ajo*_*jon 24
这是一个重复的问题:
Lukmdo对这个答案的信任:
可能不建议但工作得很好:
show index from TABLE where Key_name = 'PRIMARY' ;
Run Code Online (Sandbox Code Playgroud)
坚实的方法是使用information_schema:
SELECT k.COLUMN_NAME
FROM information_schema.table_constraints t
LEFT JOIN information_schema.key_column_usage k
USING(constraint_name,table_schema,table_name)
WHERE t.constraint_type='PRIMARY KEY'
AND t.table_schema=DATABASE()
AND t.table_name='owalog';
Run Code Online (Sandbox Code Playgroud)
Joã*_*lva 12
对于Oracle,您可以在ALL_CONSTRAINTS表中查找:
SELECT a.COLUMN_NAME
FROM all_cons_columns a INNER JOIN all_constraints c
ON a.constraint_name = c.constraint_name
WHERE c.table_name = 'TBL'
AND c.constraint_type = 'P';
Run Code Online (Sandbox Code Playgroud)
演示.
对于SQL Server,它已在这里得到解答,对于MySQL,请查看@ ajon的答案.
Rom*_*net 10
对于MySQL:
SELECT GROUP_CONCAT(COLUMN_NAME), TABLE_NAME
FROM INFORMATION_SCHEMA.KEY_COLUMN_USAGE
WHERE
TABLE_SCHEMA = '**database name**'
AND CONSTRAINT_NAME='PRIMARY'
GROUP BY TABLE_NAME;
Run Code Online (Sandbox Code Playgroud)
警告具有两列的主键将它们用逗号(,)分隔
小智 5
在SQL Server中尝试此查询:
SELECT X.NAME AS INDEXNAME,
COL_NAME(IC.OBJECT_ID,IC.COLUMN_ID) AS COLUMNNAME
FROM SYS.INDEXES X
INNER JOIN SYS.INDEX_COLUMNS IC
ON X.OBJECT_ID = IC.OBJECT_ID
AND X.INDEX_ID = IC.INDEX_ID
WHERE X.IS_PRIMARY_KEY = 1
AND OBJECT_NAME(IC.OBJECT_ID)='YOUR_TABLE'
Run Code Online (Sandbox Code Playgroud)