如何使用TinyXml解析特定元素

本文介绍了如何使用TinyXml解析特定元素的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！

问题描述

我想从TinyXml输出中解析出一组元素.本质上，我需要选择端口状态为"open"(端口23如下所示)的任何端口元素的"portid"属性.

I would like to parse a group of elements out of a TinyXml output. Essentially, I need to pick out any port element's "portid" attribute of the port has a state of "open" (shown below for port 23).

执行此操作的最佳方法是什么?这是TinyXml的输出的(简化的)清单:

What's the best way to do this? Here's the (simplified) listing for the output from TinyXml:

<?xml version="1.0" ?>
<nmaprun>
    <host>
        <ports>
            <port protocol="tcp" portid="22">
                <state state="filtered"/>
            </port>
            <port protocol="tcp" portid="23">
                <state state="open "/>
            </port>
            <port protocol="tcp" portid="24">
                <state state="filtered" />
            </port>
            <port protocol="tcp" portid="25">
                <state state="filtered" />
            </port>
            <port protocol="tcp" portid="80">
                <state state="filtered" />
            </port>
        </ports>
    </host>
</nmaprun>

推荐答案

这大致可以做到:

    TiXmlHandle docHandle( &doc );

    TiXmlElement* child = docHandle.FirstChild( "nmaprun" ).FirstChild( "host" ).FirstChild( "ports" ).FirstChild( "port" ).ToElement();

    int port;
    string state;
    for( child; child; child=child->NextSiblingElement() )
    {

        port = atoi(child->Attribute( "portid"));

        TiXmlElement* state_el = child->FirstChild()->ToElement();

        state = state_el->Attribute( "state" );

        if ("filtered" == state)
            cout << "port: " << port << " is filtered! " << endl;
        else
            cout << "port: " << port << " is unfiltered! " << endl;
    }

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tinyxml