c++ - 如何使用 Attribute 作为 TinyXML2(C++) 的搜索关键字

Question

我正在尝试使用 Attribute 作为关键字来查找我想要的元素。它可以工作，但只有当它是第一个元素时。

bool readXML(){
string gateWay_str="",user_Input="";
XMLDocument XML_file;
XML_file.LoadFile("exception_ip.xml");                      //XML file Name
XMLHandle docHandle( &XML_file );                           //XMLHandle
XMLNode* node_ctrl;                                         //Node pointer
XMLElement* gateWay_Ele = docHandle.FirstChildElement("exception_ip").FirstChildElement("ip").ToElement();  //Get Node by XMLHandle and turn to Element 
cout<<"Test ip3=";                                          //show to user 
cin>>user_Input;                                            //user input

if(gateWay_Ele){                                            //is gateWay_Ele null?
    gateWay_str=gateWay_Ele->Name();                        //get Element name and show
    cout<< "Got gateWay_Ele = "<<gateWay_str<<endl;
}
if(gateWay_Ele ->Attribute("ip3",(user_Input.c_str()))){    //find Attribute where ip3 = "user input"
        node_ctrl=gateWay_Ele->FirstChild();                //make node_ctrl point FirstChild
        if(node_ctrl==nullptr){                             //is nullptr?
            cout<<"node_ctrl = nullptr";
            return false;
        }
        else{                                               
            gateWay_Ele=node_ctrl->ToElement();             //turn node_ctel to Element
            gateWay_str = gateWay_Ele->GetText();           //get Text
            cout<<"GateWay = "<<gateWay_str<<endl;          //show
            return true;                                    //return true
        }
    }
return false;

我的 XML 是

<?xml version="1.0" ?> 
<exception_ip>  
	<ip ip3="23">	
      <gateway>123.123.23.1</gateway>
      <dnsp>dnsp23</dnsp>
      <dnss>dnss23</dnss>
	</ip>
	<ip ip3="24">	
      <gateway>123.123.24.1</gateway>
      <dnsp>dnsp24</dnsp>
      <dnss>dnss24</dnss>
	</ip>
</exception_ip>

只有输入为23时才有效

我曾尝试使用 NextSiblingElement("ip")为了使点继续转到下一个兄弟元素。但它只是无限循环

do{
if(gateWay_Ele ->Attribute("ip3",(user_Input.c_str()))){    //find Attribute where ip3 = "user input"
        node_ctrl=gateWay_Ele->FirstChild();                //make node_ctrl point FirstChild
        if(node_ctrl==nullptr){                             //is nullptr?
            cout<<"node_ctrl = nullptr";
            return false;
        }
        else{                                               
            gateWay_Ele=node_ctrl->ToElement();             //turn node_ctel to Element
            gateWay_str = gateWay_Ele->GetText();           //get Text
            cout<<"GateWay = "<<gateWay_str<<endl;          //show
            return true;                                    //return true
        }
    }
else{
    gateWay_Ele->NextSiblingElement("ip");
}
}while(true);

谢谢!

Answer 1

您需要遍历所有“ip”元素，测试“ip3”属性，直到找到您想要的元素。但是这里有一个简单的方法来使用 my tinyxml2 extension :

#include <string>
#include <iostream>
#include <conio.h>  // for _getch()
// include the header for tinyxml2ex which includes tinyxml2, remember to put them on your include path
#include "tixml2ex.h"

using namespace std;

int main()
{
    // put the XML into a string for simplicity
    string testXml{ R"-(
<?xml version="1.0" ?>
<exception_ip>
    <ip ip3="23">
        <gateway>123.123.23.1</gateway>
        <dnsp>dnsp23</dnsp>
        <dnss>dnss23</dnss>
    </ip>
    <ip ip3="24">
        <gateway>123.123.24.1</gateway>
        <dnsp>dnsp24</dnsp>
        <dnss>dnss24</dnss>
    </ip>
</exception_ip>
)-"s };

    auto doc = tinyxml2::load_document (testXml);
    // find the required element by XPath and list its member elements
    if (auto ip = find_element (*doc, "/exception_ip/ip[@ip3='24']"s))
    {
        for (auto e : ip)
            cout << e->Name() << " " << text (e) << endl;
    }
    else
        cout << "could not find ip element" << endl;

   return 0;
}