嗨,大家好想问我怎么能得到一个字符串的最后一个字母并检查它是否是元音或辅音。顺便说一下我使用oracle 10g。谢谢那些帮助我的人。和平!
这就是我的想法:
SELECT last_name,
Substr(last_name, -1, 1) "Last letter",
Substr(last_name, 1, 1) "First letter",
CASE
WHEN Substr(last_name, -1, 1) IN ( 'a', 'e', 'i', 'o', 'u' ) THEN
'ends with a vowel'
WHEN Substr(last_name, -1, 1) IN ( 'b', 'c', 'd', 'f',
'g', 'h', 'j', 'k',
'l', 'm', 'n', 'p',
'q', 'r', 's', 't',
'v', 'w', 'x', 'y', 'z' ) THEN
'ends with a consonant'
END "Last Letter Description",
CASE
WHEN Substr(last_name, 1, 1) IN ( 'a', 'e', 'i', 'o', 'u' ) THEN
'starts with a consonant'
WHEN Substr(last_name, 1, 1) IN ( 'b', 'c', 'd', 'f',
'g', 'h', 'j', 'k',
'l', 'm', 'n', 'p',
'q', 'r', 's', 't',
'v', 'w', 'x', 'y', 'z' ) THEN
'starts with a consonant'
END "First Letter Description"
FROM employees
GROUP BY first_name,
last_name
现在当您在oracle 10g上执行此操作时,“First Letter Description”为空!请帮助我,您认为我的代码有什么问题?
答案 0 :(得分:13)
试试这个,不完整,但通过简单的调整,您可以按照自己的方式工作:
FUNCTION last_is_vowel (string_in VARCHAR2)
RETURN BOOLEAN
IS
BEGIN
RETURN CASE WHEN LOWER(SUBSTR(string_in, -1)) IN ('a', 'e', 'i', 'o', 'u')
THEN TRUE
ELSE FALSE
END;
END last_is_vowel;
答案 1 :(得分:4)
查看您的数据。机会是employees.last_name中的第一个字符大写。请记住,Oracle区分大小写。您可以使用UPPER()或LOWER()来帮助找到您的匹配。
另外,如João建议的那样,只搜索元音并使用else语句来查找排除项更有效。
SELECT last_name,
Substr(last_name, -1, 1) "Last character",
Substr(last_name, 1, 1) "First character",
CASE
WHEN lower(Substr(last_name, -1, 1)) IN ( 'a', 'e', 'i', 'o', 'u' ) THEN
'ends with a vowel'
ELSE
'does not end with a vowel'
END "Last Letter Description",
CASE
WHEN lower(Substr(last_name, 1, 1)) IN ( 'a', 'e', 'i', 'o', 'u' ) THEN
'starts with a vowel'
ELSE
'does not start with a vowel'
END "First Letter Description"
FROM employees
GROUP BY first_name,
last_name