xQuery LIKE 运算符？
有没有办法以与 SQL 相同的方式使用 XQuery 执行 LIKE 操作？
我不想构造一些“startswith”、“endswith”和“contains”表达式。
我想要实现的示例：
for $x in /user where $x/firstname LIKE '%xxx' return $x
for $x in /user where $x/middlename LIKE 'xxx%' return $x
for $x in /user where $x/lastname LIKE '%xxx%' return $x
有没有办法在 XQuery 中实现此目的？
编辑：
得到了上述问题的答案。新问题：
有没有什么办法可以反其道而行之？我想使用 sql 等效的 NOT LIKE 运算符运行这些查询。这可能吗？它必须在 FLWOR 表达式中
EDIT2：
解决了问题。您可以运行 fn:not(starts-with('123', '1')) 并返回 false。
Is there any way to perform a LIKE operation with XQuery in the same way as with SQL?
I wan't to construct some "startswith", "endswith" and "contains"-expressions.
Example of what I want to achieve:
for $x in /user where $x/firstname LIKE '%xxx' return $x
for $x in /user where $x/middlename LIKE 'xxx%' return $x
for $x in /user where $x/lastname LIKE '%xxx%' return $x
Is there any way to achieve this in XQuery?
EDIT:
Got the answer to the question above. New problem:
Would there be any way to do this the opposite way around? I would like to run those queries with the sql equivalent NOT LIKE operator. Is this possible? It has to be in an FLWOR-expression
EDIT2:
Solved the problem. You can run fn:not(starts-with('123', '1')) and it returns false.
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XPath 2.0 和 XQuery 1.0（由 W3C 标准化）通过
matches函数提供正则表达式支持 http://www.w3.org/TR/xpath-functions/#func-matches：当然还有像
starts-with和contains（均在 XPath 1.0/2.0 中）和ends-with（仅在 XPath 2.0 中）可能就足够了。XPath 2.0 and XQuery 1.0 (as standardized by the W3C) have regular expression support with the
matchesfunction http://www.w3.org/TR/xpath-functions/#func-matches:And of course there are functions like
starts-withandcontains(both in XPath 1.0/2.0), andends-with(only in XPath 2.0) that might suffice.