0

我有一个存储在文本文件中的文件列表。我逐行读取文件并将它们存储在字符串数组中。文件列表如下所示：

04_02_1310.csv
04_03_1350.csv
04_04_0421.csv
04_05_0447.csv

等等。让我们调用我的字符串数组

filelist[i]

假设我正在尝试打开列表中的第一个文件：

inputFile.open(filelist[0].c_str()); // This cannot open file

该文件无法打开。如果我将文件名放在引号中，一切正常：

inputFile.open("04_02_1310.csv"); // This works perfectly

如果我打印 filelist[i] 的内容，那么它也可以正常工作：

cout << filelist[0] << endl; // This outputs 04_02_1310.csv on screen.

有人可以告诉我上述方法有什么问题吗？在过去的 2 天里，这让我发疯了，我要手动输入所有内容以完成它（一个接一个地完成 100 多个文件）。

我也愿意以任何其他方式来完成这项简单的任务。

谢谢！！！

编辑：如果您想了解它是如何实现的，我将添加代码的相关部分：

#include <cstdlib>
#include <iostream>
#include <time.h>
#include <string>
#include <sstream>
#include <fstream>
#include <vector>
#include <algorithm>
#include <iterator>

using namespace std;

//Declarations for I/O files
ifstream inputFile;

//Declare other variables (forgot to add these in my previous EDIT, sorry)
int number_of_files;
string line;
string *filelist = NULL;

//Open file list and count number of files
inputFile.clear();
inputFile.open("filelist.txt", ios::in);

//exit and prompt error message if file could not be opened
if (!inputFile){
    cerr << "File list could not be opened" << endl;
    exit(1);
}// end if

// count number of lines in the data file and prompt on screen
number_of_files = 0;

while (getline(inputFile, line))        
    number_of_files++;

cout << "Number of files to be analyzed: " << number_of_files << endl;

filelist = new string[number_of_files];
inputFile.close();

//Re-open file list and store filenames in a string array
inputFile.clear();
inputFile.open("filelist.txt", ios::in);

//exit and prompt error message if file could not be opened
if (!inputFile){
    cerr << "File list could not be opened" << endl;
    exit(1);
}// end if

// store filenames
i = 0;
while (getline(inputFile, line)){
    filelist[i] = line;
    //cout << filelist[i] << endl;
    i = i + 1;
}        

inputFile.close();

//open first file in the list, I deleted the loop to focus on the first element for now

inputFile.clear();
inputFile.open(filelist[0].c_str(), ios::in);

//exit and prompt error message if file could not be opened
if (!inputFile){
    cerr << "Data file could not be opened" << endl;
    exit(1);
}// end if

输出是：

Data file could not be opened

再次感谢！

4

3 回答 3

2

该字符串中的文本文件中可能仍然存在 '\n' 字符（或 EOF,'\0' ），您应该尝试检查字符串是否“干净”。

于 2012-07-13T16:26:44.890 回答
2

我也愿意以任何其他方式来完成这项简单的任务。

#include <cstdlib>
#include <iostream>
#include <time.h>
#include <string>
#include <sstream>
#include <fstream>
#include <vector>
#include <algorithm>
#include <iterator>

using namespace std;

int main () {

  std::ifstream inputFile("filelist.txt");
  std::vector<std::string> fileList;
  std::string line;

  if(!inputFile) {
    std::cerr << "File list could not be opened\n";
    return 1;
  }

  while(std::getline(inputFile, line)) {
    fileList.push_back(line);
  }

  std::cout << "Number of files to be analyzed: " << fileList.size() << "\n";

  std::vector<std::string>::const_iterator it(fileList.begin());
  std::vector<std::string>::const_iterator end(fileList.end());
  for(;it != end;++it) {
    std::ifstream inputTxt(it->c_str());

    if(!inputTxt) {
      std::cerr << "Data file could not be opened:" << *it << "\n";
      return 1;
    }
    while(std::getline(inputTxt, line)) {
      std::cout << line << "\n";
    }
  }
}
于 2012-07-13T16:37:17.577 回答
0

我建议使用更“C++ - ish”的方法，尽管与 Blood 发布的优雅解决方案不同：

int main()
{
    std::vector<std::string> fileList;

    // ...

    std::ifstream inputFile("filelist.txt");

    // ...

    std::string line;
    while(inputFile >> line)
        filelist.push_back(line);

    inputFile.close();

    // ...

    for(size_t t = 0; t < filelist.size(); t++)
    {
        std::ifstream dataFile(filelist[t].c_str());

        // ...

        dataFile.close();
    }
}
于 2012-07-13T16:37:17.407 回答