如何使equal_range迭代器在 Boost 多索引中按不同的索引排序?

How to make the equal_range iterator sorted by a different index in Boost Multi-index?

本文关键字：索引 排序 equal 何使 range 迭代器 Boost      更新时间：2023-10-16

我在employee类上有一个提升多索引容器(取自提升官方文档(：

typedef multi_index_container<
employee,
indexed_by<
ordered_unique<
tag<id>,  BOOST_MULTI_INDEX_MEMBER(employee,int,id)>,
ordered_non_unique<
tag<name>,BOOST_MULTI_INDEX_MEMBER(employee,std::string,name)>,
ordered_non_unique<
tag<age>, BOOST_MULTI_INDEX_MEMBER(employee,int,age)> >
> employee_set;

下面是由 (id、name、age( 打印的容器中的数据示例：

0 Joe 31
1 Robert 27
2 John 40
3 Albert 20
4 John 57
5 John 58
6 John 22

我想有一个迭代器，其中包含名称为 John 的所有项目，按年龄排序(最后一个字段(。我尝试了equal_range方法：

auto iter1 = boost::make_iterator_range(es.get<name>().equal_range("John"));

它返回一个迭代器，其中包含名为 John 的所有记录。如何使此迭代器按第三个索引(即年龄(排序？

输出应为：

6 John 22
2 John 40
4 John 57
5 John 58

好的。这是复制人Live On Coliru

输出确实是

2 "John" 40
4 "John" 57
5 "John" 58
6 "John" 22

现在，要按年龄排序(注意单词的选择(，您可以使用组合键。所以代替：

bmi::ordered_non_unique<
bmi::tag<struct name>,
bmi::member<employee, std::string, &employee::name>
>,

用

bmi::ordered_non_unique<
bmi::tag<struct name_age>,
bmi::composite_key<employee,
bmi::member<employee, std::string, &employee::name>,
bmi::member<employee, int, &employee::age>
>
>,

现在你可以

for (employee const& emp : boost::make_iterator_range(es.get<name_age>().equal_range("John"))) {
std::cout << emp.id << " " << std::quoted(emp.name) << " " << emp.age << "n";
}

指纹

6 "John" 22
2 "John" 40
4 "John" 57
5 "John" 58

完整示例

住在科里鲁

#include <boost/multi_index/composite_key.hpp>
#include <boost/multi_index/member.hpp>
#include <boost/multi_index/ordered_index.hpp>
#include <boost/multi_index_container.hpp>
#include <boost/range/iterator_range.hpp>
namespace bmi = boost::multi_index;
struct employee {
int id;
std::string name;
int age;
};
typedef bmi::multi_index_container<
employee,
bmi::indexed_by<
bmi::ordered_unique<
bmi::tag<struct id>,
bmi::member<employee, int, &employee::id>
>,
bmi::ordered_non_unique<
bmi::tag<struct name>,
bmi::member<employee, std::string, &employee::name>
>,
bmi::ordered_non_unique<
bmi::tag<struct name_age>,
bmi::composite_key<employee,
bmi::member<employee, std::string, &employee::name>,
bmi::member<employee, int, &employee::age>
>
>,
bmi::ordered_non_unique<
bmi::tag<struct age>,
bmi::member<employee, int, &employee::age>
>
> > employee_set;
#include <iostream>
#include <iomanip>
int main() {
employee_set es {
{0, "Joe",    31},
{1, "Robert", 27},
{2, "John",   40},
{3, "Albert", 20},
{4, "John",   57},
{5, "John",   58},
{6, "John",   22},
};
std::cout << "name index:n";
for (employee const& emp : boost::make_iterator_range(es.get<name>().equal_range("John"))) {
std::cout << emp.id << " " << std::quoted(emp.name) << " " << emp.age << "n";
}
std::cout << "name_age index:n";
for (employee const& emp : boost::make_iterator_range(es.get<name_age>().equal_range("John"))) {
std::cout << emp.id << " " << std::quoted(emp.name) << " " << emp.age << "n";
}
}

指纹

name index:
2 "John" 40
4 "John" 57
5 "John" 58
6 "John" 22
name_age index:
6 "John" 22
2 "John" 40
4 "John" 57
5 "John" 58
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