显示表命令的约束

Question

我有一些表，我尝试在其上设置 PK FK 关系，但我想验证这一点。如何显示 PK/FK 约束？我看到了this 手册页，但它没有显示示例，我的谷歌搜索也没有结果。我的数据库是credentialing1，我的约束表是practices 和cred_insurances。

Answer 1

尝试做：

SHOW TABLE STATUS FROM credentialing1;

外键约束列在输出的 Comment 列中。

Answer 2

我用

SHOW CREATE TABLE mytable;

这向您显示了以当前形式接收mytable 所需的 SQL 语句。您可以看到所有列及其类型（如 DESC），但它还显示约束信息（以及表类型、字符集等）。

Answer 3

只需查询 INFORMATION_SCHEMA：

USE INFORMATION_SCHEMA;
SELECT TABLE_NAME,
       COLUMN_NAME,
       CONSTRAINT_NAME,
       REFERENCED_TABLE_NAME,
       REFERENCED_COLUMN_NAME
FROM KEY_COLUMN_USAGE
WHERE TABLE_SCHEMA = "<your_database_name>" 
      AND TABLE_NAME = "<your_table_name>" 
      AND REFERENCED_COLUMN_NAME IS NOT NULL;

Answer 4

你可以用这个：

select
    table_name,column_name,referenced_table_name,referenced_column_name
from
    information_schema.key_column_usage
where
    referenced_table_name is not null
    and table_schema = 'my_database' 
    and table_name = 'my_table'

或者为了更好的格式化输出使用这个：

select
    concat(table_name, '.', column_name) as 'foreign key',  
    concat(referenced_table_name, '.', referenced_column_name) as 'references'
from
    information_schema.key_column_usage
where
    referenced_table_name is not null
    and table_schema = 'my_database' 
    and table_name = 'my_table'

Answer 5

验证答案的主要问题是您必须解析输出以获取信息。这是一个查询，可让您以更有用的方式获取它们：

SELECT cols.TABLE_NAME, cols.COLUMN_NAME, cols.ORDINAL_POSITION,
cols.COLUMN_DEFAULT, cols.IS_NULLABLE, cols.DATA_TYPE,
    cols.CHARACTER_MAXIMUM_LENGTH, cols.CHARACTER_OCTET_LENGTH,
    cols.NUMERIC_PRECISION, cols.NUMERIC_SCALE,
    cols.COLUMN_TYPE, cols.COLUMN_KEY, cols.EXTRA,
    cols.COLUMN_COMMENT, refs.REFERENCED_TABLE_NAME, refs.REFERENCED_COLUMN_NAME,
    cRefs.UPDATE_RULE, cRefs.DELETE_RULE,
    links.TABLE_NAME, links.COLUMN_NAME,
    cLinks.UPDATE_RULE, cLinks.DELETE_RULE
FROM INFORMATION_SCHEMA.`COLUMNS` as cols
LEFT JOIN INFORMATION_SCHEMA.`KEY_COLUMN_USAGE` AS refs
ON refs.TABLE_SCHEMA=cols.TABLE_SCHEMA
    AND refs.REFERENCED_TABLE_SCHEMA=cols.TABLE_SCHEMA
    AND refs.TABLE_NAME=cols.TABLE_NAME
    AND refs.COLUMN_NAME=cols.COLUMN_NAME
LEFT JOIN INFORMATION_SCHEMA.REFERENTIAL_CONSTRAINTS AS cRefs
ON cRefs.CONSTRAINT_SCHEMA=cols.TABLE_SCHEMA
    AND cRefs.CONSTRAINT_NAME=refs.CONSTRAINT_NAME
LEFT JOIN INFORMATION_SCHEMA.`KEY_COLUMN_USAGE` AS links
ON links.TABLE_SCHEMA=cols.TABLE_SCHEMA
    AND links.REFERENCED_TABLE_SCHEMA=cols.TABLE_SCHEMA
    AND links.REFERENCED_TABLE_NAME=cols.TABLE_NAME
    AND links.REFERENCED_COLUMN_NAME=cols.COLUMN_NAME
LEFT JOIN INFORMATION_SCHEMA.REFERENTIAL_CONSTRAINTS AS cLinks
ON cLinks.CONSTRAINT_SCHEMA=cols.TABLE_SCHEMA
    AND cLinks.CONSTRAINT_NAME=links.CONSTRAINT_NAME
WHERE cols.TABLE_SCHEMA=DATABASE()
    AND cols.TABLE_NAME="table"

Answer 6

afaik 向information_schema 提出请求，您需要特权。如果您需要简单的键列表，可以使用以下命令：

SHOW INDEXES IN <tablename>

Answer 7

oracle还有一个工具叫mysqlshow

如果您使用--k keys $table_name 选项运行它，它将显示密钥。

SYNOPSIS
   mysqlshow [options] [db_name [tbl_name [col_name]]]
.......
.......
.......
·   --keys, -k
   Show table indexes.

示例：

╰─➤  mysqlshow -h 127.0.0.1 -u root -p --keys database tokens
Database: database  Table: tokens
+-----------------+------------------+--------------------+------+-----+---------+----------------+---------------------------------+---------+
| Field           | Type             | Collation          | Null | Key | Default | Extra          | Privileges                      | Comment |
+-----------------+------------------+--------------------+------+-----+---------+----------------+---------------------------------+---------+
| id              | int(10) unsigned |                    | NO   | PRI |         | auto_increment | select,insert,update,references |         |
| token           | text             | utf8mb4_unicode_ci | NO   |     |         |                | select,insert,update,references |         |
| user_id         | int(10) unsigned |                    | NO   | MUL |         |                | select,insert,update,references |         |
| expires_in      | datetime         |                    | YES  |     |         |                | select,insert,update,references |         |
| created_at      | timestamp        |                    | YES  |     |         |                | select,insert,update,references |         |
| updated_at      | timestamp        |                    | YES  |     |         |                | select,insert,update,references |         |
+-----------------+------------------+--------------------+------+-----+---------+----------------+---------------------------------+---------+
+--------+------------+--------------------------+--------------+-------------+-----------+-------------+----------+--------+------+------------+---------+---------------+
| Table  | Non_unique | Key_name                 | Seq_in_index | Column_name | Collation | Cardinality | Sub_part | Packed | Null | Index_type | Comment | Index_comment |
+--------+------------+--------------------------+--------------+-------------+-----------+-------------+----------+--------+------+------------+---------+---------------+
| tokens | 0          | PRIMARY                  | 1            | id          | A         | 2           |          |        |      | BTREE      |         |               |
| tokens | 1          | tokens_user_id_foreign   | 1            | user_id     | A         | 2           |          |        |      | BTREE      |         |               |
+--------+------------+--------------------------+--------------+-------------+-----------+-------------+----------+--------+------+------------+---------+---------------+

Answer 8

类似于@Resh32，但不需要使用USE 语句：

SELECT TABLE_NAME,
       COLUMN_NAME,
       CONSTRAINT_NAME,
       REFERENCED_TABLE_NAME,
       REFERENCED_COLUMN_NAME
FROM INFORMATION_SCHEMA.KEY_COLUMN_USAGE
WHERE TABLE_SCHEMA = "database_name" 
      AND TABLE_NAME = "table_name" 
      AND REFERENCED_COLUMN_NAME IS NOT NULL;

有用的，例如使用 ORM。