如何在不知道文件名的情况下从C++应用程序自动打开和读取给定目录中文件的内容?
例如(程序的粗略描述):
#include iomanip
#include dirent.h
#include fstream
#include iostream
#include stdlib.h
using namespace std;
int main()
{
DIR* dir;
struct dirent* entry;
dir=opendir("C:\\Users\\Toshiba\\Desktop\\links\\");
printf("Directory contents: ");
for(int i=0; i<3; i++)
{
entry=readdir(dir);
printf("%s\n",entry->d_name);
}
return 0;
}
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这将打印该目录中第一个文件的名称.我的问题是如何读取该特定文件的内容并将其保存在.txt文档中.能ifstream做到吗?(对不起,我的英语不好.)
这应该做到这一点
#include <iostream>
#include <boost/filesystem/operations.hpp>
#include <boost/filesystem/fstream.hpp>
using namespace boost::filesystem;
using namespace std;
void show_files( const path & directory, bool recurse_into_subdirs = true )
{
if( exists( directory ) )
{
directory_iterator end ;
for( directory_iterator iter(directory) ; iter != end ; ++iter )
if ( is_directory( *iter ) )
{
cout << iter->native_directory_string() << " (directory)\n" ;
if( recurse_into_subdirs ) show_files(*iter) ;
}
else
cout << iter->native_file_string() << " (file)\n" ;
copyfiles(iter->native_file_string());
}
}
void copyfiles(string s)
{
ifstream inFile;
inFile.open(s);
if (!inFile.is_open())
{
cout << "Unable to open file";
exit(1); // terminate with error
}
//Display contents
string line = "";
//Getline to loop through all lines in file
while(getline(inFile,line))
{
cout<<line<<endl; // line buffers for every line
//here add your code to store this content in any file you want.
}
inFile.close();
}
int main()
{
show_files( "/usr/share/doc/bind9" ) ;
return 0;
}
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