, por lo que solicité los subíndices de esa matriz.

Answer 16

Solo hazlo: /d table_name

Pero no estoy seguro de qué quiere decir que la información sobre las columnas no está allí.

Por ejemplo:

# /d pg_class
       Table "pg_catalog.pg_class"
     Column      |   Type    | Modifiers
-----------------+-----------+-----------
 relname         | name      | not null
 relnamespace    | oid       | not null
 reltype         | oid       | not null
 reloftype       | oid       | not null
 relowner        | oid       | not null
 relam           | oid       | not null
 relfilenode     | oid       | not null
 reltablespace   | oid       | not null
 relpages        | integer   | not null
 reltuples       | real      | not null
 reltoastrelid   | oid       | not null
 reltoastidxid   | oid       | not null
 relhasindex     | boolean   | not null
 relisshared     | boolean   | not null
 relistemp       | boolean   | not null
 relkind         | "char"    | not null
 relnatts        | smallint  | not null
 relchecks       | smallint  | not null
 relhasoids      | boolean   | not null
 relhaspkey      | boolean   | not null
 relhasexclusion | boolean   | not null
 relhasrules     | boolean   | not null
 relhastriggers  | boolean   | not null
 relhassubclass  | boolean   | not null
 relfrozenxid    | xid       | not null
 relacl          | aclitem[] |
 reloptions      | text[]    |
Indexes:
    "pg_class_oid_index" UNIQUE, btree (oid)
    "pg_class_relname_nsp_index" UNIQUE, btree (relname, relnamespace)

Muestra claramente qué columnas dado índice está en esta tabla.

Answer 17

/d table_name muestra esta información de psql , pero si desea obtener dicha información de la base de datos utilizando SQL, eche un vistazo a la extracción de información META de PostgreSQL .

Utilizo esa información en mi utilidad para informar algo de información del esquema db para comparar bases de datos PostgreSQL en entornos de prueba y producción.

Answer 18

/d tablename muestra los nombres de columna para mí en la versión 8.3.8.

 "username_idx" UNIQUE, btree (username), tablespace "alldata1"

Answer 19

PostgreSQL ( pg_indexes ):

SELECT * FROM pg_indexes WHERE tablename = ''mytable'';

MySQL ( MOSTRAR ÍNDICE ):

SHOW INDEX FROM mytable;