oracle SQL Server T-SQL 中的 REGEXP_LIKE 转换
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REGEXP_LIKE conversion in SQL Server T-SQL
提问by Zakerias
I have come across this line in an old report that needs converting to SQL Server.
我在需要转换为 SQL Server 的旧报告中遇到了这一行。
REGEXP_LIKE (examCodes, learner_code)
examCodes being the source and learner_code being the pattern. I know that SQL Server doesn't have REGEXP_LIKE and most places tell you to use PATINDEX.
ExamCodes 是源代码，learner_code 是模式。我知道 SQL Server 没有 REGEXP_LIKE 并且大多数地方都告诉您使用 PATINDEX。
Here's me thinking that this would work:
我认为这会起作用：
PATINDEX(learner_code, examCodes)
But I get the error:
但我收到错误：
Msg 4145, Level 15, State 1, Line 54
An expression of non-boolean type specified in a context where a condition is expected, near 'WHERE'.
On MSDN the syntax is specified as,
在 MSDN 上，语法被指定为，
PATINDEX ('%pattern%',expression)
But learner_code is a field and I can't specify a pattern?
但是 learner_code 是一个字段，我无法指定模式？
I did not write this report in the first place so I'm puzzled to what the pattern it's looking for anyway.
这份报告一开始并不是我写的，所以我很困惑它到底在寻找什么模式。
Many thanks
非常感谢
回答by Mudassir Hasan
WHERE PATINDEX ('%pattern%',expression) !=0
If pattern is found , PATINDEX returns non zero value and you need to do a comparison in WHERE clause.
A WHEREclause must be followed by comparison operation that returns true / false.
May be you are using PATINDEX without doing the comparison and that is why error message shows non boolean expression near WHERE clause.
如果找到模式，PATINDEX 返回非零值，您需要在 WHERE 子句中进行比较。一个WHERE子句必须跟比较的操作，返回真/假。
可能是您在没有进行比较的情况下使用 PATINDEX，这就是错误消息在 WHERE 子句附近显示非布尔表达式的原因。
To search for pattern learner_codewith wildcard character
使用通配符搜索模式learner_code
WHERE PATINDEX ('%' + CAST(learner_code AS VARCHAR) +'%',examCodes) !=0
回答by Ken Thompson
ORACLE'S REGEXP_LIKE supports actual regular expressions, PATINDEX only supports the % and _ wildcards, the [] list/range, and the ^ list/range negation.
ORACLE 的 REGEXP_LIKE 支持实际的正则表达式，PATINDEX 仅支持 % 和 _ 通配符、[] 列表/范围和 ^ 列表/范围否定。
