从最流行的答案改进查询

因为对于postgresql 12+ information_schema是非常慢的

它帮助了我:

SELECT sh.nspname AS table_schema,
  tbl.relname AS table_name,
  col.attname AS column_name,
  referenced_sh.nspname AS foreign_table_schema,
  referenced_tbl.relname AS foreign_table_name,
  referenced_field.attname AS foreign_column_name
FROM pg_constraint c
    INNER JOIN pg_namespace AS sh ON sh.oid = c.connamespace
    INNER JOIN (SELECT oid, unnest(conkey) as conkey FROM pg_constraint) con ON c.oid = con.oid
    INNER JOIN pg_class tbl ON tbl.oid = c.conrelid
    INNER JOIN pg_attribute col ON (col.attrelid = tbl.oid AND col.attnum = con.conkey)
    INNER JOIN pg_class referenced_tbl ON c.confrelid = referenced_tbl.oid
    INNER JOIN pg_namespace AS referenced_sh ON referenced_sh.oid = referenced_tbl.relnamespace
    INNER JOIN (SELECT oid, unnest(confkey) as confkey FROM pg_constraint) conf ON c.oid = conf.oid
    INNER JOIN pg_attribute referenced_field ON (referenced_field.attrelid = c.confrelid AND referenced_field.attnum = conf.confkey)
WHERE c.contype = 'f'

选择的答案不为我工作，所以张贴我的sql工作。

select 
    con.conname as constraint_name,
    src_schema.nspname as source_schema,
    source.relname as source_table,
    source_col.attname as source_column,
    trg_schema.nspname as target_schema,
    target.relname as target_table,
    target_col.attname as target_column
from 
    pg_constraint con
inner join 
    pg_class source on source.oid = con.conrelid
inner join
    pg_attribute source_col on source_col.attrelid = con.conrelid and source_col.attnum = con.conkey[1] and source_col.attisdropped = false
inner join
    pg_namespace src_schema on src_schema.oid = source.relnamespace
inner join 
    pg_class target on target.oid = con.confrelid
inner join
    pg_attribute target_col on target_col.attrelid = con.confrelid and target_col.attnum = con.confkey[1] and source_col.attisdropped = false    
inner join
    pg_namespace trg_schema on trg_schema.oid = target.relnamespace 

最快的验证直接在bash答案完全基于这个答案

IFS='' read -r -d '' sql_code << EOF_SQL_CODE
      SELECT
      o.oid
      , o.conname AS constraint_name
      , (SELECT nspname FROM pg_namespace WHERE oid=m.relnamespace) AS source_schema
      , m.relname AS source_table
      , (SELECT a.attname FROM pg_attribute a
      WHERE a.attrelid = m.oid AND a.attnum = o.conkey[1] AND a.attisdropped = false) AS source_column
      , (SELECT nspname FROM pg_namespace
      WHERE oid=f.relnamespace) AS target_schema
      , f.relname AS target_table
      , (SELECT a.attname FROM pg_attribute a
      WHERE a.attrelid = f.oid AND a.attnum = o.confkey[1] AND a.attisdropped = false) AS target_column
      , ROW_NUMBER () OVER (ORDER BY o.oid) as rowid
      FROM pg_constraint o
      LEFT JOIN pg_class f ON f.oid = o.confrelid
      LEFT JOIN pg_class m ON m.oid = o.conrelid
      WHERE 1=1
      AND o.contype = 'f'
      AND o.conrelid IN (SELECT oid FROM pg_class c WHERE c.relkind = 'r')
EOF_SQL_CODE

psql -d my_db -c "$sql_code"

您可以通过information_schema表来实现这一点。例如:

SELECT
    tc.table_schema, 
    tc.constraint_name, 
    tc.table_name, 
    kcu.column_name, 
    ccu.table_schema AS foreign_table_schema,
    ccu.table_name AS foreign_table_name,
    ccu.column_name AS foreign_column_name 
FROM 
    information_schema.table_constraints AS tc 
    JOIN information_schema.key_column_usage AS kcu
      ON tc.constraint_name = kcu.constraint_name
      AND tc.table_schema = kcu.table_schema
    JOIN information_schema.constraint_column_usage AS ccu
      ON ccu.constraint_name = tc.constraint_name
      AND ccu.table_schema = tc.table_schema
WHERE tc.constraint_type = 'FOREIGN KEY' AND tc.table_name='mytable';

使用Key所引用的主键的名称并查询information_schema:

select table_name, column_name
from information_schema.key_column_usage
where constraint_name IN (select constraint_name
  from information_schema.referential_constraints 
  where unique_constraint_name = 'TABLE_NAME_pkey')

这里'TABLE_NAME_pkey'是外键引用的主键的名称。

以下是Andreas Joseph Krogh从PostgreSQL邮件列表提供的解决方案:http://www.postgresql.org/message-id/200811072134.44750.andreak@officenet.no

SELECT source_table::regclass, source_attr.attname AS source_column,
    target_table::regclass, target_attr.attname AS target_column
FROM pg_attribute target_attr, pg_attribute source_attr,
  (SELECT source_table, target_table, source_constraints[i] source_constraints, target_constraints[i] AS target_constraints
   FROM
     (SELECT conrelid as source_table, confrelid AS target_table, conkey AS source_constraints, confkey AS target_constraints,
       generate_series(1, array_upper(conkey, 1)) AS i
      FROM pg_constraint
      WHERE contype = 'f'
     ) query1
  ) query2
WHERE target_attr.attnum = target_constraints AND target_attr.attrelid = target_table AND
      source_attr.attnum = source_constraints AND source_attr.attrelid = source_table;

这个解决方案处理引用多个列的外键，并避免重复(其他一些答案无法做到这一点)。我唯一改变的是变量名。

下面是一个示例，返回所有引用权限表的雇员列:

SELECT source_column
FROM foreign_keys
WHERE source_table = 'employee'::regclass AND target_table = 'permission'::regclass;