SQL ORA-00920: 无效的关系运算符
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ORA-00920: invalid relational operator
提问by Ryan_W4588
In a database, I am trying to pull information that is later than a specified date. I should note beforehand that the date is in an odd format: YYYYMMDDHH24MISS##where ##is a two letter string which defines something useless to my query. Thus, I am using substrto just remove them.
在数据库中，我试图提取晚于指定日期的信息。我应该事先注意日期的格式很奇怪：YYYYMMDDHH24MISS##where##是一个两个字母的字符串，它定义了对我的查询无用的东西。因此，我substr只是用来删除它们。
My query, below, throws the following error, and I canot find out why:
我的查询，下面，抛出以下错误，我无法找出原因：
[Error Code: 920, SQL State: 42000] ORA-00920: invalid relational operator
[错误代码：920，SQL 状态：42000] ORA-00920：关系运算符无效
My Query:
我的查询：
SELECT *
FROM table_name
WHERE to_date(substr(COLUMN_NAME,1,14), 'YYYYMMDDHH24MISS')) >=
to_date('MIN_DATE', 'YYYYMMDDHH24MISS')
I have checked to make sure the dates are being defined correctly, and they are.
我已经检查过以确保正确定义了日期，并且确实如此。
Example of what I have used for MIN_DATEis: 20140101000000
我使用的示例MIN_DATE是：20140101000000
回答by John Maillet
You have an extra parenthesis at the end of the first to_date
在第一个 to_date 的末尾有一个额外的括号
回答by Gordon Linoff
You get this error in Oracle when you are missing a comparison operation, such as =-- as John Maillet already noted.
当您缺少比较操作时，您会在 Oracle 中收到此错误，例如=-- 正如 John Maillet 已经指出的那样。
My concern is the second part of the whereclause:
我担心的是该where条款的第二部分：
where to_date(substr(COLUMN_NAME, 1, 14), 'YYYYMMDDHH24MISS') >=
to_date('MIN_DATE', 'YYYYMMDDHH24MISS')
You have MIN_DATEin single quotes. This is interpreted as a stringwith eight letters in it, starting with 'M'and ending with 'E'. This is not interpretedas a variable. Presumably you mean:
你有MIN_DATE单引号。这被解释为一个包含八个字母的字符串，'M'以'E'. 这不被解释为变量。想必你的意思是：
where to_date(substr(COLUMN_NAME, 1, 14), 'YYYYMMDDHH24MISS') >=
to_date(MIN_DATE, 'YYYYMMDDHH24MISS')
You should only use single quotes for string and date constants.
您应该只对字符串和日期常量使用单引号。
I should add that you should be able to do this comparison without having to convert to dates:
我应该补充一点，您应该能够进行此比较而无需转换为日期：
where left(COLUMN_NAME, 14) = MIN_DATE
