C++ 打开文本对话框，获取选中文件的路径、名称和后缀名

skye1221

于 2018-12-25 10:06:55 发布

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CC 4.0 BY-SA版权

分类专栏： c++

本文链接：https://blog.youkuaiyun.com/skye1221/article/details/85243465

#include<iostream>
#include<Windows.h>
#include<commdlg.h>
#include<string>
using namespace std;
/******************************************************************************************
Function:        TCHAR2STRING
Description:     TCHAR转string
Input:           str:待转化的TCHAR*类型字符串
Return:          转化后的string类型字符串
*******************************************************************************************/
string TCHAR2STRING(TCHAR* str){
	std::string strstr;
	try
	{
		int iLen = WideCharToMultiByte(CP_ACP, 0, str, -1, NULL, 0, NULL, NULL);

		char* chRtn = new char[iLen * sizeof(char)];

		WideCharToMultiByte(CP_ACP, 0, str, -1, chRtn, iLen, NULL, NULL);

		strstr = chRtn;
	}
	catch (exception e)
	{
	}

	return strstr;
}

int main(){
	while (1){
		OPENFILENAME ofn = { 0 };
		TCHAR strFileName[MAX_PATH] = { 0 };	//用于接收文件名
		ofn.lStructSize = sizeof(OPENFILENAME);	//结构体大小
		ofn.hwndOwner = NULL;					//拥有着窗口句柄
		ofn.lpstrFilter = TEXT("All