XML 解析程序没有可行的重载
No viable overloaded on XML parsing program
我有以下代码比较 XML 文件,然后使用映射根据节点名称定义内容类型,并使用 if 命令尝试根据标签是什么采取行动找到了。
#include "pugi/pugixml.hpp"
#include <iostream>
#include <string>
#include <map>
int main()
{
const std::map<std::string, std::string> tagMap {
{"description", "content"}, {"url", "web_address"}
};
pugi::xml_document doca, docb;
std::map<std::string, pugi::xml_node> mapa, mapb;
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("data").children("entry")) {
const char* id = node.child_value("id");
mapa[id] = node;
}
for (auto& node: docb.child("data").children("entry")) {
const char* idcs = node.child_value("id");
if (!mapa.erase(idcs)) {
mapb[idcs] = node;
}
}
for (auto& eb: mapb) {
for (auto& kv : tagMap) {
kv.first = eb.second.child_value(kv.second.c_str());
if (kv.first == "id") {
// Do work on id
}
if (kv.first == "description") {
// Do work on description
}
if (kv.first == "url") {
// Do work on URL data (I.e validate it)
}
if (kv.first == "location") {
// Do work on location data
}
}
}
}
我遇到的问题是当我尝试编译这个程序时出现这个错误:
g++ -g -Wall -std=c++11 -I include -o main src/main.cpp include/pugi/pugixml.cpp
src/main.cpp:46:19: error: no viable overloaded '='
kv.first = eb.second.child_value(kv.second.c_str());
这是一个示例输入:
<data>
<entry>
<id>1</id>
<content>Test</content>
<web_address>test.com</web_address>
</entry>
</data>
您无法覆盖 std::map
等关联容器中的密钥——它总是会作为 const
参考返回给您。实际上,如果您被允许覆盖密钥,那么您很容易违反容器的不变量。通常的解决方案是删除该元素并使用更新后的键插入一个新元素。
我可能在这里偏离基地,但根据你的描述,我认为你可能正在寻找这样的东西。
我怀疑你的标签映射是倒退的,但我无法在你的示例数据上使用它,所以可能还缺少其他东西。
希望我所做的可能有用。
int main()
{
// I think maybe your map is backwards?
const std::map<std::string, std::string> tagMap {
// {"description", "content"}, {"url", "web_address"}
{"content", "description"}, {"web_address", "url"}
};
pugi::xml_document doca, docb;
std::map<std::string, pugi::xml_node> mapa, mapb;
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("data").children("entry")) {
std::string id = node.child_value("id");
mapa[id] = node;
}
for (auto& node: docb.child("data").children("entry")) {
std::string idcs = node.child_value("id");
if (!mapa.erase(idcs)) {
mapb[idcs] = node;
}
}
// Use this to find mapped nodes
std::map<std::string, std::string>::const_iterator found;
for (auto& eb: mapb) {
// try to find the node in the map
found = tagMap.find(eb.second.child_value());
if(found == tagMap.end()) // - FAIL
continue; // unknown tag - ignore
// test the corresponding (mapped) tag name
if (found->second == "id") {
std::cout << "// Do work on id" << '\n';
}
if (found->second == "description") {
std::cout << "// Do work on description" << '\n';
}
if (found->second == "url") {
std::cout << "// Do work on URL data (I.e validate it)" << '\n';
}
if (found->second == "location") {
std::cout << "// Do work on location data" << '\n';
}
// }
}
}
我有以下代码比较 XML 文件,然后使用映射根据节点名称定义内容类型,并使用 if 命令尝试根据标签是什么采取行动找到了。
#include "pugi/pugixml.hpp"
#include <iostream>
#include <string>
#include <map>
int main()
{
const std::map<std::string, std::string> tagMap {
{"description", "content"}, {"url", "web_address"}
};
pugi::xml_document doca, docb;
std::map<std::string, pugi::xml_node> mapa, mapb;
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("data").children("entry")) {
const char* id = node.child_value("id");
mapa[id] = node;
}
for (auto& node: docb.child("data").children("entry")) {
const char* idcs = node.child_value("id");
if (!mapa.erase(idcs)) {
mapb[idcs] = node;
}
}
for (auto& eb: mapb) {
for (auto& kv : tagMap) {
kv.first = eb.second.child_value(kv.second.c_str());
if (kv.first == "id") {
// Do work on id
}
if (kv.first == "description") {
// Do work on description
}
if (kv.first == "url") {
// Do work on URL data (I.e validate it)
}
if (kv.first == "location") {
// Do work on location data
}
}
}
}
我遇到的问题是当我尝试编译这个程序时出现这个错误:
g++ -g -Wall -std=c++11 -I include -o main src/main.cpp include/pugi/pugixml.cpp
src/main.cpp:46:19: error: no viable overloaded '='
kv.first = eb.second.child_value(kv.second.c_str());
这是一个示例输入:
<data>
<entry>
<id>1</id>
<content>Test</content>
<web_address>test.com</web_address>
</entry>
</data>
您无法覆盖 std::map
等关联容器中的密钥——它总是会作为 const
参考返回给您。实际上,如果您被允许覆盖密钥,那么您很容易违反容器的不变量。通常的解决方案是删除该元素并使用更新后的键插入一个新元素。
我可能在这里偏离基地,但根据你的描述,我认为你可能正在寻找这样的东西。
我怀疑你的标签映射是倒退的,但我无法在你的示例数据上使用它,所以可能还缺少其他东西。
希望我所做的可能有用。
int main()
{
// I think maybe your map is backwards?
const std::map<std::string, std::string> tagMap {
// {"description", "content"}, {"url", "web_address"}
{"content", "description"}, {"web_address", "url"}
};
pugi::xml_document doca, docb;
std::map<std::string, pugi::xml_node> mapa, mapb;
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("data").children("entry")) {
std::string id = node.child_value("id");
mapa[id] = node;
}
for (auto& node: docb.child("data").children("entry")) {
std::string idcs = node.child_value("id");
if (!mapa.erase(idcs)) {
mapb[idcs] = node;
}
}
// Use this to find mapped nodes
std::map<std::string, std::string>::const_iterator found;
for (auto& eb: mapb) {
// try to find the node in the map
found = tagMap.find(eb.second.child_value());
if(found == tagMap.end()) // - FAIL
continue; // unknown tag - ignore
// test the corresponding (mapped) tag name
if (found->second == "id") {
std::cout << "// Do work on id" << '\n';
}
if (found->second == "description") {
std::cout << "// Do work on description" << '\n';
}
if (found->second == "url") {
std::cout << "// Do work on URL data (I.e validate it)" << '\n';
}
if (found->second == "location") {
std::cout << "// Do work on location data" << '\n';
}
// }
}
}