在c++中使用curl从url下载图像

Download an image from an url using curl in c++

本文关键字：url 下载 图像 curl c++      更新时间：2023-10-16

我正在做一个项目，第一步包括从url下载图像并将其保存在某个位置。稍后将对该图像进行进一步处理。为此，我在visualstudio中使用了curl库和opencv。我是使用curl库的新手。我在这里看到了答案。但是，我不能理解。这是我的代码

#include <stdio.h>
#include <curl/curl.h>
#include <curl/types.h>
#include <curl/easy.h>
#include <string>
size_t write_data(void* ptr, size_t size, size_t nmemb, FILE* stream)
{
    size_t written;
    written = fwrite(ptr, size, nmemb, stream);
    return written;
}
int main(void)
{
    CURL* curl;
    FILE* fp;
    CURLcode res;
    char* url = "http://pimg.tradeindia.com/01063301/b/1/CRO-Oscilloscope.jpg";
    char outfilename[FILENAME_MAX] = "C:\bbb.jpg";
    curl = curl_easy_init();
    if (curl)
    {
        fp = fopen(outfilename, "wb");
        curl_easy_setopt(curl, CURLOPT_URL, url);
        curl_easy_setopt(curl, CURLOPT_WRITEFUNCTION, write_data);
        curl_easy_setopt(curl, CURLOPT_WRITEDATA, fp);
        curl_easy_setopt(curl, CURLOPT_FOLLOWLOCATION, 1L);
        res = curl_easy_perform(curl);
        /* always cleanup */
        curl_easy_cleanup(curl);
        fclose(fp);
    }
    return 0;
}

谢谢你的帮助。

#include <stdio.h>
#include <curl/curl.h>
size_t callbackfunction(void *ptr, size_t size, size_t nmemb, void* userdata)
{
    FILE* stream = (FILE*)userdata;
    if (!stream)
    {
        printf("!!! No streamn");
        return 0;
    }
    size_t written = fwrite((FILE*)ptr, size, nmemb, stream);
    return written;
}
bool download_jpeg(char* url)
{
    FILE* fp = fopen("out.jpg", "wb");
    if (!fp)
    {
        printf("!!! Failed to create file on the diskn");
        return false;
    }
    CURL* curlCtx = curl_easy_init();
    curl_easy_setopt(curlCtx, CURLOPT_URL, url);
    curl_easy_setopt(curlCtx, CURLOPT_WRITEDATA, fp);
    curl_easy_setopt(curlCtx, CURLOPT_WRITEFUNCTION, callbackfunction);
    curl_easy_setopt(curlCtx, CURLOPT_FOLLOWLOCATION, 1);
    CURLcode rc = curl_easy_perform(curlCtx);
    if (rc)
    {
        printf("!!! Failed to download: %sn", url);
        return false;
    }
    long res_code = 0;
    curl_easy_getinfo(curlCtx, CURLINFO_RESPONSE_CODE, &res_code);
    if (!((res_code == 200 || res_code == 201) && rc != CURLE_ABORTED_BY_CALLBACK))
    {
        printf("!!! Response code: %dn", res_code);
        return false;
    }
    curl_easy_cleanup(curlCtx);
    fclose(fp);
    return true;
}
int main(int argc, char** argv)
{
    if (argc < 2)
    {
       printf("Usage: %s <url>n", argv[0]);
       return -1;
    }
    if (!download_jpeg(argv[1]))
    {
        printf("!! Failed to download file: %sn", argv[1]);
        return -1;
    }
    return 0;
}

这个怎么样？？

我不理解argc和argv方面的代码。你能不用它们解释吗约翰·史密斯

它们用于在启动程序时传递命令行参数。argv是指向cstrings数组的双指针，argc是cstrings的数量。在这种情况下，argv[0]将是可执行文件的名称，argv[1]将是要下载的映像的URL。即foo.exe http://path/to/image.jpg

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