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使用正则表达式根据oracle运算符解析字符串

[英]parsing string according to oracle operators with regex

原文 2010-03-30 12:29:59 9 2 java/ regex/ oracle/ operators

Basically I was trying to replace the part of string with its actual value which comes immediately after oracle operators. 基本上，我试图用字符串的实际值替换字符串的一部分，该值在oracle运算符之后立即出现。 I can do this for limited operators list like {=,>,<} but I wonder that is there any way out to gather all the operators rather than giving them by hands? 我可以对有限的运算符列表（例如{=,>,<}执行此操作{=,>,<}但我想知道是否有办法收集所有运算符，而不是手工提供？ For instance, I have this string; 例如，我有这个字符串。 "a = xyz" , then I will replace xyz with lets say 3 . "a = xyz" ，那么我将用"a = xyz" 3代替xyz 。 But as you know we have bunch of operator namely "like,in,exists etc" . 但正如您所知，我们有一堆运算符，即"like,in,exists etc" 。 So my string can also be this: "a like xyz" . 所以我的字符串也可以是： "a like xyz" 。

So what do you suggest me? 那你对我有什么建议？

Thanks. 谢谢。

2 个解决方案

So what do you suggest me? 那你对我有什么建议？

I suggest not to do this with regex (regex are not able to do so), but use an existing, proven SQL parser. 我建议不要使用正则表达式执行此操作（正则表达式无法执行此操作），而是使用现有的，经过验证的SQL解析器。

Have a look at this question on SO: SQL parser library for Java 看看SO的这个问题： Java的SQL解析器库

Make your job simpler - require a very strict syntax on behalf of the caller. 使您的工作更简单-代表调用者要求非常严格的语法。

For example, require that the string be in the form "target operator #variable#", eg " a = #xyz# ". 例如，要求字符串的格式为“目标运算符＃variable＃”，例如“ a = #xyz# ”。

Then, all you need to do is use REPLACE(input, '#xyz#', 3) . 然后，您需要做的就是使用REPLACE(input, '#xyz#', 3) 。

As noted above, you probably don't want to reinvent the Oracle SQL statement parser. 如上所述，您可能不想重新发明Oracle SQL语句解析器。