Jim*_*mmy 2 c++ xml xml-parsing c++11
我正在尝试使用地图,因此我可以将标签名称设为参考编号.当我尝试使用它时,就像在这段代码中我得到错误(每次我引用地图时总共6个):
src/main.cpp:25:45: error: no viable overloaded operator[] for type
'std::map<std::string, std::string>'
const char* idcs = node.child_value(tagMap[3]);
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这是代码:
#include "pugi/pugixml.hpp"
#include <iostream>
#include <string>
#include <map>
int main()
{
pugi::xml_document doca, docb;
std::map<std::string, pugi::xml_node> mapa, mapb;
std::map<std::string, std::string> tagMap {{"1", "data"}, {"2", "entry"}, {"3", "id"}, {"4", "content"}};
if (!doca.load_file("a.xml") || !docb.load_file("b.xml")) {
std::cout << "Can't find input files";
return 1;
}
for (auto& node: doca.child(tagMap[1]).children(tagMap[2])) {
const char* id = node.child_value(tagMap[3]);
mapa[id] = node;
}
for (auto& node: docb.child(tagMap[1]).children(tagMap[2])) {
const char* idcs = node.child_value(tagMap[3]);
if (!mapa.erase(idcs)) {
mapb[idcs] = node;
}
}
}
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Run Code Online (Sandbox Code Playgroud)const char* idcs = node.child_value(tagMap[3]);
是不正确的,tagMap只能通过keytype索引,这是std::string
你需要的是:
Run Code Online (Sandbox Code Playgroud)const std::string& idcs = node.child_value(tagMap["3"]);