我有一个文件,其中包含一些"实体" Valve格式的数据。它基本上是一项关键价值交易,看起来像这样:
{
"world_maxs" "3432 4096 822"
"world_mins" "-2408 -4096 -571"
"skyname" "sky_alpinestorm_01"
"maxpropscreenwidth" "-1"
"detailvbsp" "detail_sawmill.vbsp"
"detailmaterial" "detail/detailsprites_sawmill"
"classname" "worldspawn"
"mapversion" "1371"
"hammerid" "1"
}
{
"origin" "553 -441 322"
"targetname" "tonemap_global"
"classname" "env_tonemap_controller"
"hammerid" "90580"
}
每对{}
计为一个实体,其中的行计为KeyValues。如您所见,它相当简单。
我想将这些数据处理成C ++中的vector<map<string, string> >
。为此,我尝试使用Boost附带的正则表达式。以下是我到目前为止的情况:
static const boost::regex entityRegex("\\{(\\s*\"([A-Za-z0-9_]+)\"\\s*\"([^\"]+)\")+\\s*\\}");
boost::smatch what;
while (regex_search(entitiesString, what, entityRegex)) {
cout << what[0] << endl;
cout << what[1] << endl;
cout << what[2] << endl;
cout << what[3] << endl;
break; // TODO
}
易于阅读的正则表达式:
\{(\s*"([A-Za-z0-9_]+)"\s*"([^"]+)")+\s*\}
我还不确定正则表达式是否适用于我的问题,但它似乎至少打印了最后一个键值对(hammerid,1)。
我的问题是,我将如何提取&#34; nth&#34;表达式中匹配的子表达式?或者是否真的没有实际的方法来做到这一点?编写两个嵌套while
- 循环,一个搜索{}
模式,然后一个搜索实际键值对,可能会更好吗?
谢谢!
答案 0 :(得分:1)
使用解析器生成器,您可以编写适当的解析器。
例如,使用Boost Spirit,您可以将语法规则内联定义为C ++表达式:
start = *entity;
entity = '{' >> *entry >> '}';
entry = text >> text;
text = '"' >> *~char_('"') >> '"';
这是一个完整的演示:
<强> Live On Coliru 强>
#include <boost/spirit/include/qi.hpp>
#include <boost/fusion/adapted/std_pair.hpp>
#include <map>
using Entity = std::map<std::string, std::string>;
using ValveData = std::vector<Entity>;
namespace qi = boost::spirit::qi;
template <typename It, typename Skipper = qi::space_type>
struct Grammar : qi::grammar<It, ValveData(), Skipper>
{
Grammar() : Grammar::base_type(start) {
using namespace qi;
start = *entity;
entity = '{' >> *entry >> '}';
entry = text >> text;
text = '"' >> *~char_('"') >> '"';
BOOST_SPIRIT_DEBUG_NODES((start)(entity)(entry)(text))
}
private:
qi::rule<It, ValveData(), Skipper> start;
qi::rule<It, Entity(), Skipper> entity;
qi::rule<It, std::pair<std::string, std::string>(), Skipper> entry;
qi::rule<It, std::string()> text;
};
int main()
{
using It = boost::spirit::istream_iterator;
Grammar<It> parser;
It f(std::cin >> std::noskipws), l;
ValveData data;
bool ok = qi::phrase_parse(f, l, parser, qi::space, data);
if (ok) {
std::cout << "Parsing success:\n";
int count = 0;
for(auto& entity : data)
{
++count;
for (auto& entry : entity)
std::cout << "Entity " << count << ": [" << entry.first << "] -> [" << entry.second << "]\n";
}
} else {
std::cout << "Parsing failed\n";
}
if (f!=l)
std::cout << "Remaining unparsed input: '" << std::string(f,l) << "'\n";
}
打印哪些(显示输入):
Parsing success:
Entity 1: [classname] -> [worldspawn]
Entity 1: [detailmaterial] -> [detail/detailsprites_sawmill]
Entity 1: [detailvbsp] -> [detail_sawmill.vbsp]
Entity 1: [hammerid] -> [1]
Entity 1: [mapversion] -> [1371]
Entity 1: [maxpropscreenwidth] -> [-1]
Entity 1: [skyname] -> [sky_alpinestorm_01]
Entity 1: [world_maxs] -> [3432 4096 822]
Entity 1: [world_mins] -> [-2408 -4096 -571]
Entity 2: [classname] -> [env_tonemap_controller]
Entity 2: [hammerid] -> [90580]
Entity 2: [origin] -> [553 -441 322]
Entity 2: [targetname] -> [tonemap_global]
答案 1 :(得分:1)
我认为使用一个正则表达式表达式很难,因为每个实体{}
内的条目数量可变。我个人认为只需使用std::readline
进行解析。
#include <map>
#include <vector>
#include <string>
#include <sstream>
#include <iostream>
std::istringstream iss(R"~(
{
"world_maxs" "3432 4096 822"
"world_mins" "-2408 -4096 -571"
"skyname" "sky_alpinestorm_01"
"maxpropscreenwidth" "-1"
"detailvbsp" "detail_sawmill.vbsp"
"detailmaterial" "detail/detailsprites_sawmill"
"classname" "worldspawn"
"mapversion" "1371"
"hammerid" "1"
}
{
"origin" "553 -441 322"
"targetname" "tonemap_global"
"classname" "env_tonemap_controller"
"hammerid" "90580"
}
)~");
int main()
{
std::string skip;
std::string entity;
std::vector<std::map<std::string, std::string> > vm;
// skip to open brace, read entity until close brace
while(std::getline(iss, skip, '{') && std::getline(iss, entity, '}'))
{
// turn entity into input stream
std::istringstream iss(entity);
// temporary map
std::map<std::string, std::string> m;
std::string key, val;
// skip to open quote, read key to close quote
while(std::getline(iss, skip, '"') && std::getline(iss, key, '"'))
{
// skip to open quote read val to close quote
if(std::getline(iss, skip, '"') && std::getline(iss, val, '"'))
m[key] = val;
}
// move map (no longer needed)
vm.push_back(std::move(m));
}
for(auto& m: vm)
{
for(auto& p: m)
std::cout << p.first << ": " << p.second << '\n';
std::cout << '\n';
}
}
<强>输出:强>
classname: worldspawn
detailmaterial: detail/detailsprites_sawmill
detailvbsp: detail_sawmill.vbsp
hammerid: 1
mapversion: 1371
maxpropscreenwidth: -1
skyname: sky_alpinestorm_01
world_maxs: 3432 4096 822
world_mins: -2408 -4096 -571
classname: env_tonemap_controller
hammerid: 90580
origin: 553 -441 322
targetname: tonemap_global
答案 2 :(得分:0)
我会这样写的:
^\{(\s*"([A-Za-z0-9_]+)"\s*"([^"]+)")+\s*\}$
或者将正则表达式分成两个字符串。首先匹配花括号,然后遍历花括号线的内容为行。
匹配大括号:^(\{[^\}]+)$
匹配线:^(\s*"([A-Za-z0-9_]+)"\s*"([^"]+)"\s*)$