如何在文件中搜索字符串并打印包含该字符串的行？

Question

我必须在名为record.txt的文件中搜索字符串look_for，但是代码不起作用。

每次我给文件中存在的look_for值时，都会说找不到记录

string look_for, line;
    in.open("record.txt");
    cout<<"what is registration no of student ?";
    cin>>look_for;
    while(getline(in,line))
    {
        if(line.find(look_for)!= string::npos)
        {
            cout<<" record found "<<endl<<endl;
            break;
        }
        else cout<<"record not found ";
    }

Answer 1

您的代码可以正常工作，但是您无需检查文件是否可以实际打开。

像这样修改您的代码：

  ...
  in.open("record.txt");

  if (!in.is_open())
  {
    cout << "Could not open file" << endl;
    return 1;
  }

  cout << "what is registration no of student ?";
  ...

无法打开文件的原因可能包括：

• 该文件不存在
• 该文件不在可执行文件运行的目录中

Answer 2

确保已打开文件，并且getline返回的line具有正确的值，并检查文件是否具有UTF-8编码。

Answer 3

#include <iostream>
#include <fstream>
#include <string>`
using namespace std;

int main()
{
   string look_for, line;
   int lineNumber = 0;
   ifstream in("record.txt");
   if (!in.is_open())
   {
       cout << "Couldn't open file" << endl;
       return -1;
   }

   cout << "what is registration no of student ?\t";
   cin >> look_for;
   while (getline(in, line))
   {
       if (line.find(look_for) != string::npos)
       {
           cout << "Line:\t" << lineNumber << "\t[ " << look_for << " ] found in line [ " << line << " ]" << endl;
           lineNumber = 0;
           break;
       }
       lineNumber++;
   }

   if (lineNumber != 0)
       cout << "[ " << look_for << " ] not found" << endl;

   return 0;
 }