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本帖最后由 newkid 于 2011-11-7 22:52 编辑
2011-11-3 答案E
D不是答案因为每行都会触发,不满足条件。
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2011-11-4 LEAST函数
作者:Steven Feuerstein
难度:低
下面的选项都实现了这个过程:
CREATE OR REPLACE PROCEDURE plch_show_lowest (
number1_in IN NUMBER
, number2_in IN NUMBER
, number3_in IN NUMBER)
IS
哪一种能够显示传入的三个数中最小的一个?假设传给plch_show_lowest的实际值都不为空。
(A)
BEGIN
DBMS_OUTPUT.put_line (
LEAST (number1_in, LEAST (number2_in, number3_in)));
END;
(B)
BEGIN
DBMS_OUTPUT.put_line (
SMALLEST (number1_in, number2_in, number3_in));
END;
(C)
BEGIN
DBMS_OUTPUT.put_line (
MIN (number1_in, number2_in, number3_in));
END;
(D)
l_lowest NUMBER;
BEGIN
SELECT LEAST (number1_in, number2_in, number3_in)
INTO l_lowest
FROM DUAL;
DBMS_OUTPUT.put_line (l_lowest);
END;
(E)
l_lowest NUMBER := number1_in;
BEGIN
IF number2_in < l_lowest
THEN
l_lowest := number2_in;
END IF;
IF number3_in < l_lowest
THEN
l_lowest := number3_in;
END IF;
DBMS_OUTPUT.put_line (l_lowest);
END;
(F)
BEGIN
DBMS_OUTPUT.put_line (
LEAST (number1_in, number2_in, number3_in));
END;
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