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PL/SQL嵌套表的MEMBER OF 操作符(挑战题编号000007)

2015年4月14日 发表评论 阅读评论

 首先我创建一个嵌套表:

CREATE OR REPLACE TYPE plch_strings_t IS TABLE OF VARCHAR2 (100);

/

问题:

执行以下哪几个选项后,屏幕会显示”out”

选项1

DECLARE

   l_strings   plch_strings_t := plch_strings_t (1, NULL, 2);

BEGIN

   IF '1' MEMBER OF l_strings

   THEN

      DBMS_OUTPUT.put_line ('in');

   ELSE

      DBMS_OUTPUT.put_line ('out');

   END IF;

END;

/

选项2

DECLARE

   l_strings   plch_strings_t := plch_strings_t (1, NULL, 2);

BEGIN

   IF NULL MEMBER OF l_strings

   THEN

      DBMS_OUTPUT.put_line ('in');

   ELSE

      DBMS_OUTPUT.put_line ('out');

   END IF;

END;

/

选项3

DECLARE

   l_not_much   VARCHAR2 (100);

   l_strings    plch_strings_t := plch_strings_t (1, l_not_much, 2);

BEGIN

   IF l_not_much MEMBER OF l_strings

   THEN

      DBMS_OUTPUT.put_line ('in');

   ELSE

      DBMS_OUTPUT.put_line ('out');

   END IF;

END;

/

选项4:

DECLARE

   l_strings   plch_strings_t := plch_strings_t (1, NULL, 2);

   l_found     BOOLEAN;

BEGIN

   FOR indx IN 1 .. l_strings.COUNT

   LOOP

      l_found := l_strings (indx) IS NULL;

      EXIT WHEN l_found;

   END LOOP;

   IF l_found

   THEN

      DBMS_OUTPUT.put_line ('in');

   ELSE

      DBMS_OUTPUT.put_line ('out');

   END IF;

END;

/

 

答案:http://www.anguora.com/plsql挑战答案/

 

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