C++ 如何对向量中的字符串进行模式匹配？

我有一个向量中的数据，我需要看看第三个元素vrecord[2]中是否有Buy或Sell这个词

查找向量中字符串出现的最直接的方法是什么

数据：

代码：

vectorvrecords；
while（std:：fgets（buff，sizeof buff，fp）！=NULL）{
vrecords.向后推（buff）；
}
对于（int t=0；t首先，使用C++中的C++ I/O系统是一个坏主意。最好使用C++函数<代码> STD:：GETLION<代码>或成员函数<代码> GETLION>代码>和/或<代码>获取<代码> >类：代码：ST:
考虑到C函数
fgets
也将新行字符存储在字符串中。您应该删除它。例如

while ( std::fgets( buff, sizeof buff, fp ) != NULL )
{
    size_t n = std::strlen( buff );
    if ( n && buff[n-1] == '\n' ) buff[n-1] = '\0';    
    if ( buff[0] != '\0' ) vrecords.push_back( buff );
}

#include <algorithm>
#include <iterator>

//...

auto it = std::find( vrecords.begin(), vrecords.end(), "Buy" );

if ( it != vrecords.end() ) 
{
    std::cout << "Word \"" << "Buy"
              << "\" is found at position " 
              << std::distance( vrecords.begin(), it )
              << std::endl;  
}

#include <algorithm>
#include <iterator>

//...

const char * s[] = { "Buy", "Sell" };

auto it = std::find_first_of( vrecords.begin(), vrecords.end(), 
                              std::begin( s ), std::end( s ) );

if ( it != vrecords.end() ) 
{
    std::cout << "One of the words \"" << "Buy and Sell"
              << "\" is found at position " 
              << std::distance( vrecords.begin(), it )
              << std::endl;  
}

const char * s[] = { "Buy", "Sell" };

auto n = std::count_if( vrecords.begin(), vrecords.end(),
                        [&]( const std::string &record )
                        { 
                            return record == s[0] || record == s[1];
                        } );

如果向量声明为
std:：vector
（我希望它不是声明为例如
std:：vector
），那么您可以改为编写

std::string record;
while ( std::getline( YourFileStream, record ) )
{
    if ( !record.empty() ) vrecords.push_back( record );
}

在这种情况下，使用标题
中声明的标准算法
std:：find
查找单词“Buy”很简单

while ( std::fgets( buff, sizeof buff, fp ) != NULL )
{
    size_t n = std::strlen( buff );
    if ( n && buff[n-1] == '\n' ) buff[n-1] = '\0';    
    if ( buff[0] != '\0' ) vrecords.push_back( buff );
}

#include <algorithm>
#include <iterator>

//...

auto it = std::find( vrecords.begin(), vrecords.end(), "Buy" );

if ( it != vrecords.end() ) 
{
    std::cout << "Word \"" << "Buy"
              << "\" is found at position " 
              << std::distance( vrecords.begin(), it )
              << std::endl;  
}

#include <algorithm>
#include <iterator>

//...

const char * s[] = { "Buy", "Sell" };

auto it = std::find_first_of( vrecords.begin(), vrecords.end(), 
                              std::begin( s ), std::end( s ) );

if ( it != vrecords.end() ) 
{
    std::cout << "One of the words \"" << "Buy and Sell"
              << "\" is found at position " 
              << std::distance( vrecords.begin(), it )
              << std::endl;  
}

const char * s[] = { "Buy", "Sell" };

auto n = std::count_if( vrecords.begin(), vrecords.end(),
                        [&]( const std::string &record )
                        { 
                            return record == s[0] || record == s[1];
                        } );

如果需要计算向量中有多少这样的单词，那么可以在循环中使用上述方法，或者使用标准算法
std:：count
，
std:：count\u If
，
std:：acculate
或基于范围的for循环。
比如说

while ( std::fgets( buff, sizeof buff, fp ) != NULL )
{
    size_t n = std::strlen( buff );
    if ( n && buff[n-1] == '\n' ) buff[n-1] = '\0';    
    if ( buff[0] != '\0' ) vrecords.push_back( buff );
}

#include <algorithm>
#include <iterator>

//...

auto it = std::find( vrecords.begin(), vrecords.end(), "Buy" );

if ( it != vrecords.end() ) 
{
    std::cout << "Word \"" << "Buy"
              << "\" is found at position " 
              << std::distance( vrecords.begin(), it )
              << std::endl;  
}

#include <algorithm>
#include <iterator>

//...

const char * s[] = { "Buy", "Sell" };

auto it = std::find_first_of( vrecords.begin(), vrecords.end(), 
                              std::begin( s ), std::end( s ) );

if ( it != vrecords.end() ) 
{
    std::cout << "One of the words \"" << "Buy and Sell"
              << "\" is found at position " 
              << std::distance( vrecords.begin(), it )
              << std::endl;  
}

const char * s[] = { "Buy", "Sell" };

auto n = std::count_if( vrecords.begin(), vrecords.end(),
                        [&]( const std::string &record )
                        { 
                            return record == s[0] || record == s[1];
                        } );

如果（vrecords[2]==“Buy”| | vrecords[2]==“Sell”）
？在我看来，regex会有点过分。@bro显示向量是如何定义的。vector vrecords；您的代码显示1次购买，但它应该显示3次。该数据流中有3次购买。（@bro请阅读您自己编写的内容：“我有向量中的数据，我需要看看它是否有“买入”或“卖出”这个词”