如何测试Oracle中的列是否等于empty_clob()？

天真FOO = empty_clob()抱怨不兼容的类型.我尝试使用谷歌搜索,但(再一次)在寻找甲骨文帮助方面收效甚微.谢谢.

你只是想检查一个没有任何长度的CLOB吗？虽然不完全是你的要求,但它基本上是一回事吗？

select *
  from bar
 where dbms_lob.getlength(foo) = 0;

这是完整的测试:

SQL> create table bar (foo clob);

Table created.

SQL> insert into bar values (empty_clob());

1 row created.

SQL> select *
  2    from bar
  3  where dbms_lob.getlength(foo) = 0;

FOO
--------------------------------------------------------------------------------

如果您尝试在PL/SQL中进行比较,您可以像Igor的解决方案那样测试相等性

SQL> ed
Wrote file afiedt.buf

  1  DECLARE
  2     dummy  clob;
  3  BEGIN
  4       dummy := empty_clob();
  5        IF dummy = empty_clob() THEN
  6           dbms_output.put_line( 'Dummy is empty' );
  7        ELSE
  8           dbms_output.put_line( 'Dummy is not empty' );
  9        END IF;
 10* END;
SQL> /
Dummy is empty

PL/SQL procedure successfully completed.

如果您尝试在SQL中执行此操作,那么您需要使用DBMS_LOB.COMPARE函数.表中的LOB列实际上是一个LOB定位器(即指针),因此您真正关心的是LOB指向的值与EMPTY_CLOB()函数返回的LOB定位符指向的值相当.

SQL> desc bar
 Name                                      Null?    Type
 ----------------------------------------- -------- ------------------------

 FOO                                                CLOB

SQL> insert into bar values ('123');

1 row created.

SQL> insert into bar values( empty_clob() );

1 row created.

SQL> insert into bar values( empty_clob() );

1 row created.

SQL> ed
Wrote file afiedt.buf

  1  select count(*)
  2    from bar
  3*  where dbms_lob.compare( foo, empty_clob() ) = 0
SQL> /

  COUNT(*)
----------
         2

SQL> ed
Wrote file afiedt.buf

  1  select count(*)
  2    from bar
  3*  where dbms_lob.compare( foo, empty_clob() ) != 0
SQL> /

  COUNT(*)
----------
         1