Oracle 用拼接字符串更新表 测试
先看我们的数据:
SQL> select count(1) from dave where cust_tel like '0551%';
COUNT(1)
----------
2723
在我们的Dave表里,cust_tel 电话以0551 开头的有2723条记录,现在我们把这些记录换成0556.
我们使用如下方法测试一下:
SQL> select * from v$version; BANNER ---------------------------------------------------------------------- Oracle Database 11g Enterprise EditionRelease 11.2.0.1.0 - Production PL/SQL Release 11.2.0.1.0 - Production CORE 11.2.0.1.0 Production TNS for 32-bit Windows: Version 11.2.0.1.0- Production NLSRTL Version 11.2.0.1.0 – Production SQL> UPDATE dave SET cust_tel = '0556'|| SUBSTR (cust_tel, 5) WHERE cust_tellike '0551%'; 2723 rows updated. SQL> rollback; Rollback complete. SQL> UPDATE dave SET cust_tel = '0556'|| SUBSTR (cust_tel, 5) WHERE cust_telin (select cust_tel from dave where cust_tel like '0551%'); 2723 rows updated. SQL> rollback; Rollback complete.
写法很多,关键要注意效率问题,尤其是要更新的数据量比较大时。
另测试发现Oracle10g 和11g 在语法支持上的一些区别:
在Oracle 10g上可以使用如下语法:
BEGIN FOR cl IN (SELECT object_name from d1 WHEREobject_type='TABLE') LOOP UPDATE d1 SET object_name = 'D' || SUBSTR (object_name, 2) WHERE object_name = cl.object_name; END LOOP; END;
但如果在Oracle 11g中使用如下语法:
BEGIN FOR cl IN (SELECT id from dave WHERE cust_tel LIKE '0551%') LOOP UPDATE dave SET cust_tel = '0556' || SUBSTR (cust_tel, 5) WHERE id = cl.id; END LOOP; END;
则会一致处于等待状态。
但是在Oracle 11g中,可以使用如下的游标来处理,这种方法也是我们之前的提高的根据rowid 来完成大量的Update操作:
DECLARE CURSOR cur IS SELECT rowid as ROW_ID from dave WHEREcust_tel LIKE '0551%' ORDER BY ROWID; ---如果表的数据量不是很大,可以不用 order by rowid V_COUNTER NUMBER; BEGIN V_COUNTER := 0; FOR row IN cur LOOP UPDATE dave SET cust_tel = '0556' || SUBSTR (cust_tel, 5) WHERE ROWID = row.ROW_ID; V_COUNTER := V_COUNTER + 1; IF (V_COUNTER>= 1000) THEN COMMIT; V_COUNTER := 0; END IF; END LOOP; COMMIT; END;
有关这种方法的具体说明参考:
http://blog.csdn.net/tianlesoftware/article/details/6576156
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