读取行的一部分（getline（））

Question

Basically this program searches a .txt file for a word and if it finds it, it prints the line and the line number. Here is what I have so far.

Code:

#include "std_lib_facilities.h"

int main()
{
    string findword;
    cout << "Enter word to search for.\n";
    cin >> findword;

    char filename[20];
    cout << "Enter file to search in.\n";
    cin >> filename;
    ifstream ist(filename);

    string line;
    string word;
    int linecounter = 1;
    while(getline(ist, line))
    {
     if(line.find(findword) != string::npos){
             cout << line << " " << linecounter << endl;}
     ++linecounter;
     }

     keep_window_open();
}

Solved.

Answer 1

您正在寻找find ：

if (line.find(findword) != string::npos) { ... }

Answer 2

我会按照您的建议进行操作，然后将行分成由空格分隔的单词或标记，然后在标记列表中搜索所需的关键字。

Answer 3

Make sure there are no spaces around the names in your text file. Otherwise, let ist take care like the following :

while(ist >> line)
{
 if(line == findword){
         cout << line << " " << linecounter << endl;}
 ++linecounter;
 }

I believe that your names file contains a name on every line. so using >> , ist will take care if there are extra spaces.

Answer 4

You could use a regular expression to find the word in the line. Don't know enough C++ to help you with the details.