在表格上显示约束命令

Question

我有我尝试设置PK FK关系的表格，但我想验证这一点。我如何显示PK/FK限制？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。我的数据库是凭证1，而我受约束的表是实践和cred_insurances。

Answer 1

我用

SHOW CREATE TABLE mytable; 

这说明你目前的形式，以receate mytable必要的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（以及表格类型，字符集等）。

Answer 2

尝试做：

SHOW TABLE STATUS FROM credentialing1; 

外键约束被列在输出的评论列。

Answer 3

简单的查询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

您可以使用此：

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' 

还是为了更好的格式化输出使用这样的：

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

的主要问题与验证答案你将不得不解析输出以获取信息。下面是一个查询让你让他们在一个更有用的方式：

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

据我所知做出information_schema请求你需要的权限。如果你需要的键简单的列表，你可以使用这个命令：

SHOW INDEXES IN <tablename> 

Answer 7

还有一个是Oracle提出所谓mysqlshow

如果用--k keys $table_name选项，将显示键运行它的工具。

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

例如：

╰─➤ 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，但无需使用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。