sql - SQL Oracle 错误 - 作业任务

Question

请帮助我使用 Oracle 数据库解决以下 SQL 问题

PROMPT：按PK升序列出仓库表的所有列。• 仅列出地址以 RD 或 ST 结尾的行

SELECT *
FROM warehouse 
WHERE address ='% rd' 
OR WHERE address ='% st',
ORDER BY whid ASC;

错误信息：

OR WHERE 地址='%st', *

第 4 行出错：ORA-00936：缺少表达式

Answer 1

您不需要第二个WHERE，逗号也应该删除。此外，您可能希望有LIKE之前的地址，并且您可能不希望有空格。

我想你的意思是：

SELECT *
FROM warehouse 
WHERE address LIKE '%rd' 
OR address LIKE '%st'
ORDER BY whid ASC;

或者，你可能很聪明：

-- You don't say this explicitly, but I think it a good idea to make sure that 
-- you are searching for rd and st in the right tense. That is why I have 'lower'
SELECT *
FROM warehouse 
WHERE lower(substr(address,-2,2)) in ('rd', 'st')
ORDER BY whid ASC;

Answer 2

摆脱第二个WHERE语句，并将您的等价语句更改为LIKE.

SELECT *
FROM warehouse 
WHERE address LIKE '% rd' 
OR address LIKE '% st'
ORDER BY whid ASC;

如果那里有等价语句，您将只匹配完全等于% rdor的字符串% st。

Answer 3

摆脱第二个WHERE，使用LIKE代替=并尝试删除%字符后的空格。

SELECT *
FROM warehouse 
WHERE address LIKE '%rd' 
OR address LIKE '%st'
ORDER BY whid ASC;

Answer 4

使用 LIKE 关键字：

SELECT * FROM warehouse WHERE address LIKE '%RD' OR address LIKE '%ST' ORDER BY whid ASC;