Lua XML从模式中提取 [英] Lua XML extract from pattern

问题描述

一个应用程序正在向我的脚本发送像这样的流:

An application is sending my script an Stream like this one:

<?xml version="1.0" encoding="UTF-8"?>
<root>
   <aRootChildNode>
    <anotherChildNode>
     <?xml version="1.0">
     <TheNodeImLookingFor>
       ... content ...
     </TheNodeImLookingFor>
    </anotherChildNode>
   </aRootChildNode>
</root>

我想提取TheNodeImLookingFor部分. 到目前为止，得到了:

I want to extract the TheNodeImLookingFor section. So far, got:

data = string.match(Stream, "^.+\<TheNodeImLookingFor\>.+\<\/TheNodeImLookingFor\>.+$")

模式在流中可以识别，但不会提取节点及其内容.

Pattern is recognized in the Stream, but it doesn't extract the node and its content.

推荐答案

通常，使用模式匹配(Lua模式或正则表达式)提取XML并不是一个好主意.使用XML解析器.

In general, it's not a good idea to use pattern matching (either Lua pattern or regex) to extract XML. Use a XML parser.

对于此问题，您无需转义\或<(即使这样做，Lua模式也会使用%转义魔术字符).并使用方括号获取节点及其内容:

For this problem, you don't need to escape \ or <(even if you do, Lua pattern uses % to escape magic characters). And use brackets to get the node and its content:

data = string.match(Stream, "^.+(<TheNodeImLookingFor>.+</TheNodeImLookingFor>).+$")

或者仅获取内容:

data = string.match(Stream, "^.+<TheNodeImLookingFor>(.+)</TheNodeImLookingFor>.+$")

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