如何在OpenFileDialog中单击打开后读取多个文件？

Question

我在我的代码中设置了一个多选函数，允许我打开多个“.txt”格式的文件。这里是问题，我打算如何通过OpenFileDialog打开所有这些选定的文件？下面的代码和“for each”这行代码，当我使用System :: Diagnostics :: Debug时，它只显示来自文件的数据，而其他文件的数据丢失。我应该如何修改“for each”之后的代码？任何人都可以提供一些建议或建议？所选文件为1_1.txt，2_1.txt，3_1.txt。感谢您的回复，并提前致谢。

这是我写的代码，

Stream^ myStream; 
OpenFileDialog^ openFileDialog1 = gcnew OpenFileDialog; 
openFileDialog1->InitialDirectory = "c:\\"; 
openFileDialog1->Title = "open captured file"; 
openFileDialog1->Filter = "CP files (*.cp)|*.cp|All files (*.*)|*.*|txt files (*.txt)|*.txt"; 
openFileDialog1->FilterIndex = 2; 
openFileDialog1->Multiselect = true; 

if (openFileDialog1->ShowDialog() == System::Windows::Forms::DialogResult::OK) 
{ 
    array<String^>^ lines = System::IO::File::ReadAllLines(openFileDialog1->FileName); 
    for each (String^ line in lines) { 
    //????? 
    System::Diagnostics::Debug::WriteLine("",line); 
    } 
} 

Answer 1

你需要看看OpenFileDialog.FileNames财产，如果你允许选择多个文件：

if (openFileDialog1->ShowDialog() == System::Windows::Forms::DialogResult::OK) 
{ 
    for each (String^ file in openFileDialog1->FileNames) 
    { 
    array<String^>^ lines = System::IO::File::ReadAllLines(file); 
    for each (String^ line in lines) 
    { 
     System::Diagnostics::Debug::WriteLine("",line); 
    } 
    } 
} 

Answer 2

使用FileNames财产。

C＃版本（应该很容易对C适应++）：

foreach (var file in openFileDialog1.FileNames) 
{ 
    foreach (var line in File.ReadAllLines(file) 
    { 
    ... 
    } 
} 

Answer 3

使用openFileDialog1->FileNames。它返回所选

这里 http://msdn.microsoft.com/en-us/library/system.windows.forms.openfiledialog.multiselect.aspx 读取多个文件名及其在C＃中，但它会很容易推断到C++。