使用opencv c ++每1分钟创建一个新的视频文件

Question

我正在使用opencv开发一个小“Dashcam”。 我想创建一个以当前日期和时间命名的新视频文件，每分钟原理很简单。 这些视频文件的内容是网络摄像头的框架。

但是，在第一分钟过后，不再生成视频文件。

这是我评论的代码：

#include <iostream>
#include <windows.h>
#include <ctime>
#include <time.h>
#include <fstream>
#include <opencv2/core/core.hpp>
#include <opencv2/opencv.hpp>
using namespace std;
using namespace cv;


int record_video()
{
    VideoCapture cam(0);
    if (!cam.isOpened())
    {
        cout << "Error from webcam" << endl;
        return -1;
    }

    time_t t = time(0);
    struct tm* now = localtime( & t );
    char buffer[80];

    strftime(buffer, 80, "%F_%Hh%M.wmv", now); //get the date and time for my file

    VideoWriter video(buffer, CV_FOURCC('W', 'M', 'V', '2'), 30, Size(640, 480), true); //Creation of my video file from my webcam

    int chrono = 0; //start the chrono at 
    bool record = false;

    while (1)
    {
        Mat frame;
        cam >> frame;
        if (frame.empty())
        {
            cout << "frame is empty" << endl;
            return -1;
        }
        cout << chrono << endl;
        if ((chrono == 1) && (record == false)) //we start from the begining at 0 seconds and the recording has to be false to record again with the last condition (3rd condition)
        {
            cout << "new recording in file: " << buffer << endl; //display the name of video file on the console
            record = true;
        }
        if (record == true) //The condition here is to activate 
        {
            video.write(frame); //copy the frame from the webcam to the video file
            imshow("webcam", frame); //display what we record on a window
        }
        if (chrono == 1800) //the 1800 chrono = 60 seconds
        {
            record = false; //record is equal at false to repeat the first condition
            chrono = 1; //record is equal at false to repeat the first condition
        }

        chrono++; //incrementation of the chrono

        char c = (char)waitKey(1);
        if (c == 27)
            video.write(frame);     
    }
    cam.release();
    destroyAllWindows();
    return 0;
}

int main()
{
    record_video();
    return 0;
}

Answer 1

    if (chrono == 1800) //the 1800 chrono = 60 seconds
    {
        record = false; //record is equal at false to repeat the first condition
        chrono = 1; //record is equal at false to repeat the first condition
    }

    chrono++; //incrementation of the chrono

在此chrono将为2并且record将为false ，因此您的代码将永远不会再次记录（因为(chrono == 1) && (record == false)将永远不会true ）。 你需要在if-body中将chrono设置为0 ，然后将其增加到1 ，或者将增量移动到if (record == true)的主体中，这样当你不记录时它不会增加。

或者：

    if (record == true) //The condition here is to activate 
    {
        video.write(frame); //copy the frame from the webcam to the video file
        imshow("webcam", frame); //display what we record on a window
        chrono++; //incrementation of the chrono
    }

（并进一步删除chrono++ ）

要么：

    if (chrono == 1800) //the 1800 chrono = 60 seconds
    {
        record = false; //record is equal at false to repeat the first condition
        chrono = 0; //record is equal at false to repeat the first condition
    }

    chrono++; //incrementation of the chrono

---

在站点节点上，您当前不是每分钟都创建一个新视频，而是添加到同一视频。 您可能希望将创建VideoWriter的行移动到循环中，特别是if ((chrono == 1) && (record == false)) 。 在循环之外，只需保留VideoWriter video; 并使用open初始化一个新的视频文件：

    if ((chrono == 1) && (record == false)) //we start from the begining at 0 seconds and the recording has to be false to record again with the last condition (3rd condition)
    {
        time_t t = time(0);
        struct tm* now = localtime( & t );
        char buffer[80];

        strftime(buffer, 80, "%F_%Hh%M.wmv", now); //get the date and time for my file

        video.open(buffer, CV_FOURCC('W', 'M', 'V', '2'), 30, Size(640, 480), true); //Creation of my video file from my webcam

        cout << "new recording in file: " << buffer << endl; //display the name of video file on the console
        record = true;
    }

你的代码也会跳过第一帧，不确定是否有意。 您可以通过将条件更改为if ((chrono == 0) && (record == false))然后修改上述修复程序来重置chrono以确保它重置为0 （而不是1或2 ）来解决此问题。