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我正在开发一个程序,并且遇到了一些问题,这些问题正在发生。

基本上,该程序旨在为用户提供程序列表,一旦用户选择它,然后为用户提供具有相关扩展名的所有文件的列表。选择文件后,它将使用所需程序打开文件。我现在的问题是我的启动程序和我的扫描程序似乎在两个单独的目录中运行。现在我只是在我的桌面上有包含该程序的文件夹。当它运行时,它会搜索文件夹,但在打开文件时,它会尝试从同一文件夹的子文件夹目录中打开它们。我认为问题出在搜索上,因为当我将 .exe 复制到另一个位置(例如桌面)时,如果我的文档中碰巧有一个名称匹配的文件,它将从桌面打开文件夹。然而它不会

这是我现在拥有的代码:

#include <iostream>
#include <filesystem>
#include <vector>
#include <string>
#include <algorithm>
#include <fstream>
#include <Windows.h>
using namespace std;
using namespace std::tr2::sys;


bool ends_with(std::string& file, std::string& ext)
{
    return file.size() >= ext.size() && // file must be at least as long as ext
    // check strings are equal starting at the end
    std::equal(ext.rbegin(), ext.rend(), file.rbegin());
}

void wScan( path f, unsigned i = 0 )
{
directory_iterator d( f );
directory_iterator e;
vector<string>::iterator it2;
std::vector<string> extMatch;

//iterate through all files
for( ; d != e; ++d )
{
    string file = d->path();
    string ext = ".docx";
    //populate vector with matching extensions
    if(ends_with(file, ext))
    {
        extMatch.push_back(file);
    }

}
//counter to appear next to file names
int preI = -1;
//display all matching file names and their counter
for(it2 = extMatch.begin(); it2 != extMatch.end(); it2++)
{
    preI += 1;
    cout << preI << ". " << *it2 << endl;
}
//ask for file selection and assign choice to fSelection
cout << "Enter the number of your choice (or quit): ";
int fSelection;
cin >> fSelection;

cout << extMatch[fSelection] << endl;

//assign the selected file to a string, launch it based on that string and record and output time
    string s = extMatch[fSelection];
    unsigned long long start = ::GetTickCount64();
    system(s.c_str());
    unsigned long long stop = ::GetTickCount64();
    double elapsedTime = (stop-start)/1000.0;
    cout << "Time: " << elapsedTime << " seconds\n";

}


int main()
{

string selection;
cout << "0. Microsoft word \n1. Microsoft Excel \n2. Visual Studio 11 \n3. 7Zip \n4. Notepad \n Enter the number of your choice (or quit): ";

cin >> selection;

path folder = "..";

    if (selection == "0")
{
    wScan ( folder );
}
    else if (selection == "1")
{
    eScan ( folder );
}
    else if (selection == "2")
{
    vScan ( folder );
}
    else if (selection == "3")
{
    zScan ( folder );
}
    else if (selection == "4")
{
    tScan ( folder );
}
}

提前感谢您提供的任何建议。

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1 回答 1

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如果目录不是当前目录,则必须在文件名之前添加目录路径才能访问目录中的文件。

假设程序在C:/top-level; 假设检查的目录是C:/another-one. 当目录迭代器返回时somefile.docx,您需要使用C:/another-one/somefile.docx来访问文件。

于 2012-11-06T01:37:13.507 回答