Answers:
SELECT cols.table_name, cols.column_name, cols.position, cons.status, cons.owner
FROM all_constraints cons, all_cons_columns cols
WHERE cols.table_name = 'TABLE_NAME'
AND cons.constraint_type = 'P'
AND cons.constraint_name = cols.constraint_name
AND cons.owner = cols.owner
ORDER BY cols.table_name, cols.position;
由于Oracle将表名存储为大写,因此请确保“ TABLE_NAME”为大写。
q
。
与“理智”的答案相同,但更为简洁。
仅查询用户约束
SELECT column_name FROM all_cons_columns WHERE constraint_name = (
SELECT constraint_name FROM user_constraints
WHERE UPPER(table_name) = UPPER('tableName') AND CONSTRAINT_TYPE = 'P'
);
查询所有约束
SELECT column_name FROM all_cons_columns WHERE constraint_name = (
SELECT constraint_name FROM all_constraints
WHERE UPPER(table_name) = UPPER('tableName') AND CONSTRAINT_TYPE = 'P'
);
user_constraints
为all_constraints
。
SELECT owner, column_name, position FROM all_cons_columns WHERE (owner, constraint_name) in (SELECT owner, constraint_name FROM all_constraints WHERE UPPER(table_name) = UPPER('&tableName') AND CONSTRAINT_TYPE = 'P') order by owner, position;
Select constraint_name,constraint_type from user_constraints where table_name** **= ‘TABLE_NAME’ ;
(这将列出主键,然后)
Select column_name,position from user_cons_cloumns where constraint_name=’PK_XYZ’;
(这将为您提供该列,此处PK_XYZ是主键名称)
尝试此代码在这里,我创建了一个表来获取oracle中的主键列,该表称为test,然后进行查询
create table test
(
id int,
name varchar2(20),
city varchar2(20),
phone int,
constraint pk_id_name_city primary key (id,name,city)
);
SELECT cols.table_name, cols.column_name, cols.position, cons.status, cons.owner FROM all_constraints cons, all_cons_columns cols WHERE cols.table_name = 'TEST' AND cons.constraint_type = 'P' AND cons.constraint_name = cols.constraint_name AND cons.owner = cols.owner ORDER BY cols.table_name, cols.position;
将以下脚本另存为findPK.sql。
set verify off
accept TABLE_NAME char prompt 'Table name>'
SELECT cols.column_name
FROM all_constraints cons NATURAL JOIN all_cons_columns cols
WHERE cons.constraint_type = 'P' AND table_name = UPPER('&TABLE_NAME');
然后可以使用
@findPK