c – 如何使用TinyXml解析特定元素

我想从TinyXml输出中解析一组元素.本质上,我需要选择端口元素的“portid”属性,该端口的状态为“open”(如下面的端口23所示).

最好的方法是什么？这是TinyXml输出的(简化)列表：

<?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>

解决方法

这将大致如下：

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;
    }

原文地址：https://www.jb51.cc/c/120199.html