我使用different question thread中给出的答案中的代码,根据表名创建序列和触发器。
但是,我的表格非常接近(或已经达到)30个字符的限制,因此我收到此错误:
Error report:
ORA-00972: identifier is too long
ORA-06512: at line 15
00972. 00000 - "identifier is too long"
*Cause: An identifier with more than 30 characters was specified.
*Action: Specify at most 30 characters.
我的问题是,我如何连接表名,以便他们不会抛出这个错误,仍然保留标题中的一些表名?也许连续到20个字符和" _SEQ"或" _TRIG"最后?
这是我的代码:
DECLARE
CURSOR TABLES
IS
SELECT *
FROM USER_TABLES
WHERE 0 =
(SELECT COUNT(*)
FROM USER_CONSTRAINTS
WHERE USER_CONSTRAINTS.TABLE_NAME = USER_TABLES.TABLE_NAME
AND USER_CONSTRAINTS.CONSTRAINT_TYPE = 'P'
);
BEGIN
FOR T IN TABLES
LOOP
EXECUTE IMMEDIATE 'CREATE SEQUENCE '||T.TABLE_NAME||'_SEQ START WITH 1';
EXECUTE IMMEDIATE 'UPDATE '||T.TABLE_NAME||' SET ID = '||T.TABLE_NAME||'Seq.NEXTVAL';
EXECUTE IMMEDIATE 'ALTER TABLE '||T.TABLE_NAME||' ADD PRIMARY KEY (ID)';
EXECUTE IMMEDIATE 'CREATE OR REPLACE TRIGGER '||T.TABLE_NAME||'_TRIG '||CHR(10) ||'BEFORE INSERT ON '||T.TABLE_NAME||' '||CHR(10) ||'FOR EACH ROW '||CHR(10) ||'BEGIN '||CHR(10) ||':NEW.ID := '||T.TABLE_NAME||'Seq.NEXTVAL; '||CHR(10) ||'END; ';
END LOOP;
END;
/
答案 0 :(得分:0)
这将使用删除元音的表名(然后取名字的前25个字符,如果你的表名恰好超过25个字符且元音很少)。
DECLARE
CURSOR TABLES
IS
SELECT *
FROM USER_TABLES
WHERE 0 =
(SELECT COUNT(*)
FROM USER_CONSTRAINTS
WHERE USER_CONSTRAINTS.TABLE_NAME = USER_TABLES.TABLE_NAME
AND USER_CONSTRAINTS.CONSTRAINT_TYPE = 'P'
);
t_name VARCHAR2(30);
BEGIN
FOR T IN TABLES
LOOP
t_name := SUBSTR( REGEXP_REPLACE( T.TABLE_NAME, '[aeiou]', NULL, 1, 0, 'i' ), 1, 25 );
EXECUTE IMMEDIATE 'CREATE SEQUENCE '||t_name||'_SEQ START WITH 1';
EXECUTE IMMEDIATE 'UPDATE '||T.TABLE_NAME||' SET ID = '||t_name||'_SEQ.NEXTVAL';
EXECUTE IMMEDIATE 'ALTER TABLE '||T.TABLE_NAME||' ADD CONSTRAINT '||t_name||'_UNQ PRIMARY KEY (ID)';
EXECUTE IMMEDIATE 'CREATE OR REPLACE TRIGGER '||t_name||'_TRIG BEFORE INSERT ON '||T.TABLE_NAME||' FOR EACH ROW BEGIN :NEW.ID := '||t_name||'_SEQ.NEXTVAL; END;';
END LOOP;
END;
/